A brief history of problems that ruined my world
A different (maybe better) title for this might be: the most frequently banned topics on the internet. Below I present a set of questions that are so counter-intuitive that their very mention can set off the right audience into a complete flame war. Each of them has been known to paralyze various online communities resulting in monster threads and lots of insults (and in one case, the company stepping in to solve the dispute). You can tell the intellectual average of a community based upon the sophistication of the brain-exploding questions that give it fits.
.9~ and 1
Is .9999~ = 1 or is .9999~ < 1?
When I was 12, this problem ruined my world. The answer is, of course, .9~ is exactly equal to 1. In elementary school, they tried to placate me with algebraic trickery, but I wasn’t buying that. Obviously, .9~ is the number right before 1, but not equal to 1. Obviously! It took me awhile to come to terms with the fact that any two distinct real numbers have an infinite amount of numbers between. If .9~ and 1 were distinct, they’d have an infinite amount of numbers between them. Since they don’t, they cannot be distinct. And so I moved on.
The monty hall problem
Suppose you’re on a game show and you’re given the choice of three doors. Behind one door is a car; behind the others, goats. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?
When I was in high school, this problem ruined my world. The answer is, of course, you should switch doors. At first, you might believe, that you are still choosing between two doors, and so it doesn’t matter, it’s 50-50. After you think a little bit more carefully you realize that the rules that Monty plays by gives him no choice but to give you information, making the choices not so equal. Intuition squares with this, eventually.
Airplane on a treadmill
A plane is standing on a large treadmill or conveyor belt. The plane moves in one direction, while the conveyor moves in the opposite direction. This conveyor has a control system that tracks the plane speed and tunes the speed of the conveyor to be exactly the same (but in the opposite direction). Can the plane take off?
When I was in college, this problem ruined my world. The answer is, of course, the plane will have no trouble taking off. Alright, that’s not quite true. The problem with this problem is that it’s open to a bit of interpretation. A more precise answer is if the pilot wants to take off, he will, regardless of what the treadmill does. This is basically because the airplane wheels are not where the power is, the engines are. And the engines are pushing against the stationary air, not the moving treadmill.
The second ace paradox
There are two games of bridge going on somewhere in the world, right now. In one of those games, Alice announces, “I have an ace in my hand.” In the other of those games, Bob announces, “I have the ace of spades in my hand.” Whose hand is more likely to contain a second ace, Alice or Bob?
In grad school, this problem ruined my world. The answer is, of course, Bob is much more likely than Alice to have a second Ace. I’ve run the math on this and it’s absolutely true. I still don’t have a good intuition for why.
The two envelopes problem
You are given two envelopes, one of which contains twice as much money as the other. You asked to choose one to keep. No matter which envelope you choose, the other one has a higher expected value. So you should switch your choice. But now… the original has a higher expected value.. so you should switch back. But now…
This problem, and its variants, still ruin my world. The answer is, of course, well, uhm. I actually heard this problem in college and not only do I not have a good answer to what’s going on, apparently no one does. That makes me feel better.
In the simple version above, the flaw in the logic comes at the very beginning where it is presumed that the initial value is an unbounded and uniform random number. An unbounded, uniform distribution is impossible, and so the number chosen cannot be from that distribution. Certain possible distributions have solutions. However, there are variants of this problem that have possible distributions and still present the paradox. Good luck with that.
July 27th, 2009 at 10:26 am
Holy crizzap. I’ve heard most of those before, and can grok the ace problem by analogy to other oddities; but I thought we were about to step into the much more tractable Newcomb’s Paradox with those envelopes. That’s not at all the case. Near as I can tell, we could solve the world financial crisis by repeated applications of the wallet swap variant.
July 27th, 2009 at 1:18 pm
Your answer to the airplane problem is wrong. Basically, you’ve invented a situation where the airplane’s engines produce thrust which move the airplane forward IN THE AIR and a conveyor belt which matches that movement. Hence, the airplane, despite the engine’s performance, remains in the same physical location relative to the ground the conveyor belt sits on.
Since the airplane doesn’t move relative to the ground there is no air blowing over the wings, apart from wind. That’s because it also doesn’t move relative to the air. The conveyor belt cancels out that forward movement IN THE AIR. So the engines produce thrust which moves the airplane forward but the conveyor belt at the same time keeps the airplane locked in the same physical location.
The function of an airplane engine is to let airplanes move relative to the surrounding air mass. That’s why you’re at your destination quicker when there’s a tail wind and why it takes longer when there’s a head wind. Since the conveyor belt cancels out that movement relative to the air mass the plane will never take off.
You can always turn airplane engines into rockets if you which to, but let’s not forget that rockets don’t depend on lift generated by wings to fly and airplanes do.
July 27th, 2009 at 1:31 pm
I agree with the second commenter about the airplane never taking off, but I have a different ( somewhat weaker ) argument.
just think of what the whole thing looks like:
the engines are working but the plane is standing in place ( with its wheels moving). all of a sudden as the nose points up, it will burst up in a speed of 400 miles an hour ( or whatever speed is needed for take off), when a second ago it was doing 0 miles an hour(both speeds are taken relative to the earth).
July 27th, 2009 at 1:32 pm
Dear god! For the last time, the airplane DOES move relative to the ground. Without friction between the wheel and it’s axle the treadmill can move however fast it damn well pleases and the plane won’t budge an inch. The motion of the airplane has NOTHING to do with the treadmill (except overcoming said friction). Go put a skateboard on a treadmill and sit on it with a fan to see what happens
In all honesty though most of the argument about this one is between people who believe that the treadmill is moving SO fast that the friction between the plane’s wheels and axle prevents it’s motion and people who realize that’s totally ridiculous.
July 27th, 2009 at 1:36 pm
@Steven: actually the article’s author is correct and you’re mistaken and for the same reason that many other people have been. They all assume an airplane on the ground behaves, relative to the ground, the same way a car does. Alas, it does not. The speed of the conveyor belt is irrelevant to the plane: should you make the conveyor move twice as fast as the plane’s propulsion, the plane will not then move backwards, its wheels will just spin that much faster.
For some thorough explanations: Straight Dope, Kottke.org.
July 27th, 2009 at 2:01 pm
The short version of the airplane problem is to apply F=ma to the airplane. The engine of the plane is putting a forward force on the aircraft (as it’s pushing on the stationary air). If you posit that the plane does not move, you must come up with a way for the conveyor belt to apply an equal and negative force to the aircraft. No amount of treadmill speed can put this force on the aircraft, meaning nothing is stopping the plane from taking off. All that you can do is make the wheels of the aircraft spin which is basically a red herring as you are not transmitting any appreciable force to the aircraft itself.
July 27th, 2009 at 2:41 pm
I’m almost sure the second ace paradox is not really a paradox.
Let’s say X are the odds of a second ace when the card is specified as being a spade (i.e. Bob’s odds). If Bob had instead specified an ace of clubs, hearts, diamonds the odds for those would also have been X. Therefore the odds of having a second ace are X regardless of the card.
I don’t see why knowing/saying the colour of the ace will affect the odds — cards are symmetric relative to colour. So my conclusion is this is a faux-paradox, and someone made a mistake somewhere in the math (which I have not seen).
Also, the two envelopes problem only looks like a paradox if one uses probabilities superficially (Wikipedia says more about it).
July 27th, 2009 at 2:48 pm
wiki article for the envelope problem actually has the solution:
“The most common way to explain the paradox is to observe that A is used inconsistently in the expected value calculation, step 7 above. In the first term A is the smaller amount while in the second term A is the larger amount. To mix different instances of a variable or parameter in the same formula like this isn’t legitimate, so step 7 is thus the proposed cause of the paradox”
July 27th, 2009 at 3:33 pm
“.9~ and 1
Is .9999~ = 1 or is .9999~ < 1?
When I was 12, this problem ruined my world. The answer is, of course, .9~ is exactly equal to 1″
Not exactly!!! Depends on your axioms… If you are using standard analysis than that is true.
But certainly is not in non-standard analysis and for another example for computable reals!
July 27th, 2009 at 3:43 pm
The two math problems that blew my mind for more than 1 day are the Monty Hall Problem, and the fact that the infinite set “integers” is exactly the same size as the infinite set “rational numbers” even though the first set is a subset of the second.
July 27th, 2009 at 4:05 pm
Airplane Problem: A better way to imagine this for those of you that still think the plane won’t take off is to imagine the treadmill being the length of the runway. Even if the treadmill were moving in the opposite direction at twice the speed it takes the plane to take off, the plane will still accelerate and take off because the wheels would just spin three times as fast as take of required to take off given the proper amount of thrust from the engines.
July 27th, 2009 at 5:07 pm
I would recommend The Paridoxicon by Nicholas Falletta. I spent many an hour when I was younger thinking about these problems. However, you have paraphrased The Second Ace.
Per the book (p121)
It is important to note that two conditions must be met for the paradox to hold: both the named ace and the person that names the ace must be specified in advance, or there is no paradox.
July 27th, 2009 at 5:27 pm
Thank you for the ace problem – I love it. Here’s my intuition – let’s limit ourselves to a game with 2-card hands to avoid complications, the solution trivially extends to bigger hands.
There are three possible hands:
* 0 aces – these are irrelevant for us
* exactly 1 ace – 100% in Alice set, 25% in Bob set
* 2 aces – 100% in Alice set, 50% in Bob set
In this formulation it’s instantly obvious that whatever is the proportion between 2 ace and exactly-1 ace situations for Alice, it is exactly twice as high for Bob.
Numbers get messier with larger hands, but it’s essentially the same reasoning.
July 27th, 2009 at 6:00 pm
RE: the plane problem,
imagine sea planes, they dont push off water, they are still pushing air to generate thrust, that generates lift.
July 27th, 2009 at 9:21 pm
Oh, you’re evil. Malevolent. Maleficent. What a diabolical trick to play on the internets. Intentionally trying to hack the hive mind into an infinite loop should be a felony.
July 27th, 2009 at 10:58 pm
In response to “Someone” who wrote the following:
============================
““.9~ and 1
Is .9999~ = 1 or is .9999~ < 1?
When I was 12, this problem ruined my world. The answer is, of course, .9~ is exactly equal to 1″
Not exactly!!! Depends on your axioms… If you are using standard analysis than that is true.
But certainly is not in non-standard analysis and for another example for computable reals!”
===========================
This is a great example of some dumb amateur math student who wants to use big words and sound smart. The truth of this statement has nothing to do with axioms, and everything to do with what a real number actually is (real number of course being a number, say, corresponding to a point on the number line). Pinning down exactly what a real number is isn’t easy. A real number is either (a) a rational number or (b) the limit of some sequence of rationals which is not a rational. i.e., real numbers are the limits of all Cauchy sequences of rationals. If this bugs you, then check out the wikipedia article:
http://en.wikipedia.org/wiki/Cauchy_sequence
If you want to truly understand why 0.999…=1, then you must understand what the number 0.999… actually is. It can be understood as the limit of the following sequence:
0.9, 0.99, 0.999, 0.9999, … . If you think hard about it, you’ll come to the conclusion that this is in fact the ONLY real meaningful definition of the symbol 0.999… (the “…” of course denoting that the sequence of 9s continues infinitely on and on).
If this is confusing then read this:
http://en.wikipedia.org/wiki/Limit_of_a_sequence
What does the above sequence converge to? Oh! 1. And so why does 0.999… =1? Because the limits of convergent sequences of numbers are unique.
July 27th, 2009 at 11:12 pm
I would also like to offer a quick explanation to the Monty Hall problem:
There are 3 doors: A, B, and C, and suppose your gut tells you that the prize you want is behind door A. The host of the show says “OK, you’ve picked door A, I’m going to show you” and then he looks at his little note card to remind himself where everything is “what’s behind door C”. He opens up door C and there’s some stupid broken down tractor or something that you don’t want. Then, of course he says “OK, you picked door A, I showed you what’s behind door C. Would you like to change your guess to door B?” And now you have to make your decision.
So. If we guess blindly, then there is certainly a 33% chance that you picked correctly, and the prize is behind door A. However… focus on the situation where you DIDN’T pick the right door, which happens 67% of the time…. If you DIDN’T pick the right door, that means that there are two other doors: one door which has the prize, and one door which doesn’t. In this situation, the host (who knows where everything is), is FORCED to show you what’s behind the door that you didn’t pick, which ALSO doesn’t have the prize behind it (right? he’s not going to show you where the prize is, so what other option does he have?). This means, that 67% of the time, if you pick the door that you didn’t initially guess, and the he didn’t show you, it will be the door that has the prize behind it.
July 27th, 2009 at 11:58 pm
Aces,
If Bob is specifying he has an ace of spades he’s unconciously indicating he has another, otherwise, why make the distinction?
July 28th, 2009 at 6:12 am
Mythbusters dealt with (an interpretation of) the plane problem: http://www.youtube.com/watch?v=YORCk1BN7QY
July 28th, 2009 at 7:57 am
FYI.. for all speculating about the airplane (@ Steven Devijver), they did an entire episode of mythbusters about this, and he is absolutely correct in the above, and the airplane does take off.
http://mythbustersresults.com/episode97
July 29th, 2009 at 9:06 am
There airplane problem is a problem of assumption. The system would have to apply a force on the airplane, through the wheels, that is equal to the force from the jet engines. That would require the conveyor belt to move incredibly fast, such that it is probably physically impossible. I’m not sure the interface between the belt and the wheels would allow that kind of force in any way before something breaks down. (I’m not experienced enough in physics to figure it out right now)
In other words: The airplane would take off.. in a physical interpretation.
August 2nd, 2009 at 5:41 pm
On Monty Hall, @John is right. Also the number of combinations is so small you can list all of them on paper and see for yourself how it works out. Sometimes that’s the only way to convince someone!
On the envelopes, I have an idea of where the problem is.
The computation starts by saying “If you envelope has A dollars, there’s a 1/2 chance the other has A/2 and 1/2 chance it has 2A dollars for a total expected value of (5/4)A, so you should switch.” Ad infinitum.
However this is NOT the situation.
First of all, the transitive step is obviously false. That is, once you take the path of “the other is A/2″ or “the other is 2A,” it is NO LONGER TRUE that having switched, the same logic applies! Because in both cases you now KNOW the other’s contents. It’s not STILL true that, having moved from A to 2A, that there’s a 1/2 chance the other has 4A!
Second, it’s false to say there’s a 50/50 chance the other is A/2 or 2A. The easiest way to see why is to recast the problem in terms of L — the lower of the two values.
Then you can say there’s L and 2L as choices. Once you do that, the expected value of switching and staying are both (3/2)L.
What appears to be just a change of variable results in a different result! How to make sense of that?
Because when you say “A” you’re not picking one situation or the other! You’re confounding the case of having picked the lesser and the greater items. When you make those cases explicit, the math comes out right.
You’re correct — it’s a great puzzle! But since no one else talked about why I felt like I should.
August 7th, 2009 at 4:42 am
A great problem set. I should note that I had earlier solved the longstanding problem
Q. Which came 1st, the chicken or the egg”?
A. Depends on whether you believe in Genesis (chicken) or Evolution (egg). Genesis says God created the first chicken, which later laid the first eggs. Evolution says that if you go back far enough, you will find the first chicken egg, which then must have been laid by something that wasn’t a chicken.
Now that’s out of the way, time for the big one, for which Monty Hall was only the primitive lead in. From above – the 2 envelope problem:
You are given two envelopes, one of which contains twice as much money as the other. You asked to choose one to keep. No matter which envelope you choose, the other one has a higher expected value. So you should switch your choice. But now… the original has a higher expected value.. so you should switch back. But now …
Interestingly, when I thought about it, I realized the answer must involve the only known example of quantum mechanics operating on the human scale (i.e. directly observable). I will be completing my article for the Physics Journals shortly. Here is the answer in a nutshell.
Suppose you randomly pick an envelope and open it. You find it contains $10. Since you know one envelope has twice as much money as the other, that must mean the other envelope has either $5 or $20, with equal probability. Should you switch? Of course – the extra payback is either +$10 or -$5. Or in probability terms:
Stay – win $10, Switch – win ($20 + $5)/2 = $12.50. Best strategy – switch.
However, note that the advice to switch is equally compelling, no matter what amount was found in the 1st envelope!
Stay – win $x, Switch – win ($2 x + $ 1/2 x)/2 = $1.25 x Best strategy – switch!!!
This means you could pick either envelope, and without looking at its contents, switch for the other one and be assured of a positive gain!! Madness – you could keep swapping the same 2 envelopes back and forth without opening them, secure in the knowledge each switch was gaining you money.
Something is badly amiss. There is a contradiction that must somehow be resolved.
* It can’t possibly matter if you switch the envelope you have selected, before you open it and see how much is in it.
* You clearly gain if you switch the envelope after you open it, no matter how much was in it
The solution of course is based upon quantum mechanics. Select an envelope. Before looking at the amount inside, the possible amount of money in each envelope is undetermined … an infinite set of possibilities. So switching doesn’t matter.
Ah, but now open the envelope. You see $10. Immediately the probability wave function for the amount of money in the 2nd envelope collapses! There are only two equal probable amounts that remain … $5 and $20. Only now does it make sense to switch.
This is very much like Schroedinger’s cat – only when you open the box do you see whether the cat is alive or dead – before that, it is both. And all it takes to replicate that famous thought experiment are two envelopes, one containing twice as much money as the other.
The remaining question is … who has to look (i.e. who can be the observer causing the probability wave to collapse, making switching envelopes the optimal strategy)?
You? A mental defective? A monkey? An automatic camera recording the amount? What if there is no film in the camera?
Remember – if you disregard my answer above (“Quantum Mechanics on the macro level my ass” as several fellow Sun employees said to me years ago when I first solved the 2 envelope problem) then you have to come up with an answer of your own.
Try this one on your co-workers … and bring some sponges to mop up the drool.
August 20th, 2009 at 11:22 am
Airplane on a treadmill – It’s the relative motion wrt air that causes lift on wings and fuselage; the ground can kiss my arse for all i care.
There are 4 ‘ideal’ scenarios:
1. No friction between wheels and axle and no friction between belt and wheels – this case plane takes off as it just slides (and not rolls) along the treadmill, coz there’s relative motion wrt air
2. No friction between wheels and treadmill and friction between axle and wheels – this case again plane takes off as it slides, coz there’s relative motion wrt air
3. Friction between wheels and belt and no friction between wheels and axle – this case plane does not takes off as there’s no motion wrt air. if you’re standing on the treadmill there’s relative motion wrt ground, otherwise not
4. Friction between wheels and treadmill and friction between wheels and axle – this case ‘ideally’ plane does not takes off
in reality (for case 4), as someone mentioned earlier, the frictional force between wheels and treadmill are no match for thrust of a plane, take off will happen.
August 27th, 2009 at 1:34 pm
XKCD explains the airplane on a treadmill and how the problem is that both sides are interpreting the problem differently.
http://blag.xkcd.com/2008/09/09/the-goddamn-airplane-on-the-goddamn-treadmill/
August 27th, 2009 at 5:41 pm
I don’t know. I think that the airplane on the treadmill thing is very intuitive. It’s just rolling resistance, man. I’m pretty sure that most planes could take off even if they had the breaks on and had to grind a strip of rubber all down the runway.
August 28th, 2009 at 12:21 am
Aces… don’t know if this was said because i don’t want to sift through all of your dribble but the woman says she has an ace in her hand. the key word is AN. so of course she only has one. if i ask for an ace, i’m only asking for one. so yeah, bob can have more than one, but the woman cannot. either way it’s stupid. bridge is for old peoples.
August 28th, 2009 at 10:14 am
I don’t know. You get those aircraft wheels spinning fast enough for a long enough time, their bearings will overheat and seize up. When that happens, all bets are off.
August 28th, 2009 at 10:17 pm
The aircraft one is a wonderful trap for the intellectually bored – generating the insane limit cases was a great laugh when I was in college. (My favorite was calling ‘no-slip’ on the conveyor, meaning there was some ridiculous speed where the air dragged by the conveyor would actually apply lift to the plane anyway, rendering the argument doubly moot!). It reminds me of the discussions on whether the earth’s core was negligible.
If I may add my own thrill, it was the day in class where we proved that when solving the quantum equations for two electrons in a hydrogen atom, you actually *can’t* label which is which. Labeling Psi[A] and Psi[B] forces the algebra to fail. We all knew that you didn’t really know which was which, but this showed you *can’t* know. The calculation scrambles the identities. Or some other philosophical nonsuch. Ahh, Good Times.
Thanks for the reminders! (Captcha: absurd)
August 31st, 2009 at 11:32 am
mythbusters tested this……
http://www.youtube.com/watch?v=YORCk1BN7QY
August 31st, 2009 at 11:33 am
For those of you too bored to read the xkcd article, there are two methods of thought for the airplane on the treadmill: Realists and theorists.
The realists say the plane will take off because the engines are separate from the treadmill and will push the plane forward no matter whats going on with the treadmill.
The theorists take the problem much more literally and spot that the stipulation that the treadmill is traveling as fast as the wheels are means that for the plane to theoretically go forwards the wheels would have to be moving at a faster speed than the treadmill, thus breaking the problem, so the plane would not take off because any external force would quickly just power the treadmill to infinity m/s and the laws of physics would break around us.
August 31st, 2009 at 2:16 pm
Airplane this…If the treadmill is going, say, 100 knots, and the plane’s thrust is 100 knots, the plane stays in the same area of reference. What a plane needs to fly is simply enough AIR speed under its wings to provide lift. The plane cannot take off until the LIFT needed is reached. Lets put it this way…
If a plane is in a wind tunnel and the wind is blown at 100 knots, and the airplane’s thrust is at 100 knots, the wheels may not even move, but enough lift is created to allow the plane to take off what seems vertically to us.
It isn’t a matter of ground movement, it is air movement.
September 2nd, 2009 at 12:36 am
To try to make sense of the second ace paradox intuitively and mathematically, consider it with combinatorics (count the total number of ways things can happen) and then think of it as a probability/statistics problem… When playing bridge, each player has 13 cards. So to start out, the total number of 13-card hands from a 52 card set is “52 choose 13″… some huge number.
But now what you really want to calculate is the probability of having the ace of spades AND another ace in your hand of 13 as opposed to the probability of having two aces in your hand of 13.
I wont go into all of the nasty details of factorials, divisions, and multiplication of independent events, but it turns out that knowing the ace of spades is in Bob’s hand “raises” the probability that one of the other 12 cards in his hand will be an ace. This is because now we have a “51 choose 12″ scenario since one card (the ace of spades) is already spelled out for us, we can ignore it in the calculation of the probability of the rest of his hand.
In Alice’s case, we just know that she has some ace, and then we must deduce whether or not she has another ace… So we end up having to calculate the chances that any two aces will show up at all, and all of this still having to determine from 13 random cards chosen from a set of 52 possible cards… She has a lot more independent events to consider.
*** There is a lot more to the math behind why the probability is higher for Bob’s hand, but this is a simple intuitive way to understand why it works. Short story, there’s less scenarios to calculate for Bob. Each scenario has it’s own probability. So when multiplying the probabilities of independent events, it yields a higher chance for the event of having two aces. (Keep multiplying more fractions and the number/chance gets smaller… like in Alice’s case)
Hope it helps,
- Scott
P.S. You can almost think of it as a much more complicated version of the goats and car problem with four types of 13 different prizes hidden behind 13 doors. Bob already showed us that the Ace of Spaces is behind one of his 13 doors, but Alice simply told us that one of the 13 doors has an ace behind it, and she hasn’t showed us which one yet.
September 2nd, 2009 at 2:47 am
as an undergraduate, i was hanging out with two of my friends (also math majors). one of them randomly asked the question “is there a continuous function f:R->R such that f(f(x)) = exp(x) for all x. the gap in the “great chain of being” between polynomials and exponentials, donchaknow.
i now tell my students about the problem that took three years and eight seconds to solve. moral of the story: don’t get married to your answer until you have a proof.
September 2nd, 2009 at 8:10 am
Aces Paradox: I’m yet to be convinced, though Scott’s probability breakdown explanation sounds vaguely plausible…
It reminds me of a similar problem:
I meet a stranger. He mentions he has 2 children.
I ask if he has any daughters and he says yes.
What is the probability that BOTH his children are girls?
The answer is of course 1/3
September 2nd, 2009 at 1:35 pm
Monty Hall Problem…
Behind Door # 1, Bob has an Ace of Spades.
Behind Door # 2, there is an Airplane on a Treadmill
Behind Door # 3, there is a Zonkey
If you pick Door # 3, Monty will offer you a choice between two envelopes…
September 3rd, 2009 at 12:46 pm
aha!
I dont know much about a card pack so I didnt actually try the 4th one…I know you ppl will be like …”waaa??” and here yet u are reading a comment from a guy that doesnt know much about a card pack other than that it has 52 cards 
..blah forget that
i think i’d have to elaborate about it more(as i understand and as true to me)
)
..hard math there
nice problems, I got the 1st three correct
others,
well because people bring out some thoughts about the air plane on treadmills
Air planes go up because the horizontally traveling wind makes an upthrust on the wings(according to Bernoulli’s theorem) Most of you know about it. Its because the shape of the wings.
Now when its on the treadmill(conveyor belt
From relativity for velocities..lets assume it doesnt accelerate
A-Air Plane
W-Wind
E-Earth
(horizontally) ->
v(A,E)=v(A,W)+v(W,E)
thats true for any case, in our case v(A,E)=0 lets apply that
-v(A,W)=v(W,E)
hmm that seems possible, and v(A,W) or v(W,E) doesnt need to be 0 so if v(A,W) isnt 0 then, v(W,A) isnt 0 either
so that means wind blows past the air planes wings at a relative speed to the plane so we can put Bernoulli now right? If that velocity provides enough upthrust…away the plane goes..but vertically! that is..
September 3rd, 2009 at 1:15 pm
k, the airplane is a tricky one. one thing we have to take into account is how we measure the speed of an airplane.
if we measure relative to ground it surely is no problem to create such a treadmill that the airplane won’t take off (as all the thrust is lost to friction of the wheels -> total velocity is 0).
if we measure the speed relative to the surrounding air the plane will most surely take off and the wheels will simply experience double the revelations they would in comparison to a “normal” takeoff (as the “floor” moves at the same velocity as the plane, but in the opposite direction)
I’m not sure about the aces though, but that may be related to me not knowing the rules of bridge.
the thing with the “0.99~ = 1″ I always explain with
3 * 1/3 = 3 * 0.33~ = 0.99~ and 3/3 = 1
September 4th, 2009 at 5:20 pm
Airplane on a treadmill:
An easy way to think of this is that you know that the connection between the plane and the ground is frictionless (along the horizontal axis), so replace the treadmill with ice and the wheels with blades.
Now it’s pretty obvious what’s going to happen.
September 5th, 2009 at 10:47 am
damn.
all those problems are brilliant.
i agree with u being screwed by these.
September 13th, 2009 at 10:33 am
x=.999~
10x=9.999~
10x-x=9
9x=8.999~
x=.999~
x=/=1
~ denotes ∞
September 13th, 2009 at 10:38 am
“the thing with the “0.99~ = 1″ I always explain with
3 * 1/3 = 3 * 0.33~ = 0.99~ and 3/3 = 1″
0.33 =/= 1/3 it’s an approximation. The only way to properly express 1/3 is 1/3, 2/6, 3/9… etc.
September 13th, 2009 at 4:46 pm
Jay:
In your first argument you go from 10x-x=9 to 9x=8.999… without any justification (i.e., why do you write 9x=8.999… rather than 9x=9, the logical next step?) You seem to be assuming that 9=8.999…, which implies that 1=0.999…, which proves the same thing you are trying to disprove.
In your second argument you say 0.33 is not equal to 1/3, which is true, but 0.333… (with an infinite number of 3s after the decimal point), the number we are talking about, is exactly equal to 1/3. To show this, we note that 0.333… is the sum of the infinite geometric series 3/10+3/100+3/1000+3/10000+…. Using the formula for an infinite geometric series a/(1-r), where a is the first term and r is the ratio between consecutive terms, we have the series equal to (3/10)/(1-(1/10))=(3/10)/(9/10)=3/9=1/3. We can do the same for 0.999…, so 0.999…=9/10+9/100+9/1000+…=(9/10)/(1-(1/10))=(9/10)/(9/10)=1
September 13th, 2009 at 4:51 pm
Whow, nice. It’s hard to not think about these questions the rest of the day..
September 13th, 2009 at 10:06 pm
Want some cheese to go with that wine?
September 14th, 2009 at 12:25 am
“0.33 =/= 1/3 it’s an approximation. The only way to properly express 1/3 is 1/3, 2/6, 3/9… etc.”
Well, he didn’t say 0.33, did he?
Personally I’m a bit stuck on the airplane. As someone mentioned before, about theorists and realists, if the treadmill matches the speed of the wheels, it can’t really take of, can it? The wheel would have to be moving faster than the treadmill. That’s what sounds right to me at least.
September 14th, 2009 at 1:33 pm
2-Aces Paradox Intuitive Explanation:
I will skip over the Combinatorial explanation as that is given, albeit vaguely, above. Consider all hands which have at least one ace in them. Now consider all the hands which have exactly one ace in them, there are four distinct categories of hands defined by the suite of the ace in question. So, the case where an ace of spades is in the hand represents 1/4 of all these hands. Now, if we allow there to be exactly two aces, more than one fourth of the hands have an ace of spades in them, don’t believe me, write it out for yourself, there are only 6 possible 2-ace combinations, you’ll find that one half have an ace of spades in them, this is intuitive since the ace of spades may appear as either of the two aces. This same reasoning extends to three aces, four is trivial. Thus, there is a better ratio between more than one ace hands and specific ace hands than between more than one ace hands and non-specific ace hands.
Hope that this helps,
PS the plane will most certainly take off since its movement against the ground is arbitrary to the problem and the envelope paradox is a result of inaccurate use of variables, it is not a calculation of an expected value since it treats the value of of the envelope picked to be a fixed amount, but then uses it as a variable in its equations, if you ignore the flawed approach and take it intuitively it becomes apparent that the expected value is 3/2 of the lesser amount, which suggests that it doesn’t matter if you switch the envelopes or not.
September 14th, 2009 at 7:16 pm
As for the airplane, I am a Mechanical Engineer and will try to give some insight. First, about jet engines: they create thrust and push the plane with air. Basically air enters through a large inlet, and is forced out of a smaller outlet. This means it leaves at a higher velocity so it accelerated… by Newtons 3rd law there is acceleration on the jet engine and therefore on the aircraft wing at the point where it is mounted. If the wheels have no mass, no speed or acceleration of the treadmill can keep the plane from taking off. If you assume the wheels have mass, then by changing the velocity of the treadmill, you can make the wheels generate a force on the airplane (changing the velocity creates a force on the wheel, some of that will go to accelerating the wheel and some will get put through the axle). This makes friction irrelevant to the absolute solution of this problem as it can basically be lumped in with this force from the wheel on the axle. What matters is that if you accelerate the treadmill to infinity you can hold the plane in the same spot. There is one problem is that this accelerating treadmill creates a huge boundary layer which will create an airspeed over the wings. Therefore even in the event of a treadmill accelerating off to infinity an airplane will take off.
September 14th, 2009 at 8:14 pm
The airplane problem is simple as soon as you remember that the wheels on the plane are completely irrelevant to speed. They could move freely in any direction whether that direction is forward of backwards. The wheels are there just to create as much of a frictionless surface as possible between the plane and the ground. In a frictionless surface the wheels will just keep the plane from hitting the ground. the plane’s propellors make the plane move forward.
its just like pulling a piece of paper from under a toy car (or anything with wheels on it that aren’t attached to fixed gears. Plane wheels are loose. They move anyway they please.). if you do it quickly, friction is mostly meaningless and the car will stay in place. If that car had a propeller on it, the it would move in the direction that the air is being pushed.
September 15th, 2009 at 1:15 am
Of course the plane can not take off. The plane crashes, and everyone in it dies.
Exhibit A: Take a small aircraft powered by a turboprop engine. The pilot is preparing for takeoff. He taxis to the runway and guns the engine. Simultaneously, some asshole, who earlier in the day replaced the runway with a treadmill that looks an awful lot like a runway (enough so to fool the pilot), turns on this massive engine that powers this massive treadmill that runs at enormous speeds. Lets say it runs at twice the takeoff speed of an aircraft.
For a second or two, the pilot doesn’t notice anything out of the ordinary. But little does he know, the landing gear is operating at twice the speed it was designed for (he’d better be praying the design factor of safety far greater than 1 (it probably isn’t)).
Worst case scenario, the landing gear experiences catastrophic failure before the plane can take off. The plane is now coming into contact with the treadmill (which moving really really fast, mind you) with its mangled mess of a landing gear, while the engines are running at full speed. I pray for you pilot, for when the treadmill and your engines violently hurl you and your tiny plane into all sorts of directions, may God’s blessing shine down on you from above.
Absolute best case scenario, the plane somehow takes off before the landing gear fails. Good luck trying to land that thing.
Exhibit B: Take a large turbojet commercial airliner. Not a chance in hell this thing can take off. If the pilot has the awareness to turn off the engines before the jet violently spins out of control on the runway, not everyone dies.
September 16th, 2009 at 12:48 am
Aces…isn’t this a misapplication of probability? The explanations listed here seem to imply that the probability of Alice having a second ace is equivalent to the probability of any hand having two aces. Shouldn’t the probability of Alice having a second ace be properly defined as the probability of having two aces conditional upon having a single ace first. Regardless of the suit of that single ace, these probabilities should be identical, no?
September 16th, 2009 at 1:00 am
Actually I found the answer in a way I understood here:
http://usna.edu/Users/physics/mungan/Scholarship/TwoAces.pdf
September 16th, 2009 at 8:30 am
A different way to explaining the Airplane on a treadmill problem to people who still believe the author’s solution is wrong:
A wheel’s purpose is to negate the effects of friction right? A plane’s wheels on a normal runway is to prevent friction interfering with the engine’s thrust. A perfect wheel has no friction.
Ice (with a layer of water on top) also has no friction (against other ice). So in physics in terms of friction, ice and wheels have the same properties. Put a slate of ice under your go-kart and slide it; put wheels under your go-kart and slide it; same result. Put nothing, friction stops you in an instant. Ice and wheels perform the some job.
Now to the problem, imagine the planes uses ice blocks instead of wheels to negate friction; ice, wheels, it does the same job. Then get the treadmill (with the plane supported by iceblocks on top) to run backwards at 10000miles per hour. Regardless of the speed of the treadmill, nothing moves. The ice just slides as the treadmill runs. ALL STATIONARY.
The plane’s engines start. Before, the ice blocks have always been stationary. They haven’t moved in relation the to ground (or the air). But now, they are forced to MOVE FORWARD by the plane’s engines, as the ice blocks are connected to the plane and engine. All the while the treadmill however is still going backwards 10000miles per hour. A quintillion miles an hour, its irrelevant. The plane itself edges forward. Then it takes off.
I used the ice thing to free people’s mind from the wheel, as they assume too much from cars and buses. The source of thrust for the plane is NOT the wheel, its the engine. If is was the wheel, things would be very different.
If you still think I’m wrong: Ever tried walking on ice? If the ice was pure frictionless, you won’t move at all. If you somehow don’t fall, you’ll never move in relation to the earth no matter how long you walk. Now get a rocket booster behind your and set it off. You’ll shoot off. If you where a plane you’ll take off.
September 16th, 2009 at 6:21 pm
LOOOOOL U R NOOB AIRPLANE NEEDS AIR FRICTION UPLIFT UNDER THE WINGS TO TAKE OFF DOESNT MATTER BOUT THE ENGINES IF THE UPPER AIR FORCE ISNT THERE, WHICH IT ISNT BECAUSE THE PLANE NEEDS TO MOCE TO CONTROL THAT SO NOOB SAUCE ALL OVER U
September 18th, 2009 at 1:53 pm
About the .9999~ and 1.0 problem . . . .
The reason they are NOT equal is because if I slow down infinitely, but never stop, I may LOOK stopped, but I am not stopped. If I have .99999999~ I may look, to the casual observer on any level, to be close enough to 1.0 to be 1.0, but to the person applying any effect that can work on .99999~ and not on 1.0, they know the difference. It is simply how you cut your bread, or define your edges. Those who think it IS 1 are defining an edge that says I will not recognize the infinite pattern as anything except 1, because I feel I hav eto round AT SOME POINT. Those who have to look at every single digit, infinitely on forever, will tell you, “I’d really love to see a 1 about now.” It’s kind of a narrative issue, though, since we work at a finite level in our infinite universe.
September 18th, 2009 at 1:54 pm
Oh, there’s a joke about the slowing thing. A cop pulls over a man for running a stop sign. The man tell the cop, “I slowed down!”. The cop pulls him out of the car and proceeds to beat the man. Then he asks the man,” Would you prefer I stop, or just slow down?”
September 18th, 2009 at 3:55 pm
0.999… = 1 vs 0.999… 9A
1 = A
and therefor 0.999… = 1
This buggs me because it does not seem to be proper and I don’t find it to be entirely true. How ever we quickly find our selves in the philosophy of the different infinities. Let me illustrate my point of view;
0.999…999 = A
9.999…990 = 10 x A
8.999…991 = 10A – A => 9A
0.999…999 = A
In the other equation above we subtract A from 10A but do not take in consideration that we actually have one less decimal.
I don’t mind that we approximate 0.999… to 1 but to claim that it de facto is one does bother my mind. Perhaps there is an important distinction we’ve ignored or perhaps I fail to wrap my head around what we mean when we say infinite?
I humbly stand my ground.
September 18th, 2009 at 3:57 pm
0.999… = 1 vs 0.999… < 1 "ruined" my day.
Spent an hour arguing with a friend who's more familiar with mathematics than I am. We seem to understand each others points of view but we do not agree. I am the one fighting against the current on this one it seems.
I appreciate the following two explanations above;
1. If .9~ and 1 were distinct, they’d have an infinite amount of numbers between them. Since they don’t, they cannot be distinct.
2. the thing with the “0.99~ = 1″ I always explain with
3 * 1/3 = 3 * 0.33~ = 0.99~ and 3/3 = 1
no 1. I do not understand why two distinct numbers need an infinite amount of numbers between them to make them distinct. Would it not suffice with a tiny fraction? Even if it is "infinitely small" ?
Please try to avoid clarifying with the same argument used to prove that 0.999… = 1 as it is the issue of the paradox. In other words, 0.000…1 = 0 is not a valid argument in clarifying the matter.
no 2 Seems to make sense in a sense. ie; I have always been tought that a third is 0.333… And three thirds make apparently both 1 and 0.999… It seems to me however that the three combined 0.333..'s make a somewhat different 0.999… than the 0.999… issued in the paradox.
The kind explanation I have come across on the net states, as above, that;
0.999… = A
9.999… = 10 x A
September 19th, 2009 at 6:14 pm
The Airplane problem:
Many people are saying that the airplane WOULD take off, citing that the force is being applied to the air and NOT to the wheels.
If I understand the problem being posited correctly, the engines are applying force equal to say, 200mph, and the conveyer belt is moving in the opposite direction at that same speed. If this is what is happening, the airplane is stationary with relation to both the surrounding air AND the earth. Since no air is flowing over the wings no lift is being generated and thus the plane cannot take off.
I watched the mythbusters video on this topic, and the large-scale experiment they do is not the problem being spoken of here. In that experiment the plane is moving at a speed suitable for lift-off (relative to the earth,) which just so happens to be on a conveyer belt moving in the opposite direction.
If there is something that I am missing here, please correct me because I desperately want to understand this problem. Thank you.
September 21st, 2009 at 8:16 am
Simplify the issues!
A plane is tied to a solid building by a rope.
It guns its engines! Does it rise vertically?
I think that some planes do and others do not, depending on wether the engine can pull or push air over the wings to create enough lift. My guess is that most engines are not designed to do this. My guess is that the main function for propellers and jets are for forward ‘point’ thrust only, and that lift is mostly a function of the wings speed relative to the air.
Correct me if I’m wrong ;p
September 21st, 2009 at 4:53 pm
Just a quick one, and i think someone might have said this earlier but it has taken me quite a while to read through all of the arguments. The fundamental reason that the Plane on a conveyor belt WOULD take off is that no force is actually applied through the wheels. The wheels are simply there to REDUCE friction between the body of the plane which is being propelled via application of force to the surrounding air and the ground which would otherwise produce friction and therefore a much greater force would have to be produced by the engines….. I think i might be rambling tell me if i am……..
September 21st, 2009 at 5:35 pm
I’ve never seen you with a rucksack Craig! The thrust of the jet engine acts on the air surrounding aircraft, not on the ground. If the thrust of the engines generated an airspeed of 200 mph and the belt moved at 200 mph the aircraft wheels would be in effect traveling at 400 mph, the fuselage & wings 200 mph. Unless there was any serious friction in the wheel bearings (there isn’t), the airspeed would generate lift under the wings and you fly. That will make me a realist then!
September 24th, 2009 at 1:14 am
Mythbusters did an episode on the aeroplane problem, and they managed to get flight.
September 24th, 2009 at 2:20 pm
the two envelopes is a logic problem, you have to make the decision to take the one with the most without ever making the decision, that way you get the more money
I’ll take the envelope you decide not to take
September 25th, 2009 at 2:02 am
Airplane Problem:
I am an airline pilot and you guys are all right…based on your very different assumptions. I might be getting a little too deep into this but screw it…its fun.
I am going to make a few of my own assumptions 1:zero wind, 2:airspeed and ground speed are equal, 3:small planes take off at 60 knots of airspeed, 4: big ones take off at 160, and 6:friction exists(because it does)
It has been said before but:the mythological treadmill would probably have to auto-adjust itself to one of two speeds 1-Ground speed (meaning the speed you are moving past the non-treadmilled ground) or 2-wheel/tire speed.
1-If the treadmill speed is adjusted to ground speed the model of the plane would matter a lot. In the case of mythbusters the small plane requires so little airspeed to leave the ground that causing the wheels to spin at double take-off speed (120 knots instead of 60) is not going to stop it from flying. In a jet where you have maximum tire speeds in the 180 range (due to friction) doubling 160 knots would liquify the tires and very quickly ruin the flight.
2-If the treadmill were “programmed” to match tire speed then it would just accelerate toward infinity as the aircraft accelerated until the plane tires or the belt or something broke.
I read a bunch of the posts and have forgotten a lot of the arguments. I meant to make a more detailed post but its bedtime.
September 27th, 2009 at 2:45 pm
As far as the airplane problem goes no one seems to be reading the original question right: the conveyor belt is set to move at the speed of the airplane. Not the wheels, or air or anything else. If the belt were moving at some astronomical speed so would the airplane and if the airplane were stationary the belt would be too. The only way for the belt to exert a significant frictional resistance would be if the airplane were moving at a significant speed. And as others have pointed out that friction would only exist at the wheel surface and axle which is probably not enough to counteract the immense thrust and inertia of the plane itself. The plane would be relatively unhindered in its attempt to move faster regardless of the fact that the belt was moving just as fast in the opposite direction. The whole argument against it taking off fails when you consider the speed the plane would have to already be going for the belt to generate enough friction to slow it down. I.e. it would already be airborne.
September 27th, 2009 at 8:35 pm
The friction IS enough to keep SOME planes from taking off, that was the point I was trying to make. Lets be really specific.
The plane I fly can produce enough lift for flight at about 160 miles per hour. In your response you seem to assume (like I would) that the belt would rotate at the same speed the airplane is traveling but in the opposite direction. If the plane is moving forward at 50 miles per hour then the belt spins backward at 50 miles an hour. In this case the plane tires would be spinning at 100 miles per hour while the plane travels through the air at 50.
In the case of my plane the belt would be traveling an extra 160 miles per hour toward the plane at takeoff speed. If the plane is going 160 that would mean that the tires would be rotating at 320 miles per hour. This is about 120 miles per hour faster than the maximum tire speed for my plane. The tires are not designed to handle the friction produced by that kind of speed. They would effectively liquefy. The disintegrating tires would add even more friction and the plane would not be able to accelerate to 160 mph. The plane would accelerate…for a time. The friction would not be a factor until the tires got going around 200mph (100 mph of actual groundspeed/airspeed) and then it would increase exponentially as the tires turned to glue.
October 4th, 2009 at 12:19 am
An interesting note on the .9999~ = 1 problem. It is only accurate in a base ten number system. However, it has it’s analogs in any other base. In haxidecimal, .FFFF~; binary, .1111~
October 26th, 2009 at 10:33 pm
Aces paradox! I Dont get it!
james said
Aces…isn’t this a misapplication of probability? The explanations listed here seem to imply that the probability of Alice having a second ace is equivalent to the probability of any hand having two aces. Shouldn’t the probability of Alice having a second ace be properly defined as the probability of having two aces conditional upon having a single ace first. Regardless of the suit of that single ace, these probabilities should be identical, no?
I agree. Ill flesh his argument out. Alice only has 2 cards (assumption). compare scenario I. with II.
scenario I. Alice has an ace (any ace). The chance she has one more ace 3/51 (remaining three aces out of remaining 51 cards)
scenario II. Alice has an ace of spade (A♠). The chance she has one more ace 3/51 (one of the three aces left A♣ A♥ or A♦ out of the remaining 51 cards)
to really flesh out I. (which i believe some may dispute)
Alice is has an ace..the ace she is referring to is either. A♠ A♣ A♥ or A♦. there is a 1/4 chance that is ace being referred to is a specific suit.
1/4 she has A♠ … chance the second is an ace? refer to II.
1/4 she has A♣ … tweak II changing suits slightly
1/4 she has A♥ … ditto
1/4 she has A♦ … ditto
each of the above has 1/4 chance being the scenario in question
each scenario has 3/51 chance she has a second ace.
total each scenario up
= 1/4 x 3/51 + 1/4 x 3/51 + 1/4 x 3/51 + 1/4 x 3/51
= 3/51 x (1/4 + 1/4 + 1/4 + 1/4) …. (factored out 3/51)
= 3/51 x 1
= 3/51 (same answer)
SO WHAT THE HELL AM I MISSING HERE?
October 26th, 2009 at 11:07 pm
As for plane… some ppl still dont get it…unbelievably
Ill try to add an explanation in my words…maybe it will help
(all speeds are relative to an observer who is just standing there)
imagine a few scenarios comparing the plane speed (P) vs conveyor speed (C)
1. P = 50mph vs C = 0mph
easy! plane moves forward
maybe get enough lift
the plane and flys
2. P = 0mph vs C = 50mph
interesting! lets break this up into two
2.a The breaks are on.. oh noes
The wheels dont spin at all!!!!!
The friction between the conyeyor and wheels pulls the plane
The plane goes backward..
the plane aint taking off! lol
2.b The breaks are NOT on!
The wheels spin on its axel
The plane is unaffected to its spinning wheels
The plane just sits there
The plane does not get lift…
The plane does not fly
(It may help to think of those people who pull table cloths off fully set dinner tables. http://www.youtube.com/watch?v=cuG8sIiV8iQ . If the plates where all on wheels, it would work even BETTER!)
3. P = 50mph vs C = 50mph
Here we go! From now on breaks are never on.
The propeller (or jet engine)
thrusts (forces) the plane forward using air.
The plane pulls its wheels forward cause they are attached
But they are allowed to spin freely,
cause the breaks are not on.
The wheels spin because they are touching the (ground)
The ground is moving because of the conveyor,
making the wheels spin faster.
The faster spinning wheels do not affect the plane at all,
because the breaks are not on
The plane continues moving forward at 50mph
Plane gets lift
Plane flys
And just for shits and giggles
4. P = 50mph vs C = 1,000,000,000 mph
The propeller (or jet engine)
thrusts (forces) the plane forward using air.
The plane pulls its wheels forward cause they are attached
But they are allowed to spin freely,
cause the breaks are not on.
The wheels spin because they are touching the (ground)
The ground is moving because of the conveyor,
making the wheels spin LIKE A MOFO.
The UBER spinning wheels do not affect the plane at all,
because the breaks are not on
The plane continues moving forward at 50mph
Plane gets lift
Plane flys
October 26th, 2009 at 11:39 pm
Real simple.
1/3=0.333~
3*1/3 = 1
3*0.333~ = 0.999~
Therefore,
0.999~ = 1
November 5th, 2009 at 12:44 pm
The airplane is only overcoming rolling resistance, not the resistance of having to displace air. Lift is created by air flowing over the wings and air is not flowing over the wings.
November 5th, 2009 at 1:12 pm
So I understand now that the airplane is actually trying to take off, that is, the pilot is acting as if s/he is performing a normal takeoff. Then yes, the plane takes off as long as the rolling resistance doesn’t exceed the ability of the engines to move the plane forward at the required takeoff speed, and assuming we can ignore effects of the wheels heating up or disturbing the air with increased rotation, and also ignoring the effect of the conveyor dragging air along with it (which would only help the airplane reach it’s required takeoff air speed anyway). So yep, plane takes off.
November 9th, 2009 at 11:46 pm
Swenick,
“Would it not suffice with a tiny fraction? Even if it is “infinitely small” ?”
If there is an infinitely small fraction between 0.999~ and 1, there is an infinite number of infinitely small fractions between 0.999~ and 1. You’re thinking that after 0.000~ infinitely, there’s a one at the end somehow, filling a gap. The thing is that that the nines go on in infinity. There is no gap between 0.999~ and 1. Not even an infinitely small one.
The thing that makes this hard for you is your intuition. Your intuition, which comes from experience with other decimal numbers, makes you do a logical leap as you assume that there is a gap between 0.999~ and 1. Essentially, as soon as you see the 0. part, your mind says “This number is obviously less than 1″.
Even though it looks as if it would be, it is not as has been presented in a couple of different ways.
November 12th, 2009 at 12:18 am
Martin, here’s what you’re missing.
Let’s group all cards into two categories: the Aces and the not-Aces. We will further limit the comparison to two-card hands. Alice is dealt two cards, which may be either Aces or not.
A-X
X-A
A-A
From this, it’s clear that (similar to the do-you-have-any-daughters problem mentioned above) if the chances of getting an Ace and a not-Ace were equal, then the chance of having two aces would be one-third. Suit is not important in this calculation.
Bob’s hand is more complicated, because it’s suit-specific. He could have any of the following two-card hands:
♠-X
X-♠
♠-A
A-♠
Again, the “daughters” problem should help illustrate why order is important here.
If the chances of getting an Ace and a not-Ace were equal, then Bob’s chances of having two aces would be one-half. Distinction between hearts, diamonds and clubs is not important in this calculation.
Now, I’ve made the obviously-incorrect assumption that the chances of getting an Ace and a not-Ace are equal. Since they aren’t, we adjust for this by multiplying by the actual probability of Ace vs. not-Ace. However, we multiply by that same probability in both Alice’s and Bob’s hands. Therefore, they cancel each other out in the ratios, and Bob is still 1.5 times more likely to have a second Ace.
By the way, the two-daughters problem goes like this: either the older or the younger child must be a girl, but we are looking for the probability that they both are. So, originally the four possibilities for the children’s genders were:
MM
MF
FM
FF
The first has been eliminated by the fact that we know there is at least one daughter, but because of the importance of order, the middle two remain separate and the chance is 1/3. Had the man been told that the ELDEST child, specifically, was female, then the first two scenarios would both be eliminated and the chance of having two daughters would be 1/2. This second setup is analogous to Bob’s hand.
November 21st, 2009 at 7:40 am
Airplane:
let’s assume that a toy airplane is on the treadmill and you push it forward with fingers (jet propulsion). nevermind the speed of wheels and the treadmill plane will go forward and will take off
November 28th, 2009 at 11:03 am
.9~ and 1
If .9999~ = 1 logic implies that 1/ꝏ=0. If 1/ꝏ=0 then according to algebra rules 1=0*ꝏ and 1/0 = ꝏ.
The two envelopes problem
It should rather be called the 3 amounts problem.
Lets take 3 amounts A B and C where B=2A and C=½A
Amount B and C are put in envelopes, the envelopes are put in a box one is randomly chosen and given to the poor.
Amount A is put in an envelope and the envelope joins the one in the box, and the game begins.
There is a 1/4 probability that you will get B.
There is a 1/4 probability that you will get C.
And there is a 1/2 probability that you will get A.
December 3rd, 2009 at 8:42 pm
After first reading the airplane problem, I was pissed because I just KNEW the airplane would NOT take off. After reading all of the net, I finally understand why it would (because I’m an ideal situation kinda guy), and I feel I’m a more complete human being because of it.
For others that will finally see the light, here’s the question that causes the disconnect: Is it an ideal situation? (i.e. Is there friction in the wheel assembly? Is the treadmill speed limitless? Is everything indestructible?)
December 10th, 2009 at 10:38 pm
let x= 0.9999…
therefore
10*x = 10* 0.99999…
multiplying by ten just means to move the decimal over one place therefore
10x = 9.99999…
subtract x from both sides
9x = (9.99999…)-(0.9999999)
The infinite series of nines cancel each other. 9-9=0
therefore
9x = 9
divide by 9
x = 1.
But at the beginning of the problem, we defined x as .999… and now it is defined as 1. Therefore, .999… = 1
December 21st, 2009 at 3:26 pm
Airplane problem:
It doesn’t matter what the wheels are doing. It doesn’t even matter that there ARE wheels. The ONLY reason there is any kind of support beneath the plane at all is because gravity causes the plane to sit on the runway. The only reason this support is wheels, that turn, is because the plane engines are going to thrust the plane forward through the surrounding air and it’d be a lot harder for them (and a pain in the ass to land…) if they had to drag giant kickstands along until the plane reached the speed required for takeoff.
The engines of a plane move the plane through the air around it, which is plentiful EVEN when the plane is on the ground. The plane will move relative to the surrounding air no matter how fast (and it’d have to be REALLY fast) the treadmill was going. If you hung the airplane from a string so that the wheels weren’t even touching anything, it’d still move forward. Except for countering the effect of friction, the wheels are completely irrelevant to the plane’s forward motion.
No more arguing. The plane would take off. Deal with it.
January 18th, 2010 at 10:08 pm
The treadmill spins the wheels backwards. The wheels move and the axle remains still. The axle is fixed to the fuselage, therefore the fuselage remains still. The turbines on the undersides of the wings rotate and push against the air. The wheels continue to be spun backwards, but because they are not fixed to the fuselage, the plane is pushed forward by the turbines. The friction between the wheels and the treadmill does not matter because the plane does not generate thrust by pushing on the ground (which is where the wheels are). As a result, the plane is pushed forward until it gains enough speed for the air currents around the wings to generate the lift needed to hoist the plane into the air.
If you could prevent a plane from taking off by preventing its wheels from spinning, you could cause a plane to drop out of the air by stopping the movement of the wheels in mid-flight, and seeing as the wheels are folded up during flight without the plane falling to the ground, it can be concluded that a plane attempting to take off from a treadmill that causes the plane’s wheels to remain stationary will take off as long as the turbines are spinning fast enough. The turbines can spin no matter what is happening to the wheels, so if the pilot starts the turbines, the plane will take flight.
It is not possible for a treadmill to render an aeroplane unable to fly.
February 9th, 2010 at 3:05 pm
Ah, thanks! This cleared up some contradictions I’ve heard.
February 23rd, 2010 at 12:39 am
As a college student I am familiar with these types of problems and the various, unscrupulous conditions that can change the computation of the information. I believe, my friend, that your memory is forgetting certain distinctions necessary to prove your points. For example in airplane problem you are forgetting that lift occurs when air molecules flow around the wing causing an upward force. Where there is curved pressure flow there is a pressure difference between the air above and below the wing. If the planes forward force is neutralized by the treadmill there would not be enough pressure cause by the standing air around the plane to cause successful lift. Perhaps a better question would be “If an airplane with an attached parachute, in order to neutralize the airplanes speed, was put in front of a large fan could it be able to take off?
February 24th, 2010 at 11:01 pm
The plane canNOT take off! If the conveyor, and the plane are moving in opposite directions at equal speeds the net is zero. Thus, although the engines are working, and the ground (conveyor) is moving, the air moving over the wheels is zero. I assure you that no plane, or helicopter can take off with no air moving relative to the airfoil. In this case, the pilot could pull on the controls all that they wish, and the weight of the aircraft would still be stuck to the surface no matter how much speed nor how fast the wheels move. Aircraft fly on airspeed and NOT on ground speed. NOTE: This is why planes crash when suddenly hit by a tail wind on landings or takeoffs as the airspeed (not the ground speed) suddenly changes to less than that required to fly. Next time that you fly, be grateful that people a lot smarter than you are in the cockpit when you think the speed of the wheels or the power of the engine is what makes things work.
March 15th, 2010 at 10:50 am
Did not read the whole posts so this is possible already clear by now. But to the plane problem, the above CollegeStudent is the closest.
In very very simple terms, the normal planes (not those with Vertical Take Off/Landing like Harrier) take off because the air hits the flaps (those moving parts from the back of the wings) and that makes an upward reaction. The motors are just to increase the speed in order to reach that moment. So because the plane stands still relatively to the ground/air it can not take off.
Think for the cars, all the spoilers are there to make the car being pushed to the ground by air at higher speeds. Some similar think is here, only in reverse.
March 25th, 2010 at 6:23 pm
The one with the cards is wrong as is the one with the envelope.
The cards one has been made more complicated than it really is and the order is also irrelevant.
In game ‘A’ taking place in LA Alice says she has an ace. Which means there are 3 aces left that she could have in her hand. Assuming we don’t know what the other players at the table have, the odds of her having another ace are 3/51 multiplied by the number of cards in her hand(don’t play bridge. is it 12 or 13 cards in a hand?) In game ‘B’ taking play in NY Bob says he has the ace of spades. Which means that there are 3 aces left that he could have in his hand. Assume again we don’t know the other players cards, the odds of him having another ace in his hand are 3/51 multiplied by cards in his hand. I play a lot of poker I’m used to working out the odds of different hands. Look at it this way…if you’re playing texas hold’em and the first 3 cards are dealt face up on the table and one is an ace, if you specify it is the ace of spades do the odds of another ace coming up in the next 2 cards to be turned up change?
In the envelope problem…if X is the lower amount then 2X is the greater amount. You have a 50/50 chance of getting the larger sum. As RonK posted earlier it changes if you can look inside and then change envelopes. He explained it perfectly, so I will not attempt explain it again.
March 26th, 2010 at 7:11 pm
With the plane problem…
I get it. I don’t care about the wheels. Or brakes. I’m comfy with that logic.
I don’t understand how lift can be achieved without moving air. If our plane is stationary relative to the air around it then no lift, right?
The argument I’ve heard is that it’s impossible to have a completely stationary airplane in these circumstances, but my counter is that it’s impossible to have a freak monster conveyor belt too and quit screwing with the rules of a thought experiment.
Somebody explain the lift so I can continue my life. And don’t talk about aerostatic motion or VTOL or anything stupid like that. Only normal airfoils using old school lift is what’s obviously assumed here.
March 27th, 2010 at 1:19 am
Just pic a fucking envelope
March 27th, 2010 at 5:34 pm
For the airplane problem, if it is a propeller-powered airplane, then the propeller will force the air across the wings, creating lift, allowing the plane to take off.
So although (under ideal theoretical conditions), there would be no motion relative to the ground (at least initially), the plane does have motion relative to the air, which is all that is needed for lift.
In fact, although not as logically intuitive, a similar process happens with a jet engine (although where jet engines push, propellers pull). But the thing is, jets don’t push against the ground, they push against the air.
So unless the air somehow remained stationary over the wings, and at the same time moved away from the jet engines at the same velocity as the thrust from the engines, the airplane would take off.
I hope this answers everyone’s questions.
March 27th, 2010 at 5:35 pm
Actually, on second thought, jets and props aren’t so similar.
March 27th, 2010 at 9:02 pm
To the airplane argument: I’m pretty sure the jet engines needs to be going a certain speed against the air to produce the resistance it needs to take off, it has nothing to do with the wheels actually propelling the plane.
March 28th, 2010 at 2:50 am
my ass, .999whatever < 1, the representation for a range between 0 and this is [0 to 1) meaning u will never ever reach 1, of course if ur measuring something on the real world where that kind of precision is usually not required , like if a piece must be 1cm radius and its .99999999999999999whatever then its considered 1cm for all what matters, but take this and store it in your mind, i said its considered, but not that it actually is 1cm. and me, u actually forgot to substract the inf series on the other side of the equation, breaking the balance.thats what we get from arranged algebra, plz note that any number can be equal to any other number when using algebra if not try this on google 2=1. and look for some really curious stuff.
March 28th, 2010 at 6:33 pm
I think whats confusing about the plane problem is that it says the conveyor belt goes at the same speed as the PLANE, not the wheels. If the plane has a velocity (relative to the ground), the conveyor will move. For lift to occur the plane must accelerate forward compared to the ground. Even if the plane and the conveyor belt go the same speed, the wheels will go that many times faster than the conveyor belt, and the plane will take off.
March 28th, 2010 at 8:12 pm
Airplane one is on Mythbusters.
March 29th, 2010 at 2:29 am
The Airplane problem had me for a second… and I must agree, the plane will take off.
The wheels have no way of gaining speed. As a matter of fact, you don’t NEED wheels. Look at any seaplanes out there(or planes with skids or skis for landing on snow covered lakes and glaciers), and you will understand better.
Wheels simply create less friction than other methods for dry, solid land, and less friction means the plane needs less force to achieve liftoff velocity. Planes do not use wheels to power any sort of acceleration, they use propellers or turbines to produce force, which has nothing to do with ground movement.
That is another reason an aircraft taking off into a headwind (a wind blowing from the front of the plane to the tail) will takeoff in a shorter distance than a plane taking off with a tailwind.
Think of the engines as bottle rockets… You light one above a treadmill and it will launch itself forward regardless of how fast the treadmill is going… Now simply put a pair of wheels on this imaginary bottle rocket… the same forces that propel the bottle rocket forward are in effect, and the wheels are simply there to hold it up, but it will go forward.
March 29th, 2010 at 8:30 am
After some deep thought on The two envelopes problem ; where one envelop contains twice the value of the second and the recipient is given one and the option to switch.
Giving values to the envelopes 2 and 4 the expectation value is 3.
No matter which envelope is considered, its expectation value is 3, switching envelopes does not increase expectation value. And thus switching choice does not increase expectation value at all.
I really appreciated these paradoxes and the comments tring to work them out.
Any thoughts on my solution?
March 29th, 2010 at 3:59 pm
Reading this thread reminds me of the effect produced when an ant mound is sprayed with insecticide. The ant hill erupts with ants in random motion. Every ant is behaving independently of the other ants and in ways that do not relate to the original stimulation.
April 1st, 2010 at 3:30 am
About .9~ and 1
1/3 = 0.3~
so 0.9~ = 3 * 0.3~ = 3/3 = 1
thus 1 = 0.9~
so 1 – 0.9~ = 0
and since
1 – 0.9 = 0.1 = 1/10 and
1 – 0.99 = 0.01 = 1/100 and
1 – 0.999 = 0.001 = 1/1000
we can say
1 – 0.9~ = 1/infinity
and so 1/infinity = 0.
So problem you have sprouts from the fact that 1/infinity = 0;
April 1st, 2010 at 11:43 pm
The plane will take off because the air displaced by the jets will create lift. The critical value for a plane taking off is lift and not forward momentum. No?
Is the “Aces” question specific to Bridge and the hands that win in Bridge?
April 2nd, 2010 at 3:05 pm
About the Ace of Spades
I think you and many others cite the problem wrong:
It should be like this:
Alice is playing Bridge, You ask Alice if she has at least on ACE, And if she responds “yes”, her probability of having a second ACE is lower than asking her if she has the ACE of SPADES and she says yes.
See also these pages in Martin Gardner’s Hexaflexagons and other Mathematical Diversions:
Maybe this wrong citing comes from the orginal form in which it was presented by Gardner:
===
Perhaps even more astounding is the paradox of the second ace. Assume that you are playing bridge and just after the cards are dealt you look over your hand and announce, “I have an ace.” The probability that you have a second ace can be calculated precisely. It proves to be 5359/14498 which is less than 1/2. Suppose, however, that all of you agree upon a particular ace, say the Ace of Spades. The play continues until you get a hand which enables you to say, “I have the Ace of Spades,” The probability that you have another ace is now 11686/20825 or slightly better than 1/2! Why should naming the ace affect the odds?
===
Note the “The play continues until you get a hand which enables you to say, “I have the Ace of Spades,”"
April 5th, 2010 at 5:47 pm
About .9~ and 1, describe the point on the line immediately before 1 in an equation….
The airplane problem says the plane is moving and the conveyor is counteracting the movement by increasing the speed. Once the plane reaches stall speed, irrespective of how fast the conveyor is trying to offset its movement, the plane will lift. But lift requires air moving across the wing. By counteracting the forward movement of the engines by increasing the speed of the conveyor, there is no lift, thus the plane never leaves the ground. The formula for lift has nothing to do with the formula for the conveyor moving or the wheels turning.
April 7th, 2010 at 6:19 am
@RonK
I was sure to die without understanding the switch problem. You are the man!
From now on, whenever I am offered a choice. I will change my mind to be sure that I now can be pissed quantum mechanics for getting the worse version.
April 8th, 2010 at 12:05 am
can someone explain the daughter one? You meet a man that has 2 children you ask if he has a daughter he says yes, what is the probability that he has two girls? answer is 1/3.
I thought the answer would be 1/2 since if you know he has 1 daughter and only 2 children the only other possibility is 1 girl and 1 boy. With only 2 choices shouldn’t the probability be 1/2?
April 8th, 2010 at 12:44 am
the aces problem is quite interesting. I have discovered my own way to think of it that may make sense. what you have to do is think about piles of cards as all of the combinations that a deck can take on. in a really big pile is all the no aces hands that are thrown completely out no matter what by this problem. in four largish piles are hands with 1 ace of each suit and no other aces, so an ace of spades pile, ace of hearts pile etc… We can all agree that when the player specifies the ace of spades we throw out the other 3 piles leaving 1/4 of the 1 ace only hands. This seems normal so far but what happens to the 2 ace piles the 3 ace piles and the 4 ace piles? with 4 aces that pile stays exactly the same size no matter what so already the proportion of multi ace hands goes up. That is to say that retaining all the 4 ace hands and 1/4 of the 1 ace hands the proportion of multi ace hands goes up. Now look at the three ace pile. It retains 3/4 of itself because only 1/4 of its hands use the 3 non spade aces. A similar thing happens with 2 aces, 1/2 of the 2 ace pile remains because 1/2 of 2 ace hands use the ace of spades as there are 6 possibilities: SH, SC, SD, HC, HD, DC. so we keep all of the 4 aces pile, 3/4 of the 3 aces pile, and 1/2 of the 2 aces pile and we lose 3/4 of the 1 ace pile. So yes specifying the suit matters quite a bit.
April 8th, 2010 at 8:20 am
The aircraft problem. As an Aircraft Engineer I can tell you this as fact. If there is no forward motion then there is No lift produced. If there is no lift produced the aircraft CANNOT take off. Regardless to it’s speed relative to the treadmill.
Ponder this. A pilot can apply his brakes at the start of the runway and get his engines up to full thrust without moving an inch. Then if he removes his brakes does he immediatly take off? NO, because there is no lift. The only reason an aircraft needs speed on the runway is to provide adequate flow over the wings to produce enough lift to counter-act Gravity.
April 8th, 2010 at 8:54 pm
Ah, how maths can muddy the mind. It’s a good brain tool but should not be used beyond its boundaries. Gordon was quite right, but lets stick the antsnest and stir for a bit of fun! Forget the maths, lets try a little thinking instead …
Airplane. Forget the treadmill. There is no treadmill.
Envelope. Solution on Wiki .. simple confusion of variables.
.9~ and 1. This problem is related to the continuuum hypothesis (see e.g. http://en.wikipedia.org/wiki/Continuum_hypothesis).
Watch out, this can be dangerous and has driven strong mathematicians mad or killed them in the past. This one aint going to be resolved quickly on a web blog, be warned…
Here goes nothing …
Multi 9’s is a rational number in base ten. There are an infinite (and uncountable) number of real numbers between it and 1. Its limit is one, and it may be asserted that it may be as close as one wishes to one for all practical purposes (FAPP) but this is not the same thing. 1 is a number which when multiplied by itself yields itself (a property it shares, interestingly with zero) (and leaving aside the small matter of what “multiplication might be or numbers are anyway). Having said this the problem simply vanishes in any base divisible by three, showing it is conceptual rather than real (no pun intended). Any rational number less than one shrinks under multiple multiplication. One doesn’t. 1 is one of the good numbers (others are 0 pi e etc …) as opposed to the made up for the convenience of ten fingered monkeys variety.
Moral: maths is not always a perfect representation of reality.
As to the aces. Martin and Josh are both right. Analyses to the contrary assume the independence of outcomes that are not independent of the information: that there is at least one ace in the hand (irrelevant if that of spades or not). Alice is cleverer than Bob: she KNOWS what kind of ace she has but just aint telling!
This is why Josh likes playing poker with mathematicians … hihi.
I liked rccolabean’s analysis and I think it exposes the essence of the mathematical mis-direction inherent in some of the earlier posts and web references. I’ll just expand on it a little and explain why the suit makes no difference to the sorted piles. The key is understanding which group should, and should not be compared to the “win” outcome. Its no good just computing the number of possibilities since these are not independent of the knowledge one has. Lets compare which one card piles should be removed from consideration in the case where one knows the ace is a spade or not. Indeed 3/4 of the piles should be rejected in the case that its a spade, however in the case that the suit is unknown 3/4 of the piles should also be rejected virtually, one just does not know which 3/4. The ace must have some suit (right Martin!), and the other three aces a different suit. Three piles have vanished as possibilities virtually but we just do not know which. This is similar to the measurement problem often cited in quantum mechanics as a paradox (poor cat), though equally, here it is not.
Another way to see it is to imagine we have two marked decks. This allows Alice and Bob to at least be honest about what they do and do not know.
One deck has marked aces, the other the aces and their suit. Alice and Bob are dealt two cards face down under a screen with one card sized slot in it, through which they and only they can see one of the two cards. They have been forced, under the rules of the arcane brotherhood of mathematics, to reveal what they know. Alice has been allocated, at random, the pack with less information. If you like, which of the two cards, the left one or the right one each can see is also random (though this does not matter either).
“Its an ace” says Alice. “Its the ace of spades” says Bob.
Clearly, the probability of the other, unseen, card being an ace is 3/51 in both cases. The only thing one can say for certain is that, in Bob’s case, the other one will not be the ace of spades!
There is a moral to this story too: do not take things too bloody seriously! I know I might be wrong!
Dear Mr (or Ms?) Brandy. Please do not let any more of these problems ruin anything at all!
April 9th, 2010 at 1:06 am
I don’t know if it has been said before, but The Airplane Problem has been BUSTED on Mythbusters. As said above, the plane does not depend on the movement on the ground, but on the propulsion of the engine, and will take off on a conveyor belt.
http://mythbustersresults.com/episode97
watch the video for ‘proof.’
April 9th, 2010 at 8:15 am
first of all .99999 recurring is a surd and therefore not a “distinct real number” after all the definition of a surd is effectivly a number which has an infinite amount of decimal places (granted most surds such as Pi are suppossed to have no pattern to there infinite digits but if you do the maths this is wrong anyway in an infinite stream there has to be recurrence of a pattern by definition).
Secondly take it from a pilot the plane will be developing thrust from it’s engines, but whilst still on the ground it has weight and will in fact act just like a car in that it will not move relative to the air around it if the treadmill speeds up as you add thrust. Think about what happens when a plane does a normal take off run. even though it is not the wheels of the aircraft which push (or pull for most prop planes) it forward the engines still must make it “gain ground” the fact is there must be friction on the wheels whilst the plane is on the ground that is basic and obvious how else does a pilot manage to stop it with brakes? For another way of looking at it consider what happens whilst a plane is taxying on a runway without friction and lots of it due to weight not only could it not turn but by the laws of physics it wouldn’t be able to move. Every action has an equal and opposite reaction. The only feasible scenario in which this aircraft could take off in this situation is if there was a strong enough headwind to provide the necessary lift (possible 50-60kts of “normal wind” for a cessna 152 for example).
April 9th, 2010 at 8:17 am
oh and as for mythbusters yes they do nice work but I think if you watch and LISTEN closely you will see the slightly unusual scenario which they used to make this happen.
April 9th, 2010 at 6:08 pm
In response to the brilliant idiots concerning the airplane problem.
Lets use a turboprop engine for my example.
Plane on CB.
Pilot turns on prop
CB turned on to match velocity of plane
Prop generates force for plane by way of creating an artificial vacuum in front of the plane.
CB moves.
The Vacuum created by Prop causes Plane to move forward, REGARDLESS of how fast the ground is moving.
CB moves faster, but at this point I think you realize that it has nothing to do with the plane.
Jet Propulsion Engines will do the same thing, but will create an area of extremely high pressure behind the plane (in effect creating a vacuum in front) pushing the plane forward.
Once the plane is moving fast enough in relation to the surrounding air, lift will be generated and the plane will take off.
Any argument that the mechanical failure of the wheels will cause the plane to crash and burn are wrong, because the wheels aren’t a part of the problem.
April 9th, 2010 at 8:43 pm
I am glad to see there is another intelligent person on this blog (Airplane Dissenter). I guess people fail to think about the logic that it is impossible to run a CB at a speed that could feasibly overcome the power of an aircraft engine. IF you people need further proof that the plane will take off, watch Mythbusters on the Discover Channel, or go to http://dsc.discovery.com/tv/mythbusters/ and look for the appropriate video of a single engine plane taking off on a giant conveyor belt.
April 10th, 2010 at 11:44 pm
Yeah, I’m gonna have to disagree on the airplane one:
The ground speed of the airplane does not matter, but what does matter is the volume of air passing both over and under the wings. If the plane has 0 ground speed, and turbines pushing at there normal take off power, the volume of air passing the wings will be less than the turbines pushing at normal power PLUS the plane moving at normal speed.
Thus, less air passes the wings, less circulation is generated = less lift is generated and the plane will fail to take off.
That being said, if you push the turbines to more than 100% of there regular take off power it would be possible but I’m pretty sure that isn’t what this paradox is getting at – because anybody would agree with you that if you ripped the turbines off a plane, and put it in a very strong wind tunnel it would lift off the ground and if it was tied at the front wouldn’t fly backwards.
kapeesh?
April 12th, 2010 at 3:43 pm
I would like say this about the plane problem…..
The speed of the stationary conveyer bel would never match the speed of the plane wheels in a “REAL TIME FASHION”.
Because the body of the plane has to move forward in response to the massive force generated by the turbines, even if the belt was moving at the speed of infinite, the the wheel speed of the plane would be infinite+a2 at any given smallest possible fraction of time (where a2 stands for squared acceleration)……
If one could ideally create a situation (as they say in the question WHICH i think is not possible) where both the speed matches in a real time fashion, it would be analogous to a situation where the wheels of a plane are locked(because the speed of the plane is 0 with respect to ground) but the turbine is trying to push it forward. And ideally the friction between the wheels and ground would not let the plane move UNTIL the force generated by the plane turbines F overcomes the gravity G. In this condition the plane would virtually fall of its own wheels and would crash.
SO, IT WOULD NOT BE ABLE TO TAKE OFF IF WE PUT IT INTO AN IDEOLOGIC SITUATION REQUIRED BY THE PROBLEM.
April 15th, 2010 at 9:31 pm
I think this is horse crap. My head hurts. Bad.
April 16th, 2010 at 4:22 am
The two envelope (or wallet swap) problem always seemed like a variant of the Montey Hall in which none of the doors were removed.
In other words, unless the contents of the envelopes are changed, the expected value is the same whether you take the swap or stick with what you have (3/2 the contents of the less valuable envelope).
It’s not like some magical gremlin flips a coin after you’ve selected your envelope in order to decide if he should put half or twice it’s value into the other envelope.
The contents of both envelopes is fixed, and have been since before the first selection.
April 17th, 2010 at 12:13 pm
Here’s how I would explain why 0.999 … = 1
If you were to subtract x (a non-zero positive value) from 1, then the resulting number would always be less than 0.999 repeating, no matter which value of x
x > 0
(1 – x) 0.9999~
since adding or subtracting ANY known number from the first part of the equation would make the resulting value more or less than the other side of the equation, its clear that the values are exactly the same, and there is no gap between them.
April 17th, 2010 at 12:16 pm
Here’s how I would explain why 0.999 … = 1
If you were to subtract x (a non-zero positive value) from 1, then the resulting number would always be less than 0.999 repeating, no matter which value of x
x > 0
(1 – x) < 0.9999~
(1 + x) > 0.9999~
since adding or subtracting ANY known number from the first part of the equation would make the resulting value more or less than the other side of the equation, its clear that the values are exactly the same, and there is no gap between them.
(EDIT: my comment was messed up as greater/less signs were processed as html tags)
April 17th, 2010 at 12:17 pm
I give up trying to use greater than or less than tags in my comment. Just replace > and < with the tags in my above comment
April 19th, 2010 at 6:17 pm
Name is in honor of 20 April 2010 (not as good as these; just a pop culture/riddle combo of my own)
.333~=1/3
.999~ is not equal to 1, it is always .00…01 away
therefore .333~*3 =1/3*3 =1
.333~ is not a real number, only theoretical, while 1/3 is real
I like the Monty Hall, very nice.
The airplain has three major forces working on it. The friction, gravity, lift, and air. Friction vs. air; gravity vs. lift. Unless the friction is greater than the the air (why use wheels then?) the plane will move forward. So basically the conveyer belt will just move faster and faster as the plane speeds up reletive to the air until the wing attains enough lift to overcome gravity and fly.
There is always 3 aces left whatever ace you have. You have to build your odds off of that.
With the envelope you can expect the value to be higher all you want, but even money is all you will have get odds-wise. You can wish in one hand and sh!t in the other and see which one gets full first.
I hope someone actually reads this some day. This was fun.
QED MFers
April 24th, 2010 at 9:46 pm
The ground speed of the plane has to be zero as the conveyor belt can hypothetically match any speed required to keep the plane still. Unless the wheels are perfectly frictionless, this would be enough to ensure it does not move across the ground as anything multiplied by an potentially infinite value will certainly equal the forward thrust of the plane
The wheels are not perfect. This, and to a lesser extent the air resistance, is why I cannot push a jumbo around with my finger. A quick search shows a Boeing 747 for example weighs 362,870kg and that pressing down on the wheels which are a friction *reduction* device, still generates a considerable amount of friction. The planes ground speed is therefore determined by the wheels ability to turn and all progress across the ground is halted by friction. So unless it can generate lift, it will not fly.
The air speed of the plane is what will determine lift and ability to take off. The only possible situation that might cause this to occur, is if the thrust of the engines was enough to generate a sufficient air flow over the entire wing to generate this lift, i.e. the engines move the air over the wing. Another search shows the take off speed of a 747 as being 180 mph, so I doubt that the engines would be able to achieve this. True, they would easily generate enough thrust localised to the engines, but they would need to average this speed over the entire wing. This combined with the friction force of the conveyor belt on the air which would act to drag the air along might be enough, though the air would be moving fast closer to the belt and slower above it which would create turbulence which would potentially reduce the lift generated by the wing as the air flow might no longer be even.
I believe that it would probably not work in theory and definitely not in practice.
The myth busters thing did not accurately reflect the problem outlined (otherwise the car would have been accelerating to infinity as the plane pulled away).
May 7th, 2010 at 2:11 pm
I don’t buy the .999~= 1, is exactly equal to 1, that algebra example is trickery not reality, I could just as easily claim if .999~ = X , and 10X = 9.999~ then 10X – 10 = X, and it’s still X = .999~ and that’s not 1 .999~ is an imaginary number that can not be written, the ~ simply represents imagination [which is not a number] where as 1 is a real number, exists in a specific and defined place.
But the fact remains that you can’t do the actual math until you get a final number, you have to actually reach the end of infinity, and who is to prove it does not have an end point, we only assume it is unending, it is imaginary, and if you attempt to write it out in full, at any point when you stop adding 9’s there is still an infinite number of them between there and 1. That sits in the same realm as the old challenge of if you move half the distance from where you are now [point A] to the destination [point B] a fixed location brick wall, and if you keep repeating that move you will never get there. When in reality, in our physical world, your nose will be squashed up against that wall in a few moves no matter how far away you start from. But no matter how hard you press you will never actually touch it, there will always be space between your atoms and those of the wall.. .
June 15th, 2010 at 1:32 am
Maybe it’s me, but why open a comment-ready page with controversial claims/ideas that, as even stated in the opening, will cause argument. Trolls have an odor, and this one has flies…
June 17th, 2010 at 1:17 am
I like how most of the people posting decided not to read the others or consider them.
Very cool, very helpful…(some of you).
June 17th, 2010 at 2:08 am
The airplane problem is a trick. It assumes that the force from the conveyor belt pushing the plane back is negligible, so no matter how fast the belt is moving it doesn’t push the plane backward, it makes the wheels spin. The wording of the problem sort of suggests that the conveyor belt applies a non-negligible force backward on the plane, thereby counteracting the force forward on the plane from its engines, which is to intentionally confuse the reader and is the only thing that makes it a problem. In the answer to the problem the wheels are perfectly lubricated and free to spin under the plane, so when the belt moves and the engines are off the wheels spin and the plane sits in place, and when the engines get turned on they propel the plane forward and the wheels spin faster. The engines don’t power the wheels, they suck in air and shoot it out the back going all Newton 3 on that a$$.
June 17th, 2010 at 9:08 am
Aces,
The obvious, non-mathematical oversight everyone seems to make is to assume the hand with the non-specified ace to be different than the one which is specified to have the ace of spades. Although you do not know which suit the card is, you know that it is A suit. Thus, you have to assume it to be 1 card and 1 card only. For example, you may imagine it to be the ace of spades.
Basically what I am saying here is, although most of the posts prior to mine have had accurate math, they are beginning on incorrect premises. A hand with an ace of spades and a hand with an ace of (suit) are equal. The ace is of one suit only, and the other 12 cards are a mystery with 51 cards remaining, 3 of which are the aces of the other 3 suits.
This is a “real world” representation of the hands, if you will. As we all know, theoretical probability can be often misleading.
June 17th, 2010 at 10:30 pm
to prove that .9999…=1, you must use the limit of an infinite series….
.999….=9(.111….)=9(.1+.01+.001+.0001…..)
=9(1/10 + 1/100 + 1/1000….)
=9{the sum from n=1 to infinity of [(1/10)^n)]}
the sum of an infinite geometric series = a/(1-r)
where a = the first term of the series (plug in n=1 to find the first term) and r is the rate (keep in mind that the series is only convergent if r<1)
a/(1-r)=(1/10)/[1-(1/10)]=(1/10)/(9/10) (a=(1/10) and r=(1/10))
this converges to 1/9
bring down the "9" that we previously divided out, and we get 9*(1/9) which =1
I can see why people are confused cause' its a slightly more complicated conundrum than some of the others, but I dont understand how people can't figure out the airplane on a treadmill….
the force causing the plane to move is caused by the propeller or jet engine or rocket or whatever the plane uses…. all of which provide force through the air (not the ground) why people are getting into specifics of infinitely fast treadmills and boeing 747s is beyond me. You are just overcomplicating the problem… if the treadmill moves at the same speed in the opposite direction, the wheels will just move twice as fast because the air doesnt move with the treadmill and the opposing force of friction due to the wheels on the axle is negligible with respect to the force provided by the prop/jet engine/rocket engine
June 18th, 2010 at 1:24 am
Okay, step back for a second. This is frustrating.
Airplane: There are two separate arguments going on. One argument proves that the plane lifts, one proves that it does not. People are huffing and puffing at each other, demanding that their argument is valid and turning red because the other people, who also claim validity, are reaching different conclusions.
The answer? YOU’RE BOTH RIGHT. You’re just arguing about two different things. My mother.
Okay, so the common elements- plane, treadmill, wheels, engines, wings. The miscommunication comes from the velocity of the plane. There are two possible ways to interpret the way the treadmill works.
In one interpretation, the plane is actually moving forward at a certain velocity, while the treadmill is spinning backwards at the same velocity, meaning the wheels are spinning twice as fast. The key factor is that the plane is still making forward motion. To a little kid at the end of the treadmill-runway (but not on the treadmill), the plane is actually growing larger, it’s closing the gap. In this scenario, THE PLANE LIFTS.
I refuse to reiterate why. It’s not even a paradox.
In the second interpretation, which more accurately reflects the original idea of the paradox, the plane is not actually moving, but the wheels are spinning and the treadmill is pushing backwards. So we have a plane, its wheels are turning, and it’s staying exactly where it is because the treadmill is matching each rotation of the wheel. To a guy on a cliff a mile away, looking through a telescope, it’s just a stationary plane. THE PLANE DOESN’T LIFT.
Assuming the plane’s odometer is connected to its wheels (which wouldn’t make sense for a plane, but whatever), when the odometer reads that the plane should be at takeoff speed, it won’t take off because no air is hitting the wings. This is the closest thing we have to a paradox.
The problem is that people can’t agree on whether or not the plane itself is moving relative to the ground. Originally, I believe it was intended that the plane be stationary, with a treadmill underneath allowing the wheels to move forward. HOWEVER, people who know planes have pointed out that a treadmill cannot prevent a plane from moving forward because the wheels don’t generate propulsion, the engines do. So the hypothetical situation that created a paradox has been proven impossible. On a treadmill, a plane would move forward anyway, no matter how fast the treadmill were going, and eventually it would achieve flight.
MYTH MOTHERFUCKING BUSTED.
June 18th, 2010 at 1:37 am
One more thing. I’m not a math wizard. I can’t hearken the mythical powers of infinity to cast amazing, geometrically accurate spells and slay the demons of the wretched base-11 dimension. I can only drop acid and dream…
My point is, I don’t completely get why .999~ should equal 1. But I resigned myself to agree with the results of this proof. In your head, take all those nines that come after that decimal, and divide this “number” by nine. You should be able to do this in your head. Unless I’m missing something crucial, we should all have .111~ in our heads. Now, take the number 1 and use the little calculators your computers have. Divide that shit by 9.
Just saying.
June 18th, 2010 at 4:23 am
Airplane. Dudes. F=ma. a=v/t. If the distance never changes, there is no acceleration. fa. no force applied, no work done, no takeoff.
June 19th, 2010 at 1:25 am
I didn’t have time to read all the posts so I don’t know if this is old information, but I believe the Mythbusters crew actually took on the airplane on a treadmill myth, so there IS an answer out there.
June 20th, 2010 at 10:44 pm
The airplane will take off. The air around the treadmill is moving because the jet engines are pushing it backwards. this means that even if the plane is stationary there is still air rushing under its wings. think about a fan. it is stationary and still moves the air. that is how propellors work.
.99999… is simply defined to equal one. there is no proof. limits of series are only limits they are not exact.
June 21st, 2010 at 11:22 am
Envelope: You are presented with two envelopes. One has twice as much as the other.
A person who is aware of the world as it is today, would assume that there is a catch (anthrax?) and walk away never picking an envelope.
A person ignorant to the world should: Pick an envelope. Walk away. Enjoy whatever amount of money (or anthrax xP)you get for the simple fact that the only work exerted to get said money was a simple decision. Then, as you spend said money, light in the fact that there are probably several morons that haven’t decided which envelope they want OR are swapping them back and forth nervously, perpetually realizing that their other decision might result in them getting twice as much money for, essentially, nothing.
June 24th, 2010 at 9:41 pm
Some of the arguments given here against the plane taking off are the same in essence to those given against Galileo et al who claimed that the earth moved.
June 27th, 2010 at 7:33 pm
1.0~ – 0.9~ =/= 0
therefore, 1 =/= 0.9~ because the difference between these two numbers would have to be 0 if the two numbers were equal in the first place.
June 27th, 2010 at 11:45 pm
All of these problems have already been worked over enough times where if people actually read through the entirety of the posts, the answers would be found. If you want to avoiding appearing foolish, look for the posts where your problem of difficulties has been addressed and then add input if your methodolgy hasn’t already been attempted. Because I’m sure your logic will absolutely outshine that of everyone who has already contributed.
July 5th, 2010 at 11:37 pm
people who don’t think 0.999… = 1
what is the difference 1 – 0.999… equal to?
reminder: 0.000…1 is not a number
August 23rd, 2010 at 4:39 am
With the plane problem:
If you imagine a plane sitting at rest on a treadmill and you switch the treadmill on, assuming negligible friction in the wheel axles the plane remains stationary relative to the ground. This is because the plane has mass and will therefore remain stationary unless a force is applied to it.
So as the treadmill / wheel system doesn’t exert any directional force on the body of the plane, if the pilot starts the planes engines (propeller or jet) these will be the only forces acting on the plane (just as in a normal take-off scenario). And so the plane takes off.
If this were not the case then planes would also not be able to take off if they were facing in the opposite direction of the Earth’s rotation!
As far as the 0.999.. = 1 problem, if you don’t believe it Wikipedia has an extensive list of proofs, and it comes down to this: for 0.999.. to not be equal to 1 it would require there to be a fundamental error underpinning all of mathematics as we know it. So if you think that they are not equal you need to supply a proof that will show a how a vast list of mathematical geniuses throughout all of history were wrong. Good luck with that.