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Although I'm not a professional mathematician by training, I felt I should have easily been able to answer straight away the following puzzle:

Three men go to a shop to buy a TV and the only one they can afford is £30 so they all chip in £10. Just as they are leaving, the manager comes back and tells the assisitant that the TV was only £25. The assistant thinks quickly and decides to make a quick profit, realising that he can give them all £1 back and keep £2.

So the question is this: If he gives them all £1 back which means that they all paid £9 each and he kept £2, wheres the missing £1?

3 x £9 = £27 + £2 = £29...??

Well, it took me over an hour of thinking before I finally knew what the correct answer to this puzzle was and, I'm embarrassed.

It reminds me of the embarrassement some professional mathematicians must have felt in not being able to give the correct answer to the famous Monty Hall problem answered by Marilyn Vos Savant:


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. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

Yes; you should switch.

It's also mentioned in the book: The Man Who Only loved Numbers, that Paul Erdos was not convinced the first time either when presented by his friend with the solution to the Monty Hall problem.

So what other simple puzzles are there which the general public can understand yet can fool professional mathematicians?

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Well, I am pretty easily fooled… The list would be too long! –  Mariano Suárez-Alvarez May 3 '11 at 0:27
I don't worry to much about being wrong becuase I'm focused on learning and really have no one I want to impress. –  futurebird May 3 '11 at 0:31
See Grothendieck Prime –  Mitch May 3 '11 at 0:45
While this is nitpickery, I should note that the Monty Hall problem was posed and answered long before vos Savant came along, and the confusion and logical debates it engenders have lingered well after her 'answering' it - I'm not sure she did anything for the problem except making it marginally more well-known. –  Steven Stadnicki May 3 '11 at 2:13
It should also be noted that the Monty Hall problem is notorious in part because slight variations change the correct answer, and vos Savant's original description of the problem was ambiguous enough to possibly include several variants with different possible solutions (e.g., not specifying clearly the host always opens a door he knows does not contain a car, always offers the choice to switch). And, how did Erdos get dragged into this? I was not aware he was part of that little imbroglio. –  Arturo Magidin May 3 '11 at 3:49

8 Answers 8

Along the same lines as the Monty Hall Problem is the following (lifted from Devlin's Angle on MAA and quickly amended):


Read a fuller analysis here.

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This, even more than the Monty Hall problem, is a problem whose trouble comes entirely from ambiguity. See for example this article. Just as the answer to Monty Hall depends entirely on the way the door was picked (whether the host made a biased choice to open only a door with a goat), the answer here depends on how "at least one of them is a boy born on a Tuesday" was generated. There is no sane reason to assume that the statement was generated from a distribution biased towards Tuesdays, so the intuitive answers of 1/2 or 1/3 are actually more right. –  ShreevatsaR May 3 '11 at 5:45
Actually, perhaps, accepting the argument for "13/27" without careful thought is the real mistake made by many mathematicians. :-) –  ShreevatsaR May 3 '11 at 6:05
ShreevatsaR, suppose there are two mathematicians, and at least one of them made this mistake, and also gave it careful thought. –  Dan Brumleve May 3 '11 at 8:11
@ShreevatsaR, I opted for the short version instead of getting bogged down in semantics. To me, this conveys the idea of the problem enough so that people understand what is being asked. That being said, this is the more rigorous version: You take a poll of $N$ families and discard all but families with exactly 2 children and that have at least one boy born on Tuesday. Of the restricted set, what is the proportion of families that have both children as boys as $\lim N \to \infty$? (and, of course, 1/2 prob. of a child being a boy with 1/7 prob. being born on a given day of the week) –  user4143 May 3 '11 at 10:05
@user4143: Yes, the whole point is that "discard all but families with exactly 2 children and that have at least one boy born on Tuesday" is not at all the same information as someone making the statement "I have two children, and at least one of them is a boy born on a Tuesday", and to assume the former when given the later is a mistake. (Depending on the sample process the statement came from, all answers between 0 and 1 may be correct.) So to say that someone who, given the latter, (rightly) does not assume the former makes a mistake is also a mistake. :-) –  ShreevatsaR May 3 '11 at 11:21

I'm not sure if this qualifies as I don't have any reports of this actually tricking any mathematician, but it's a good problem. So, at the risk of violating the criteria, the following is Robert Connelly's "Say Red" (taken from Gardner's "Fractal Music, Hypercards and more", Chapter 14):

The banker shuffles a standard deck of 52 cards and slowly deals them face up. The dealt cards are left in full view where they can be inspected at any time by the player. Whenever the player wants, he may say "Red." If the next card is red, he wins the game, otherwise he loses. He must call red before the deal ends, even if he waits to call on the last card. What odds should the banker give to make it a fair game, assuming that the player adopts his best strategy on the basis of feedback form the dealt cards? The player must announce the size of his bet before each game begins.

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Weird! I did the maths, and it seems that your probability of winning is 1/2, whatever strategy you choose. But I can't quite believe it. –  TonyK May 5 '11 at 19:06
@TonyK, that's right. That's what makes this such a wonderful puzzle. –  user4143 May 6 '11 at 7:42
@TonyK, The other way to see it is to consider flipping over the bottom card when the player says 'red'. The probability is exactly the same as flipping the top card in the remaining pile, but obviously only 1/2. –  user4143 May 10 '11 at 0:55

I guess this well known von Neumann anecdote fits the description of the question:




When this problem was posed to John von Neumann, he immediately replied, "150 miles." "It is very strange," said the poser, "but nearly everyone tries to sum the infinite series." "What do you mean, strange?" asked Von Neumann. "That's how I did it!"

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Does this demonstrate the calculating brilliance of Von Neuman or him lacking a creative intuition to see the beautiful, elegant solution? I also solved it as he did, but in a few minutes, only to be utterly humiliated by the true solution. Moral of the story - always check for the elegant solution first, before someone else humiliates you with it. –  John McVirgo Jul 4 '12 at 20:35
I'd say it was that Von Neumann saw summing the infinite series as the beautiful, elegant solution! –  Joe Z. Mar 13 '13 at 14:53

Perhaps not what you were aiming at, but have a look at this. Fabio Massacci provides a counterexample for a lower bound proved by Cook and Reckhow (that's the same Cook from Cook's theorem), and also referred to in several other papers (Section 5 of the paper).

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"the same Cook from Cook's theorem" -- I'm afraid that doesn't help me out any. –  Pete L. Clark May 10 '11 at 20:24
@Pete See here: en.wikipedia.org/wiki/Cook%E2%80%93Levin_theorem. –  Yuval Filmus May 10 '11 at 20:54
thanks. –  Pete L. Clark May 10 '11 at 22:49

I don't know if this qualifies as simple.
A train line from an airport to the Cantor Hotel operates in the following manner. There is a station at each ordinal number, and every station is assigned a unique ordinal. The train stops at each station, in order. At each station people disembark and board, in order, as follows:
i) if any passengers are on the train, exactly $1$ disembarks, and then
ii) $\aleph_0$ passengers board the train.
Station $0$ is at the airport, and the Cantor Hotel is at station $\omega_1$, the first uncountable cardinal. The train starts its journey empty. $\aleph_0$ passengers board the train to the Cantor Hotel at the airport (station $0$), and off it goes. When the train pulls into the Cantor Hotel at station $\omega_1$, how many passengers are on it?

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Surely it depends on which passenger disembarks at which station? I believe I can arrange it so that the train has anywhere between 0 and $\aleph_1$ passengers. –  Nate Eldredge Jul 4 '12 at 19:13
It's already "hard to tell" how many are on board at station $\omega$ if at each finite station two get on and one gets off the train. If odd-numbered people never leave, there are $\aleph_0$ people on on borad; but if people leave in th esame order they enter, nobody is left. –  Hagen von Eitzen Feb 21 '13 at 20:58

Erm... I don't even understand the solution, but there is this one: An ant starts to crawl along a taut rubber rope 1 km long at a speed of 1 cm per second (relative to the rubber it is crawling on). At the same time, the rope starts to stretch by 1 km per second (so that after 1 second it is 2 km long, after 2 seconds it is 3 km long, etc). Will the ant ever reach the end of the rope?

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OK, here's one which had me stumped and when I knew the answer, I was pretty embarrassed. Even more so when I told my mother and she immediately saw the solution. And she's no mathematician.

A music lover lives in a town which has two concerts every weekend. One is in the North of the town, the other in the South. The music lover lives in the centre of town and decides to take the bus to the concert. There's a bus to the North and one to the South every hour. The music lover just decides to take the first bus he sees and go to that concert. At the end of the year, he notices that he only went to the concert in the South. How come?

Here's the answer:

The clue is that there's a bus every hour, but I didn't specify at what time in the hour the buses leave for the North and South. Imagine that the bus for the South leaves only 5 minutes before the one for the North. Then, it is more likely that if the music lover arrives at a random time, he will take the bus to the South. The odds are 1 to 11 that he takes the one to the North. If they leave even closer to one another, the probability will be even smaller.

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Instead of using ROT13, why not consider using StackExchange's built-in spoiler space function? –  Willie Wong May 3 '11 at 11:11
Right, forgot about that. Also realised I made a mistake in the solution. And then I wonder why such simple stuff can stump me when I even fail at simple arithmetic. :P –  Raskolnikov May 3 '11 at 11:20
I would have said that the bus to the North remains at the stop in the center longer than for the bus for the south. –  user10389 May 5 '11 at 17:51
Because it seems you're allowed to add whatever constraint to the problem you like, we could also posit that the northbound bus is never visible from his favorite bus stop. Or perhaps he's had a hemispherectomy and can perceive motion in one direction but not the other. Or maybe he never learned that both routes pass the southern venue, but only one continues on to the northern venue. Or he always drops acid before he leaves for the concert, and the northbound bus looks like a dragon that eats people because it passes in front of the Lucky Dragon Chinese restaurant. –  MackTuesday Jul 22 '14 at 3:29

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