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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:

http://www.marilynvossavant.com/articles/gameshow.html

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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    $\begingroup$ See Grothendieck Prime $\endgroup$
    – Mitch
    Commented May 3, 2011 at 0:45
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    $\begingroup$ 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. $\endgroup$
    – Steven Stadnicki
    Commented May 3, 2011 at 2:13
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    $\begingroup$ 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. $\endgroup$
    – Arturo Magidin
    Commented May 3, 2011 at 3:49
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    $\begingroup$ @Moron: So, not part of the vos Savant little imbroglio; of course, we run into the same problem with that description: were the conditions clearly stipulated to Erdos? If the host doesn't know where the car is and just happens to open a door with a goat, switching doesn't help; if the host doesn't always offer a switch, then it may be beneficial to stay, or it may amount to the same thing, depending on how the host decides whether to offer a switch or not, etc. If the problem is not very carefully specified, there is enough ambiguity for alternative answers. $\endgroup$
    – Arturo Magidin
    Commented May 3, 2011 at 4:49
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    $\begingroup$ @Arturo The Erdős story is true. Vazsonyi describes it in detail on pp. 130-131 in Erdős on graphs: his legacy of unsolved problems. Vazsonyi failed to convince Erdos using a decision tree. He had to resort to a Monte Carlo simulation in Excel. A few days later Ron Graham convincely explained to Erdős why to switch. $\endgroup$
    – Bill Dubuque
    Commented May 4, 2011 at 23:04

9 Answers 9

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How about the Two envelopes problem?

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

The following problem can be solved either the easy way or the hard way.

Two trains 200 miles apart are moving toward each other; each one is going at a speed of 50 miles per hour. A fly starting on the front of one of them flies back and forth between them at a rate of 75 miles per hour. It does this until the trains collide and crush the fly to death. What is the total distance the fly has flown?

The fly actually hits each train an infinite number of times before it gets crushed, and one could solve the problem the hard way with pencil and paper by summing an infinite series of distances. The easy way is as follows: Since the trains are 200 miles apart and each train is going 50 miles an hour, it takes 2 hours for the trains to collide. Therefore the fly was flying for two hours. Since the fly was flying at a rate of 75 miles per hour, the fly must have flown 150 miles. That's all there is to it.

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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  • $\begingroup$ 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. $\endgroup$
    – John McVirgooo
    Commented Jul 4, 2012 at 20:35
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    $\begingroup$ I'd say it was that Von Neumann saw summing the infinite series as the beautiful, elegant solution! $\endgroup$
    – Joe Z.
    Commented Mar 13, 2013 at 14:53
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Along the same lines as the Monty Hall Problem is the following (lifted from Devlin's Angle on MAA and quickly amended):

I have two children, and (at least) one of them is a boy born on a Tuesday. What is the probability that I have two boys?

Read a fuller analysis here.

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    $\begingroup$ 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. $\endgroup$
    – ShreevatsaR
    Commented May 3, 2011 at 5:45
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    $\begingroup$ Actually, perhaps, accepting the argument for "13/27" without careful thought is the real mistake made by many mathematicians. :-) $\endgroup$
    – ShreevatsaR
    Commented May 3, 2011 at 6:05
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    $\begingroup$ ShreevatsaR, suppose there are two mathematicians, and at least one of them made this mistake, and also gave it careful thought. $\endgroup$
    – Dan Brumleve
    Commented May 3, 2011 at 8:11
  • $\begingroup$ @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) $\endgroup$
    – user4143
    Commented May 3, 2011 at 10:05
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    $\begingroup$ @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. :-) $\endgroup$
    – ShreevatsaR
    Commented May 3, 2011 at 11:21
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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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  • $\begingroup$ 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. $\endgroup$
    – TonyK
    Commented May 5, 2011 at 19:06
  • $\begingroup$ @TonyK, that's right. That's what makes this such a wonderful puzzle. $\endgroup$
    – user4143
    Commented May 6, 2011 at 7:42
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    $\begingroup$ @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. $\endgroup$
    – user4143
    Commented May 10, 2011 at 0:55
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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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  • $\begingroup$ "the same Cook from Cook's theorem" -- I'm afraid that doesn't help me out any. $\endgroup$
    – Pete L. Clark
    Commented May 10, 2011 at 20:24
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    $\begingroup$ @Pete See here: en.wikipedia.org/wiki/Cook%E2%80%93Levin_theorem. $\endgroup$
    – Yuval Filmus
    Commented May 10, 2011 at 20:54
  • $\begingroup$ link downloads 13 bytes. $\endgroup$
    – user645636
    Commented Nov 13, 2019 at 16:06
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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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    $\begingroup$ 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. $\endgroup$
    – Nate Eldredge
    Commented Jul 4, 2012 at 19:13
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    $\begingroup$ 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. $\endgroup$
    – Hagen von Eitzen
    Commented Feb 21, 2013 at 20:58
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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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  • $\begingroup$ Basically, it's the harmonic series. After the first step, it has gone $\frac{1}{100000}(1)$ of the way around. After the second, it has gone $\frac{1}{100000}(1+\frac{1}{2})$ of the way around. And so on. Because the harmonic series diverges, there exists a (huge) $n$ such that $\frac{1}{100000}(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n})>1$, and so it goes the whole way around eventually. $\endgroup$
    – Lieutenant Zipp
    Commented Jun 24, 2021 at 3:08
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Every few years the Daily Mail newspaper publishes a story of three people in the same family having the same birthday, featuring a different family each time the story is dusted off and recycled.

One one occasion in 2010, they used this version of the story, asking the mathematician Roger Heath-Brown what the odds were of this happening.

Professor Heath-Brown stated that the odds were 48,627,125 to one, ($365^3$ to one). The odds are of course 133,225 to one ($365^2$ to one), not allowing for leap years.

This means that there are usually several births each year that provide the Daily Mail with opportunities for new versions of the story.

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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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  • $\begingroup$ Instead of using ROT13, why not consider using StackExchange's built-in spoiler space function? $\endgroup$
    – Willie Wong
    Commented May 3, 2011 at 11:11
  • $\begingroup$ 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 $\endgroup$
    – Raskolnikov
    Commented May 3, 2011 at 11:20
  • $\begingroup$ 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. $\endgroup$
    – user10389
    Commented May 5, 2011 at 17:51
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    $\begingroup$ 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. $\endgroup$
    – MackTuesday
    Commented Jul 22, 2014 at 3:29

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