# Tell me problems that can trick you

I am looking for problems that can easily lead the solver down the wrong path.

For example take a circle and pick $N$, where $N>1$, points along its circumference and draw all the straight lines between them. No $3$ lines intersect at the same point inside the circle. The question is how many sectors do those lines divide the inside of the circle into. First it looks like $2^{N-1}$, which is true up to $5$ points, but with $6$ there's only $31$.

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Does this count?-John can cut a piece of wood into 2 pieces in 10 minutes.How much time will he take to divide a piece of wood into 3 pieces?Many people answer 15,which is wrong. – rah4927 Apr 7 '14 at 12:48
@rah4927 It takes five minutes to find the saw. So fifteen minutes is correct ;) – Daniel Fischer Apr 7 '14 at 14:04
@rah4927 No. You only need to find the saw once, for the second cut you have it right in your hand. Ten minutes to cut the wood into two pieces minus five minutes to find the saw makes five minutes for the actual sawing. – Daniel Fischer Apr 7 '14 at 15:10
John needs another five minutes to figure out how long it will take him. – chaosflaws Apr 7 '14 at 18:16
And an additional 20 minutes to watch this friendly banter unfold. – G. Bach Apr 7 '14 at 20:18

My favourite is the Monty Hall Problem. It's an old one but a good one. A game-show host has 3 boxes and knows that behind 1 is a car and behind the other 2 are donkeys. You pick a box and he opens one of the other 2, always being sure to reveal a donkey. He then asks if you would like to keep the box you first picked or switch to the other closed box. What should you do?

Instinctively many people will say it doesn't matter. In truth you should always swap, there is a 1/3 chance that you originally chose the correct box and therefore a 2/3 chance that the other closed box contains a car. The thing which makes the probability counter-intuitive here is that the host knows where the car is and always removes a donkey so if you don't pick the donkey immediately you will always be left with a donkey in the other box at the end.

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FWIW people can also be mislead by this presentation of the problem. They might think that it was the host's choice whether or not to open a box, and/or whether or not to offer to let you switch, and hence will be influenced by whatever factors they think might affect that choice. At least, people sometimes report being confused on that point. They might be covering for their error ;-) All that's needed is to clearly state that the events described are the rules of the game, not just what has happened in one particular play of it. – Steve Jessop Apr 7 '14 at 15:21
@Steve: It's important to clearly state what the rules of the game are (and the host's 'strategy' if relevant). e.g. if the host will always open a door to reveal a car if you have not selected it, then you had better not switch when he reveals a donkey! Similarly, if his strategy is to only offers you the chance to switch when you've picked the door with the car. – Hurkyl Apr 7 '14 at 20:52
If you've ever watched the actual show, with people in chicken costumes trying to get picked as contestants, it's easy to imagine that the host's whims determine the rules of the game. Treating Monty as a choiceless automaton doesn't really capture the spirit. With that background, if the puzzle is presented in narrative form (You pick a door. Monty opens another door!) then the reader is justified in thinking there might be more to it than a probability calculation. I have my suspicions that some people present it that way on purpose so they can watch you fail to answer, and be smug about it – Wumpus Q. Wumbley Apr 7 '14 at 22:54
This is not rigorous, but I always imagined the problem as follows: I think this problem becomes immediately clear if you think of it with many more doors. Suppose the host presents 100 doors and claims that 99 contain goats, and 1 contains a car. You choose one and the host eliminates 98 other doors. It is highly unlikely that you picked the correct door on your first guess (1% chance). So switching would make sense considering the car is in one of the three doors remaining. Just increase the amount of doors to as large as you need to convince yourself that switching is a good idea. – Joseph DiNatale Apr 24 '14 at 1:51

When my friends encountered with double factorial for the first time, almost all of them mistakenly interpreted $\color{blue}{n!!}$ as $\color{blue}{(n!)!}$.

Also this limit $$\Large\lim_{n\to\infty}e^{-n}\sum_{k=0}^n\frac{n^k}{k!}=1,$$ which is incorrect because it is equal to $\dfrac{1}{2}$. We might think the term $\displaystyle\lim_{n\to\infty}\sum_{k=0}^n\frac{n^k}{k!}$ as $e^{n}$, but unfortunately it does not equal.

The last but not the least, the most classic and idiotic internet debate: $$\Large\color{blue}{6\div2(1+2)}$$ is equal to $\Large\color{blue}1$ or $\Large\color{blue}9$??

I forgot this one, the tricky question about simple arithmetic from my past IQ test. I also encountered this problem in many different variants on internet recently with fantastic hyperbole tagline: 99.99% FAILED!!

Suppose you have $100$ pounds of potatoes and these potatoes consist of $99$% water. You decide to leave the potatoes outside and let them dehydrate until they consist of $98$% water. Now the potatoes should weigh a little less than they were before. How much do they weigh now?

The answer is, of course $\Large\color{blue}{50}$ pounds.

$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

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Two of these are just bad notation. – Ryan Reich Apr 7 '14 at 12:40
The "double factorial" is mistakenly defined to abuse the notation $n!!$, which already means $(n!)!$, so your friends are quite justified. – Marc van Leeuwen Apr 7 '14 at 19:51
That double factorial thing got me too. Me and my friend saw $5!!$ and thought it was $(5!)!$, or $120!$. – Cole Johnson Apr 8 '14 at 5:07
As for that $6 \div 2(1 + 2)$, if you replace the division by multiplying by the inverse, you get $6 \cdot 2^{-1} \cdot (1 + 2)$, which you can solve to $6 \cdot 2^{-1} \cdot 3$, which you can solve to $3 \cdot 3$ which is $9$. – Cole Johnson Apr 8 '14 at 5:07
@mardavi Please look at page 2 of your link, which explicitly calls this issue out as ambiguous. It is neither clear not accepted whether implied multiplication binds stronger than explicit, and different software will evaluate it differently. There is no correct answer, which means the question itself is wrong and should be changed to eliminate the ambiguity -- unless, of course, you're using it to annoy mathematicians. – Trevel Apr 8 '14 at 18:53

Question 1) You are building a straight fence 100 feet long. There is a fencepost every 10 feet. Fence panels are 20 feet long.

How many fence panels do you need? How many fenceposts do you need?

Question 2) You are fencing a rectangular area 100ftx100 ft. There is a fencepost every 10 feet. Fence panels are 20 feet long.

How many fence panels do you need? How many fenceposts do you need?

Folks often forget that the corners are special and count wrong, especially when presented with the first question immediately before the second.

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The 'exclusive/inclusive' dilemma. – Display Name Apr 7 '14 at 22:43
This even got famous in programming as "fencepost error" – PlasmaHH Apr 8 '14 at 11:30
Also know as an off by one error en.wikipedia.org/wiki/Off-by-one_error – Halfwarr Apr 8 '14 at 13:37
I though first about the round fence in Question 1). It's not stated that the fence is straight. – Vi0 Apr 14 '14 at 22:10
@Vi0 Good call. Edited. – kmort Apr 15 '14 at 12:35

I like tricking my friends with this one:

$S_n(\mathbb R)$ denotes the set of $n,n$ real symmetric matrices

$S_n(\mathbb R) ^{++}$ denotes the set of positive definite matrices.

$S_n(\mathbb R) ^{--}$ denotes the set of negative definite matrices.

What are the path-connected components of $S_n(\mathbb R)$?

They all answer $S_n(\mathbb R) ^{++}$ and $S_n(\mathbb R) ^{--}$.

This is clearly wrong since $S_n(\mathbb R)$ is a vector space.

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What is $S_n(\mathbb{R})$? – EuYu Apr 7 '14 at 8:57

When asked to answer fast, in my experience people usually fail at this one:

$$P\implies \neg P\,\,\,\,\text{ tautology or contradiction?}$$

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Well, to be fair, asking people to decide a false dichotomy, fast, will lead them to fail most of the time, unless they now it quite well beforehand. – tomasz Apr 7 '14 at 12:37
@tomasz OK.${}$ – Git Gud Apr 7 '14 at 12:38
Well, my point was, this wording of question simply states false information (instead of having the person asked infer it themselves), which I wouldn't consider fair game. I'd rather word the question "Is X a tautology" instead, so that they go down the rabbit hole themselves, as opposed to pushing them down into it. – tomasz Apr 7 '14 at 12:43
@tomasz Fair enough, but I have no experience with that formulation. I have no idea if people fall for it or not. – Git Gud Apr 7 '14 at 12:58
So, what is the correct answer? – Ruslan May 18 '14 at 13:22

I get this math problem from 9GAG.

It is said that over $50\%$ of Harvard, Princeton, and MIT students get this wrong.

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