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 $31$.
 A: 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.
A: 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. 
A: When asked to answer fast, in my experience people usually fail at this one:

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

A: 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.

A: 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 $\Large\color{blue}9$??

ADDENDUM :
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.
$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
A: I get this math problem from 9GAG.

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