You and a friend are standing in front of two houses. In each house lives a family with two children.
"The family on the left has a boy who loves history, but their other child prefers math," your friend tells you.
"The family on the right has a 7-year old boy, and they just had a new baby," he explains.
"Does either family have a girl?" you ask.
"I'm not sure," your friend says. "But pick the family that you think is more likely to have a girl. If they do have a girl, I'll give you $100."
Which family should you pick, or does it not matter?

You should pick the house on the left. Specifically, there is a 2/3 chance that the family on the left has a girl, whereas there's only a 1/2 chance that the house on the right has a girl.
This is a very counterintuitive riddle. It seems like there should always be a 1/2 chance that a given child is a girl. And in fact there is. The key word there is "given". Because we are not asking about a "given" child for the house on the left. We are asking about what could be either child. Whereas for the house on the right, we are asking about a "given" child...specifically, we're asking about the younger child.
There are 3 possibilities for the children in the first house:
Younger Older
Girl Boy
Boy Girl
Boy Boy
There is no "Girl, Girl" option because we know the house on the left has at least one boy. Since each of these 3 options is equally likely, and 2 of them have one girl, there is a 2/3 chance of there being a girl in the house on the left.
For the house on the right, because we already know the older child is a boy, there are only two possibilities:
Younger Older
Girl Boy
Boy Boy
And as we can see, there is a 1/2 chance for the house on the right having a girl.
Search for: Boy or Girl paradox

A bad king has a cellar of 1000 bottles of delightful and very expensive wine. A neighboring queen plots to kill the bad king and sends a servant to poison the wine.
Fortunately (or say unfortunately) the bad king's guards catch the servant after he has only poisoned one bottle.
Alas, the guards don't know which bottle but know that the poison is so strong that even if diluted 100,000 times it would still kill the king. Furthermore, it takes one month to have an effect.
The bad king decides he will get some of the prisoners in his vast dungeons to drink the wine. Being a clever bad king he knows he needs to murder no more than 10 prisoners – believing he can fob off such a low death rate – and will still be able to drink the rest of the wine (999 bottles) at his anniversary party in 5 weeks time.
Explain what is in mind of the king, how will he be able to do so?

Think in terms of binary numbers. (now don’t read the solution, give a try).
Number the bottles 1 to 1000 and write the number in binary format.
bottle 1 = 0000000001 (10 digit binary)
bottle 2 = 0000000010
bottle 500 = 0111110100
bottle 1000 = 1111101000
Now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc.
prisoner = 10 9 8 7 6 5 4 3 2 1
bottle 924 = 1 1 1 0 0 1 1 1 0 0
For instance, bottle no. 924 would be sipped by 10,9,8,5,4 and 3. That way if bottle no. 924 was the poisoned one, only those prisoners would die.
After four weeks, line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit. The number that you get is the bottle of wine that was poisoned.
1000 is less than 1024 (2^10). If there were 1024 or more bottles of wine it would take more than 10 prisoners.

There are 1 million closed school lockers in a row, labeled 1 through 1,000,000.
You first go through and flip every locker open.
Then you go through and flip every other locker (locker 2, 4, 6, etc...). When you're done, all the even-numbered lockers are closed.
You then go through and flip every third locker (3, 6, 9, etc...). "Flipping" mean you open it if it's closed, and close it if it's open. For example, as you go through this time, you close locker 3 (because it was still open after the previous run through), but you open locker 6, since you had closed it in the previous run through.
Then you go through and flip every fourth locker (4, 8, 12, etc...), then every fifth locker (5, 10, 15, etc...), then every sixth locker (6, 12, 18, etc...) and so on. At the end, you're going through and flipping every 999,998th locker (which is just locker 999,998), then every 999,999th locker (which is just locker 999,999), and finally, every 1,000,000th locker (which is just locker 1,000,000).
At the end of this, is locker 1,000,000 open or closed?

Locker 1,000,000 will be open.
If you think about it, the number of times that each locker is flipped is equal to the number of factors it has. For example, locker 12 has factors 1, 2, 3, 4, 6, and 12, and will thus be flipped 6 times (it will end be flipped when you flip every one, every 2nd, every 3rd, every 4th, every 6th, and every 12th locker). It will end up closed, since flipping an even number of times will return it to its starting position. You can see that if a locker number has an even number of factors, it will end up closed. If it has an odd number of factors, it will end up open.
As it turns out, the only types of numbers that have an odd number of factors are squares. This is because factors come in pairs, and for squares, one of those pairs is the square root, which is duplicated and thus doesn't count twice as a factor. For example, 12's factors are 1 x 12, 2 x 6, and 3 x 4 (6 total factors). On the other hand, 16's factors are 1 x 16, 2 x 8, and 4 x 4 (5 total factors).
So lockers 1, 4, 9, 16, 25, etc... will all be open. Since 1,000,000 is a square number (1000 x 1000), it will be open as well.

You have a basket of infinite size (meaning it can hold an infinite number of objects). You also have an infinite number of balls, each with a different number on it, starting at 1 and going up (1, 2, 3, etc...).
A genie suddenly appears and proposes a game that will take exactly one minute. The game is as follows: The genie will start timing 1 minute on his stopwatch. Where there is 1/2 a minute remaining in the game, he'll put balls 1, 2, and 3 into the basket. At the exact same moment, you will grab a ball out of the basket (which could be one of the balls he just put in, or any ball that is already in the basket) and throw it away.
Then when 3/4 of the minute has passed, he'll put in balls 4, 5, and 6, and again, you'll take a ball out and throw it away.
Similarly, at 7/8 of a minute, he'll put in balls 7, 8, and 9, and you'll take out and throw away one ball.
Similarly, at 15/16 of a minute, he'll put in balls 10, 11, and 12, and you'll take out and throw away one ball.
And so on....After the minute is up, the genie will have put in an infinite number of balls, and you'll have thrown away an infinite number of balls.
Assume that you pull out a ball at the exact same time the genie puts in 3 balls, and that the amount of time this takes is infinitesimally small.
You are allowed to choose each ball that you pull out as the game progresses (for example, you could choose to always pull out the ball that is divisible by 3, which would be 3, then 6, then 9, and so on...).
You play the game, and after the minute is up, you note that there are an infinite number of balls in the basket.
The next day you tell your friend about the game you played with the genie. "That's weird," your friend says. "I played the exact same game with the genie yesterday, except that at the end of my game there were 0 balls left in the basket."
How is it possible that you could end up with these two different results?

Your strategy for choosing which ball to throw away could have been one of many. One such strategy that would leave an infinite number of balls in the basket at the end of the game is to always choose the ball that is divisible by 3 (so 3, then 6, then 9, and so on...). Thus, at the end of the game, any ball of the format 3n+1 (i.e. 1, 4, 7, etc...), or of the format 3n+2 (i.e. 2, 5, 8, etc...) would still be in the basket. Since there will be an infinite number of such balls that the genie has put in, there will be an infinite number of balls in the basket.
Your friend could have had a number of strategies for leaving 0 balls in the basket. Any strategy that guarantees that every ball n will be removed after an infinite number of removals will result in 0 balls in the basket.
One such strategy is to always choose the lowest-numbered ball in the basket. So first 1, then 2, then 3, and so on. This will result in an empty basket at the game's end. To see this, assume that there is some ball in the basket at the end of the game. This ball must have some number n. But we know this ball was thrown out after the n-th round of throwing balls away, so it couldn't be in there. This contradiction shows that there couldn't be any balls left in the basket at the end of the game.
An interesting aside is that your friend could have also used the strategy of choosing a ball at random to throw away, and this would have resulted in an empty basket at the end of the game. This is because after an infinite number of balls being thrown away, the probability of any given ball being thrown away reaches 100% when they are chosen at random.

Every day, Jack arrives at the train station from work at 5 pm. His wife leaves home in her car to meet him there at exactly 5 pm, and drives him home. One day, Jack gets to the station an hour early, and starts walking home, until his wife meets him on the road. They get home 30 minutes earlier than usual. How long was he walking? Distances are unspecified. Speeds are unspecified, but constant. Give a number which represents the answer in minutes.

The best way to think about this problem is to consider it from the perspective of the wife. Her round trip was decreased by 30 minutes, which means each leg of her trip was decreased by 15 minutes. Jack must have been walking for 45 minutes.

You are standing in a pitch-dark room. A friend walks up and hands you a normal deck of 52 cards. He tells you that 13 of the 52 cards are face-up, the rest are face-down. These face-up cards are distributed randomly throughout the deck.
Your task is to split up the deck into two piles, using all the cards, such that each pile has the same number of face-up cards. The room is pitch-dark, so you can't see the deck as you do this.
How can you accomplish this seemingly impossible task?

Take the first 13 cards off the top of the deck and flip them over. This is the first pile. The second pile is just the remaining 39 cards as they started.
This works because if there are N face-up cards in within the first 13 cards, then there will be (13 - N) face up cards in the remaining 39 cards. When you flip those first 13 cards, N of which are face-up, there will now be N cards face-down, and therefore (13 - N) cards face-up, which, as stated, is the same number of face-up cards in the second pile.

You have just purchased a small company called Company X. Company X has N employees, and everyone is either an engineer or a manager. You know for sure that there are more engineers than managers at the company.
Everyone at Company X knows everyone else's position, and you are able to ask any employee about the position of any other employee. For example, you could approach employee A and ask "Is employee B an engineer or a manager?" You can only direct your question to one employee at a time, and can only ask about one other employee at a time. You're allowed to ask the same employee multiple questions if you want.
Your goal is to find at least one engineer to solve a huge problem that has just hit the company's factory. The problem is so urgent that you only have time to ask N-1 total questions.
The major problem with questioning the employees, however, is that while the engineers will always tell you the truth about other employees' roles, the managers may lie to you if they like. You can assume that the managers will do their best to confuse you.
How can you find at least one engineer by asking at most N-1 questions?

You can find at least one engineer using the following process:
Put all of the employees in a conference room. If there happen to be an even number of employees, pick one at random and send him home for the day so that we start with an odd number of employees. Note that there will still be more engineers than managers after we send this employee home.
Then call them out one at a time in any order. You will be forming them into a line as follows:
If there is nobody currently in the line, put the employee you just called out in the line.
Otherwise, if there is anybody in the line, then we do the following. Let's call the employee currently at the front of the line Employee_Front, and call the employee who we just called out of the conference room Employee_Next.
So ask Employee_Front if Employee_Next is a manager or an engineer.
If Employee_Front says "manager", then send both Employee_Front and Employee_Next home for the day.
However, if Employee_Front says "engineer", then put Employee_Next at the front of the line.
Keep doing this until you've called everyone out of the conference room. Notice that at this point, you'll have asked N-1 or less questions (you asked at most one question each time you called an employee out except for the first employee, when you didn't ask a question, so that's at most N-1 questions).
When you're done calling everyone out of the conference room, the person at the front of the line is an engineer. So you've found your engineer!
But the real question: how does this work?
We can prove this works by showing a few things.
First, let's show that if there are any engineers in the line, then they must be in front of any managers.
We'll show this with a proof by contradiction. Assume that there is a manager in front of an engineer somewhere in the line. Then it must have been the case that at some point, that engineer was Employee_Front and that manager was Employee_Next. But then Employee_Front would have said "manager" (since he is an engineer and always tells the truth), and we would have sent them both home. This contradicts their being in the line at all, and thus we know that there can never be a manager in front of an engineer in the line.
So now we know that after the process is done, if there are any engineers in the line, then they will be at the front of the line. That means that all we have to prove now is that there will be at least one engineer in the line at the end of the process, and we'll know that there will be an engineer at the front.
So let's show that there will be at least one engineer in the line. To see why, consider what happens when we ask Employee_Front about Employee_Next, and Employee_Front says "manager". We know for sure that in this case, Employee_Front and Employee_Next are not both engineers, because if this were the case, then Employee_Front would have definitely says "engineer". Put another way, at least one of Employee_Front and Employee_Next is a manager. So by sending them both home, we know we are sending home at least one manager, and thus, we are keeping the balance in the remaining employees that there are more engineers than managers.
Thus, once the process is over, there will be more engineers than managers in the line (this is also sufficient to show that there will be at least one person in the line once the process is over). And so, there must be at least one engineer in the line.
Put altogether, we proved that at the end of the process, there will be at least one engineer in the line and that any engineers in the line must be in front of any managers, and so we know that the person at the front of the line will be an engineer.

You've been placed on a course of expensive medication in which you are to take one tablet of Plusin and one tablet of Minusin daily. You must be careful that you take just one of each because taking more of either can have serious side effects. Taking Plusin without taking Minusin, or vice versa, can also be very serious, because they must be taken together in order to be effective. In summary, you must take exactly one of the Plusin pills and one of the Minusin pills at one time.
Therefore, you open up the Plusin bottle, and you tap one Plusin pill into your hand. You put that bottle aside and you open the Minusin bottle. You do the same, but by mistake, two Minusins fall into your hand with the Plusin pill.
Now, here's the problem. You weren't watching your hand as the pills fell into it, so you can't tell the Plusin pill apart from the two Minusin pills. The pills look identical. They are both the same size, same weight (10 micrograms), same color (Blue), same shape (perfect square), same everything, and they are not marked differently in any way.
What are you going to do?
You cannot tell which pill is which, and they cost $500 a piece, so you cannot afford to throw them away and start over again. How do you get your daily dose of exactly one Plusin and exactly one Minusin without wasting any of the pills?

Carefully cut each of the three pills in half, and carefully separate them into two piles, with half of each pill in each pile. You do not know which pill is which, but you are 100% sure that each of the two piles now contains two halves of Minusin and half of Plusin. Now go back into the Plusin bottle, take out a pill, cut it in half, and add one half to each stack. Now you have two stacks, each one containing two halves of Plusin and two halves of Minusin. Take one stack of pills today, and save the second stack for tomorrow.

Last week, the local Primary school was visited by the Government School Inspector who was there to check that teachers were performing well in their respective classes. He was very impressed with one particular teacher. The Inspector noticed that each time the class teacher asked a question, every child in the class put up their hands enthusiastically to answer it. More surprisingly, whilst the teacher chose a different child to answer the questions each time, the answers were always correct.
Why would this be?

The children were instructed to ALL raise their hands whenever a question was asked. It did not matter whether they knew the answer or not. If they did not know the answer, however, they would raise their LEFT hand. If they knew the answer, they would raise their RIGHT hand. The class teacher would choose a different child each time, but always the ones who had their RIGHT hand raised.