cleanfunnyprobabilityThere was once a college that offered a class on probability applied to the real world.
The class was relatively easy, but there was a catch.
There were no homework assignments or tests, but there was a final exam that would have only one question on it.
When everyone received the test it was a blank sheet of paper with a solitary question on it: "What is risk?"
Most students were able to pass, but only one student received 100% for the class!
Even stranger was that he only wrote down one word!
What did he write?

He, brilliant student, wrote down: "This".

logicprobabilityHussey has been caught stealing goats, and is brought into court for justice. The judge is his ex-wife Amy Hussey, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Hussey from close range.
To make things a little better for Hussey, Amy Hussey tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot.
If Hussey is still alive, she will then either take another shot, or spin the chamber again before shooting. Hussey is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower.
He steels himself as Amy Hussey loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Amy Hussey asks, 'Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?'
What should Hussey choose?

Hussey should have Amy Hussey pull the trigger again without spinning.
We know that the first chamber Amy Hussey fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Hussey has Amy Hussey pull the trigger again, the probability that a bullet will be fired is 1/4.
If Amy Hussey spins the chamber again, the probability that she shoots Hussey would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.

logicmathprobabilityYou are on a gameshow and the host shows you three doors. Behind one door is a suitcase with $1 million in it, and behind the other two doors are sacks of coal. The host tells you to choose a door, and that the prize behind that door will be yours to keep.
You point to one of the three doors. The host says, "Before we open the door you pointed to, I am going to open one of the other doors." He points to one of the other doors, and it swings open, revealing a sack of coal behind it.
"Now I will give you a choice," the host tells you. "You can either stick with the door you originally chose, or you can choose to switch to the other unopened door."
Should you switch doors, stick with your original choice, or does it not matter?

You should switch doors.
There are 3 possibilities for the first door you picked:
You picked the first wrong door - so if you switch, you win
You picked the other wrong door - again, if you switch, you win
You picked the correct door - if you switch, you lose
Each of these cases are equally likely. So if you switch, there is a 2/3 chance that you will win (because there is a 2/3 chance that you are in one of the first two cases listed above), and a 1/3 chance you'll lose. So switching is a good idea.
Another way to look at this is to imagine that you're on a similar game show, except with 100 doors. 99 of those doors have coal behind them, 1 has the money. The host tells you to pick a door, and you point to one, knowing almost certainly that you did not pick the correct one (there's only a 1 in 100 chance). Then the host opens 98 other doors, leave only the door you picked and one other door closed. We know that the host was forced to leave the door with money behind it closed, so it is almost definitely the door we did not pick initially, and we would be wise to switch.

logicmathprobabilityWhat is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?

Only 23 people need to be in the room.
Our first observation in solving this problem is the following:
(the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0
What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations).
With some simple re-arranging of the formula, we get:
the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday)
So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer.
The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far.
These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918
More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365)
We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%.
Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.

logicmathprobabilityYou 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.

logicprobabilityYour enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.
Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?

You are better off shooting again without spinning the barrel.
Given that the gun didn't fire the first time, it was pointing to one of the four empty slots. Because your enemy spun the cylinder randomly, it would have been pointing to any of these empty slots with equal probability. Three of these slots would not fire again after an additional trigger-pull, and one of them would. Thus, by not spinning the barrel, there is a 1/4 chance that pulling the trigger again would fire the gun.
Alternatively, if you spin the barrel, it will point to each of the 6 slots with equal probability. Because 2 of these 6 slots have bullets in them, there would be a 2/6 = 1/3 chance that the gun would fire after spinning the barrel.
Thus, you are better off not spinning the barrel.

logicprobabilityYour friend shows you two jars, one with 100 red marbles in it, the other with 100 blue marbles in it.
He proposes a game. He'll put the two jars behind his back and tell you to pick one of them at random. You'll then close your eyes, he'll hand you the jar you picked, and you'll pick a random marble from that jar.
You win if the marble you pick is blue, and you lose otherwise.
To give you the best shot at winning, your friend gives you the two jars before the game starts and says you can move the marbles around however you'd like, as long as all 200 marbles are in the 2 jars (that is, you can't throw any marbles away).
How should you move the marbles around to give yourself the best chance of picking a blue marble?

Put one blue marble in one jar, and put the rest of the marbles in the other jar. This will give you just about a 75% chance of picking a blue marble.

logicprobabilityYou 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.