What 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.
You 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.
Search: Monty Hall problem
A man who lives in Middletown has two girlfriends, one in Northtown and one in Southtown. Trains from the Middletown train station leave for Northtown once every hour. Separate trains from the station also leave for Southtown once every hour. No trains go to both Northtown and Southtown.
Each day he gets to the Middletown train station at a completely random time and gets onto the first train that is going to either Northtown or Southtown, whichever comes first.
After a few months, he realizes that he spends 80% of his days with his girlfriend from Northtown, and only 20% of his days with his girlfriend from Southtown.
How could this be?
The train to Northtown leaves every hour, on the hour (9:00AM, 10:00AM, etc...).
The train to Southtown leaves at 12 after the hour (9:12AM, 10:12AM, etc...).
So there is only a 12/60 (1/5) chance that he will end up on the train to Southtown each day, since he will usually get to the station during the 48 minutes of each hour when the train to Northtown will be the next to come.
Your 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.