Another Probability Paradox: What's your birthday?

As I said about a month and a half ago, probability is a topic that people have difficulty understanding. I showed you the Monty Hall Paradox, where the average person is almost 17% off on the odds! 17% makes a big difference in what you should do in that scenario! Let's look at another thing that completely fools people, the Birthday Paradox.

Say you walk into a room of 23 people. What do you think the odds are that two of the people have the same birthday? Maybe like 23 in 365 because there are 365 days in a year, being about 6.3%. Would you be surprised if the true answer was over 50%?

The mistake people make is they try to determine the odds that someone has the same birthday as them. This is not correct, as there is no specification in the problem as to which two people it is. If you were to calculate those odds, it would be around 0.14%, which is nowhere close to the true odds.

In the Monty Hall Problem, I showed you how we know this. For this problem, we don't need much proof, because we can figure out the odds. Not of this question, but of the opposite question.

First off, what if we want to make a list of how many possibilities there are for 23 people. There are 365 for the first person, 365 for the second, and so on. Basically, there are a total of about 85 octodecillion (8.5 x 10^58), or 365^23. Basically, a list for n people is 365^n.

Now let's cover the rest of it. How many of these different possibilities have all different birthdays? Well, the first person's birthday could be any of 365 days. Since the second must be different, it only can choose between 364 days. The third only has 363, all the way up to the 23rd, who has 343 different possibilities. For n people, this is equal to:

365 x 364 x 363 x ... x (367 - n) x (366 - n)

You can very easily shorten this expression. On your calculator, there may be a button that is an x with an exclamation point after it. All this does is takes the number you type in and multiply it by every single whole number before it. For instance, 6! is 720 because 6 x 5 x 4 x 3 x 2 x 1 = 720. So, this expression is equal to:

365!/(365-n)!

And this is out of a list of 365^n possibilities. Therefore, the total possibilities there are for no one having the same birthday is:

365!/(365^n)(365-n)!

If you subtract this decimal from one, you will know the probability that two people have the same birthday. For 23 people, we 365!/(365^n)(365-n)!]

This is about a 50.7% chance. For 30 people, we are already at 70%. 50 people is already at a 97% chance. 100 people is like a 99.99996% chance that two people will have the same birthday.