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The Birthday Problem

The birthday problem is an interesting and counterintuitive statistical problem that asks, "What is the probability that among a group of n randomly chosen people, at least two people will have the same birthday?"

At first glance, it may seem that the probability is quite low. After all, there are 365 days in a year, so the chance of two people having the same birthday should be 1/365, right? However, as the number of people in the group increases, so does the probability of two people sharing a birthday.

So, what is the probability of two people sharing a birthday in a group of n people? To solve this problem, we need to use a bit of combinatorics. Let's start by considering the case of just two people. The first person has a birthday on any day of the year with probability 1, but the second person must have the same birthday as the first person, which has a probability of 1/365. Thus, the probability of two people sharing a birthday in a group of two is 1/365.

Now, let's consider the case of three people. The first person still has a birthday on any day of the year with probability 1. The second person must have the same birthday as the first person, which has a probability of 1/365. The third person must have a birthday that is the same as either the first or second person, which has a probability of 2/365 (since there are two people to choose from). Thus, the probability of three people sharing a birthday in a group of three is:

1365×1365×2365=23653\frac{1}{365} \times \frac{1}{365} \times \frac{2}{365} = \frac{2}{365^3}

As we can see, the probability is quite low. But what happens as the number of people in the group increases? We can derive a general formula for the probability of two people sharing a birthday in a group of n people using the principle of inclusion-exclusion. The probability that no two people share a birthday is:

365365×364365×363365×⋯×365−n+1365=365!365n(365−n)!\frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \cdots \times \frac{365-n+1}{365} = \frac{365!}{365^n(365-n)!}

Therefore, the probability that at least two people share a birthday in a group of n people is:

1−365!365n(365−n)!1 - \frac{365!}{365^n(365-n)!}

Let's see how this formula works for different values of n:

  • For n=2, the probability is approximately 0.0027
  • For n=10, the probability increases to approximately 0.116
  • For n=23, the probability reaches 0.507, which means that in a group of 23 people, there is a greater than 50% chance that at least two people will share a birthday!
  • For n=50, the probability is almost 1, which means that in a group of 50 people, it is almost certain that at least two people will share a birthday.

So, why does the probability increase so rapidly with the number of people in the group? It turns out that the key factor is not the number of days in a year, but rather the number of pairs of people in the group. For example, in a group of 23 people, there are 253 pairs of people (since 23 choose 2 equals 253), and each pair has a 1/365 chance of sharing a birthday. Therefore, the probability that at least one of those pairs shares a birthday is much higher than the probability of any individual pair sharing a birthday.

In conclusion, the birthday problem is a fascinating example of how probability can be counterintuitive and surprising. While it may seem unlikely that two people would share a birthday in a group of only a few dozen people, the mathematics tells us otherwise. So next time you're at a party with a lot of people, try asking around to see if anyone has the same birthday as you - you might be surprised at the results!

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