## Sunday, January 20, 2013

### 23

How many people do you need to choose at random if you want the probability that there will be two of them that have the same birthday to be over 50%?

23

1. Can you describe the math behind getting this number? I'm curious.

1. Start from the beginning. First, we make the simplifying assumptions that there are 365 days in a year, that births are uniformly distributed through the year, and that birthdays are all independant.

Suppose you have a room with 1 person in it. Clearly, the probability that there are two or more people who share a birthday is zero.

Now, suppose you have two people in the room. Line them up. The order doesn't matter. The first one has a birthday. The probability of the second one having a different birthday is 364/365. The 364 in the numerator is the number of birthdays not already taken by the first guy. That is 99.7260%. That means that the probability that they have the same birthday is 0.2740%. Note that this is 1 - 364/365.

Now, suppose you have three people in the room. Line them up. As shown above, the probability that the first two don't share a birthday is 99.7260%. Assuming that's the case, the probability that the third has a birthday that is different from the first two is 363/365, or 99.4521%. So the probability of all three having different birthdays is 0.997260 × 0.994521 = 0.991796, or 99.1796%. So the probability of there being a shared birthday is 0.8204%. Note that this is 1 - (364/365) × (363/365).

Continuing the pattern, if you have four people in the room, the probability of there being a shared birthday is 1 - (364/365) × (363/365) × (362/365) = 1.6356%.

Skipping ahead, the probability of there being a shared birthday with twenty-two people in the room is 1 - (364/365) × (363/365) × (362/365) × (361/365) × ... × (344/365) = 47.5695%, and the probability of there being a shared birthday with twenty three people is 1 - (364/365) × (363/365) × (362/365) × (361/365) × ... × (343/365) = 50.7297%.

2. What Moish said

3. Thank you. I understand it. I'm really enjoying picking up math information. I'm trying not to be angry that I was moved from Virginia, where they had a wonderful high school math program, to Shelley, ID, where, in my sophomore year, when I took Algebra II, I was the only non-senior in the class. Oh well, never too late to learn, I guess. Thanks for bearing with me. I appreicate the explanations.