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%.

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.

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

ReplyDeleteStart 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.

DeleteSuppose 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%.

What Moish said

DeleteThank 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.

Delete