Question

what are the odds against two or more people having the same birthday in a group of 6 people? I'll give a medal

Answer 1

First let's look at the probability that none of the 6 people in the group share the same birthday. For two people the probability that the second person has a different birthday from the first is 364/365. Then the probability those two have different birthdays and a third person has a different birthday from either of them is 364/365 * 363/365. Similarly the probability that those three have different birthdays and a fourth has a different birthday from any of the first three is 364/365 * 363/365 * 362/365. Continuing in the same way, the probability that none of the six people in the group share the same birthday is given by:
\[\large P(none\ share\ birthday)=\frac{364}{365}\times\frac{363}{365}\times\frac{362}{365}\times\frac{361}{365}\times\frac{360}{365}=0.9595\]
Therefore the probability that two or more share the same birthday is given by:
P(2 or more share birthday) = 1 - 0.9595 = 0.0405
The odds against 2 or more people sharing a birthday are 0.0405 : 0.9595 or
1:23.69