Lecture 3. Birthdays.

During class on Friday, for a little “icebreaker” (not that there is much ice to break in Colombia), we discovered that the 26 people in the room all have different birthdays.

1. What’s the probability of this? (Assume that all birthdays are equally likely and Feb. 29 doesn’t exist.)

2. What’s the probability that all 45 participants in the class have different birthdays?

3. Do all 45 students have different birthdays? 


One comment

  1. Well if I’m not wrong this probability should be \frac{\frac{365!}{(365-n)!}}{365^n} (From the 365 days select n days to place the students and multiply by the total number of combinations, which is n!, then divide by the total number of possibilities). If I’m correct, then for n=26 we get 0.401759179864061 which tells us that paying ten times the bet if loosing wasn’t a good idea. But… if n=45 we got 0.05902410053422507
    with a 94% chance of winning the bet, good way of winning money if you are a teacher in a public school with a mean of 50 students per classroom! Is this correct?

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