What is the minimum number of students for a likely win in a birthday bet?

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SUMMARY

The minimum number of students required for a lecturer to have a greater than 50% chance of at least one pair sharing a birthday is 23. This conclusion is derived from calculating the probability of no shared birthdays using the formula Pm = (n^m).(e^-n)/m!, where n represents the mean number of pairs. The probability decreases as more students are added, specifically calculated as 364/365 for 2 students and 364/365 * 363/365 for 3 students, continuing this pattern until the probability of shared birthdays exceeds 50%.

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Hi there, I'm a bit stuck and was hoping somebody could give me a couple of pointers...

A lecturer wages that at least one pair of students in his class have birthdays on the same day. What is the minimum number of students in his class for him to be likely to win the bet?

I have assumed a Poisson formula for Pm = (n^m).(e^-n)/m!
for the probability of m pairs having their birthdays on the same day, when n is the mean number of such pairs.

I know the answer is 23 but I am really struggling to obtain this.
Any pointers would be much appreciated. Cheers
 
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(To greatly simplify the problem, I'll assume February 29th is a myth.)

With 1 person in the room, there is a 100% chance that there would be no shared birthdays.

With 2 people, the chance of having no shared birthdays is 364/365. With 3 people, the chance is 364/365 * 363/365.

Continue this and you should get the answer. Remember, you're looking for the chance of having no shared birthdays to fall under 50%.
 
OK, that's nice and clear now. Thanks for your reply CRGreathouse!
 

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