How to Calculate the Probability of 13 Out of 408 Guessing a Birthday Correctly?

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To calculate the probability of exactly 13 out of 408 people guessing a birthday correctly, use the binomial distribution formula: C(408, 13) * (1/365)^13 * (364/365)^395. Each individual's chance of guessing correctly is 1/365, assuming independence and ignoring leap years. The C function, or binomial coefficient, is calculated as C(n, k) = n! / (k! * (n-k)!). The final probability of exactly 13 correct guesses is approximately 1 in 5,370,675,393. This highlights the rarity of such an event occurring.
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If 408 people try to guess someone's birthday, how do you calculate the chance of 13 of them being right? http://www.greasypalm.co.uk/gpforum/forum14/363.html
 
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Well, the chance of each individually being right is 1/365 (ignoring leap years and assuming people's guesses are independent of the person's actual birthday, which is probably not true), and if they are independent then it follows the binomial distribution, so the answer is
C(408, 13) * (1/365)^13 * (364/365)^395. Unless you're interested in the probability of at least 13 being right.
 
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Sorry, statistics represents a bit of a gap in my education. What is this C function, and how do I calculate it/look it up?

Let's go with exactly 13 of them being right.
 
C(n, k) = \frac{n!}{(k!)(n-k)!}
 
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Thanks. 1 in 5,370,675,393 I make it.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks

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