2 of the same bday problem

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# 2 of the same bday problem

Submitted by M3n0 on Sun, 01/09/2011 - 17:55

Forums:

What is the probability that among n=30 people two have been born on the same day?

The way I solved this problem is like this:

Let's assume the 29 people were born on 29 different days(dates), which is the worst possible scenario. The last person (30th) has to be born on one of those 29 days, which can be done on 29 ways. He has the probability to be born on each day 1/365 which all equals:

29\cdot \frac{1}{365}\cdot 100 = 7.94 which seems like a reasonable solution.

Hence, if there are n=366 people the probability is:

365\cdot \frac{1}{365}\cdot 100 = 7.94 which is obvious, and for all n>366 the probability is still 100%.

Am I right about this one?

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## Re: 2 of the same bday problem

I did the calculus:

Q=364*363*...*336/(365)^29 = 0.29368 meaning that P=1-Q=0.70631 or 70.61%

Yes that approach seems much easier. Thank you.

----problem solved----

## Re: 2 of the same bday problem

The simplest approach is calculate the probability(Q)that they all have different birthdays and P=1-Q is the desired result.

Q=364.363......336/(365)^29.

To see this line the people up. The second person's prob. of diff. date from the first is 364/365. The third person's prob. of diff date from the first 2 is 363/365. The prob. for the 30th person of diff. date from the first 29 is 336/365. When you get to 366 people, the last term is 0.

## Nice solution

Nice solution

## Mangammal primes

Definition: These are the impossible prime factors

^{}of 3^n - 2 (n belongs to N). This is identical with the sequence A123239 (OEIS ).