## Probability of a given birthday?

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

### Probability of a given birthday?

Hi all!

So, my friend came upon this problem in her Maths textbook the other day.
From a group of 5 people, what is the chance that at least one of them was born on the 15th of April?

At first, it seemed to be just
[(1/365)+(1/365)+(1/365)+(1/365)+(1/365)] = 5/365 = 1/73

But what if the group has 365 people? This reasoning says that there is a 100% chance of a certain birthday.
And for a group of 400 people, the probability rises to 110%. This logic is obviously flawed.

So what is the answer? How do you get there? I can't stop thinking about it

Posts: 10
Joined: Wed May 11, 2011 12:01 am UTC
Location: Hm?

### Re: Probability of a given birthday?

That method is actually pretty close to correct for small numbers of people, with the benefit of being far quicker to calculate. But of course as you noticed, it isn't exactly correct, and gets increasingly inaccurate as the number of people increases.

To get the probability that at least one of them was born on a given date (and with the additional assumption that none of them was born in a leap year), it's easiest to first calculate the probability that none of them was born on that date. This is just (364/365)^5, since there's a 364/365 chance for each one to have been born on some other day, and we're assuming all five birthdays are independent.

Then, the probability that at least one of them *was* born on that day, we just subtract this result from 1, getting approximately 1.36%. Your estimate was 1.37% which is, as I said, pretty close.
---
If you want to do more complicated things than just getting the probability of all or at least one of them having something true of their birthdays, you need to use some of the other traits of the binomial distribution. For example, if you want to know the probability of *exactly* one person having an April 15 birthday, you need to compute the probability that person A has that birthday and all the others don't, plus the probability that person B has that birthday and all the others don't, and so on.

(This time we can add the probabilities directly, because they are mutually exclusive, meaning no two of them can both happen together. That is, you can't have A born on April 15 and B, C, D, and E, born on other days, AND B born on April 15 and A, C, D, and E born on other days. In your initial problem, on the other hand, two or more of the situations could happen together, because you weren't considering B, C, D, or E's birthday when noting the 1/365 chance that A was born on April 15. So you ended up overcounting all the times when two or more of the people shared an April 15 birthday.)
In the future, there will be a global network of billions of adding machines.... One of the primary uses of this network will be to transport moving pictures of lesbian sex by pretending they are made out of numbers.
Spoiler:
gmss1 gmss2

gmalivuk
Archduke Vendredi of Skellington the Third, Esquire

Posts: 19295
Joined: Wed Feb 28, 2007 6:02 pm UTC
Location: Here, There, Everywhere (near Boston, anyway)