# Probability of birthdays shared in a group

1. Mar 10, 2010

### IniquiTrance

In a group of 23 people, the probability that any 2 people share a birthday is:

1 - $$\frac{365!}{342!365^{23}}$$

Why can't I just do the following?

$$(23\mathbf{C}2)(\frac{1}{365})$$

Thanks!

Last edited: Mar 10, 2010
2. Mar 10, 2010

### tiny-tim

Hi IniquiTrance!

Do you mean 1 - 23C2/365 ?

because there are "overlaps" that you aren't subtracting …

the individual events (of one particular pair not sharing the same birthday) are not independent … eg the pairs Tom and Dick, Tom and Harry, and Dick and Harry, are not independent.

3. Mar 10, 2010

### IniquiTrance

Hmm, I kind of see what you're saying...

But why is say P(Tom and Harry|Dick and Harry) different than P(Tom and Harry)?

4. Mar 10, 2010

### tiny-tim

Yes, the probabilities are the same, but the events are different.

5. Mar 10, 2010

### IniquiTrance

Thanks for your response. I'm on the verge of uynderstanding it, can you think of any other way to explain it?

I thought the definition of independence is that $$E_{i}$$ and $$E_{j}$$ are independent events so long as $$P(E_{i}|E_{j}) = E_{i}$$

and $$P(E_{j}|E_{i})= E_{j}$$

A bit confused...

6. Mar 10, 2010

### tiny-tim

I prefer to write it P(Ei and Ej) = P(Ei)P(Ej).

(because, that way, you can string more than two together)

But it's much easier just to use common-sense, and to say that the three events of Tom and Dick, Tom and Harry, and Dick and Harry, sharing (or not sharing) a birthday are obviously not independent.

7. Mar 11, 2010