Questions around this would be like if someone bets with you that in a room of n people, for each person with different birthday, he gets $x and you get $y otherwise. Would you bet?

First, it says

**any**two people and

**not with just you**(and so the paradox). There are 23 people and so 23C2 i.e. 253 pairs and not just 22 pairs.

e = event that at least two people have same birthday.

e' = event that all people have different birthdays.

And p(e) = 1 - p(e')

For p(e'):

For 1 person to have different birthday,

p(e') = 365/365 = 1

For 2 persons to have different birthdays,

p(e') = (

**364/365**) * 365/365 = 365! / (365 -

**2**)! * 1/365^

**2**

For n persons to have different birthdays,

p(e') = 365! / (365 -

**n**)! * 1/365^

**n**

And p(e) = 1 - 365! / (365 -

**n**)! * 1/365^

**n**

With n as 23, p(e) = 50.7%

p(e) is 100%, when n = 366.

Update: The probability of

**you**having the same birthday as someone else is (n-1)/365 and you would need 134 people for ~50% and 366 for 100%.

## No comments:

## Post a Comment