suppose a group of n people is randomly selected. For each value v of n, find the probability that everyone in the group has a differnet birthda;y. (assume no one is born on february 29 of a leap year, so there are 365 equally likely burthdays possible.)

Answer 1

Denote the probability by P(n). Clearly:

P(n) = 0 for all n > 365.

For n smaller than 365, we can argue as follows. Since everyone's birthday is assumed to be random, selecting n people and comparing the birthdays is equivalent to randomly selecting a string of n integers between 1 and 365.

The total number of ways one can select a string of n integers between 1 and 365 is:

365^n

These will include strings containing integers that are all different and strings in which some integers occure more than once. All these strings are equally likely, so they all have a probability of 1/365^n.

The total number of strings containing numbers that are all different is:

365!/(365 - n)!

you can also write this as:

365*(365 - 1)*(365 - 2)*...*(365 - n +1)

For the first integer you have 365 possibilities, for the next integer there are 365 - 1 possibilites left etc. etc.

Since each of these strings has a probability of 1/365^n, the total probability of selecting a string in which non of the integers are repeated is:

P(n) = 365!/(365 - n)! 365^(-n)

Answer 2

If there are two people, what is the probability that the second will have a different birthday?
That would be 364/365

Now if there are three people, what is the probability that the third has a different birthday?

That would be 363/365
so the probability of all three having different birthdays is
364 * 363/365^2

etc
so I get for n people

p no coincidence = { 364!/(365-n)!} /365^(n-1)

so for example for 26 people:
p no coincidence = (364!/339!)/365^25

Do this by doing on our calculator
364/365*363/365*362/365 etc to 340/365

I got about .403
Which means that in a class with 26 people, there is about a 60% probability that at least two will have the same birthday.

Answer 3

Whew.
It turns out the Count and I gave you the same equation, but he did it much more neatly.
he wrote

P(n) = 365!/(365 - n)! 365^(-n)

I wrote

p no coincidence = { 364!/(365-n)!} /365^(n-1)

which can be written
(365!/365) /(365-n)! / 365^(n-1)
which is
(365! / (365-n)!) / 365^n