easy question... 12 possible outcomes, how long til all 12 happen?

[RiverTheNuts]

12 events, each has P=1/12, how do I find the average number of trials before all 12 events occur?

[Siegmund]

You don't care which order you see the events, so:

First, you see an event, any event, with probability 1.

Then, you get an event you haven't seen with probability 11/12, which will take, on average, 12/11 trials.

Then you've seen two and want one of the other 10, getting a new event with probability 10/12, which takes an average of 12/10 trials to get....

So on until you get them all, so the total EV is 1 + 12/11 + 12/10 + 12/9 + 12/8 + 12/7 + 12/6 + 12/5 + 12/4 + 12/3 + 12/2 + 12.

[RiverTheNuts]

Duh, ty, I was doing some crazy [censored] with 1/(n!/n^n), which is the probability of getting them in exactly n tries or something similar

[uDevil]

Are these events independant or mutually exclusive? How do you define a trial?

Siegmunds' solution assumes the events are mutually exclusive-- like rolling a 12-sided die, no?

If they are independent-- like rolling 12 12-sided dice until you have gotten, say a "1" on each individual die, then it would take an average of 12 rolls/die to get a "1", so it would take 12^12 (about 8.9E12) rolls to get all 12 events. If a trial were defined as simultaneously rolling all 12 dice, then it would take 1/12 fewer trials.

Edit: I guess your post title makes it clear these events are mutually exclusive, so nevermind... /images/graemlins/blush.gif

[BeerMoney]

no

[AaronBrown]

You can approximate this as:

N*[ln(N+0.5) + 0.577]

for N=12 that gives 12*[ln(12.5) + 0.577] = 12*[2.526 + 0.577] = 12*3.103 = 37.233 versus the exact answer of 32.239.