I'm having trouble wrapping my head around a probability question, and I know you guys love these.

If you roll a set of N N-sided dice, the chances of them all rolling a 1 is 1 : N^N
The chances of rolling the same number is N : N^N

Now the chances of rolling all the numbers 1-N is N! : N^N (I think)
What happens as you add more N-sided dice?
How many dice would you have to roll to be at least 99% certain of rolling at least one of each number 1-N?

It's bugging me that I can't figure this out. If you want to use a specific case, let N=8

Stick with Mafia Wars, it's easier.

You're correct that with N N-sided dice, the probability of rolling all of the numbers from 1 to N is:
N!/N^N.

As you add dice to a total of M dice, you add a probability to this of:
Sum from i=N+1 to M {N! x Sum from j=1 to N-1 [j^(i-N)] / N^i

QED

So for the 8-sided die example, in adding one more die you add a probability of:
8! x (1+2+3+4+5+6+7) / 8^9

Adding another die after that adds an additional probability of:
8! x (1^2+2^2+3^2+4^2+5^2+6^2+7^2) / 8^10

Dave:
Are you trying to out evil Wec?

Just go back to your room, pack your things, and hightail it out of AC realizing your blew your life savings at the craps tables.

Rehashing the odds gets you nowhere but bitter.

quote:

Originally posted by SD-Wineaux:
You're correct that with N N-sided dice, the probability of rolling all of the numbers from 1 to N is:
N!/N^N.

As you add dice to a total of M dice, you add a probability to this of:
Sum from i=N+1 to M {N! x Sum from j=1 to N-1 [j^(i-N)] / N^i}

QED

So for the 8-sided die example, in adding one more die you add a probability of:
8! x (1+2+3+4+5+6+7) / 8^9

Adding another die after that adds an additional probability of:
8! x (1^2+2^2+3^2+4^2+5^2+6^2+7^2) / 8^10

Ugh! I missed a number of permutations that lead to a successful roll of the dice, so the portion in red is incorrect. This portion actually needs to capture all of the permutations of dice rolls that fail to provide a unique value among the extra dice. So the example formula for 9 dice is fine, but for 10 dice it becomes:
8! x (1*1 + 1*2 + 1*3 + 1*4 + 1*5 + 1*6 + 1*7 + 2*2 + 2*3 + 2*4 + 2*5 + 2*6 + 2*7 + 3*3 + 3*4 + 3*5 + 3*6 + 3*7 + 4*4 + 4*5 + 4*6 + 4*7 + 5*5 + 5*6 + 5*7 + 6*6 + 6*7 + 7*7) / 8^10

Good luck finding a simplified form for this - I give up!