[euhlmann]

Originally Posted by Ether

If a fair die is thrown 420 times, what's the probability of getting exactly 70 of each number (1 thru 6)?

Alright, not looking at any of the posts yet, this question looks like it's essentially "how many ways can you arrange 70 each of 6 options"

The formula for permutations with repetition is

n!/(n1! * n2! * ... * ni!)

So we have

n!/(n1! * n2! * ... * n6!) = (420!)/(70!)^6

So that's the total number of ways to get 70 exactly of 1 to 6.
To find the probability, divide by total number of ways to throw the dice, which is 6^420 since there are 6 options on each of 420 steps.

I get 5.99605*10^-7
In conclusion, not likely to happen

Edit: looks like I came a bit late to this thread
But it also looks like other people got the same thing. I guess this is correct then.