How about this:
Throwing the die 420 times, there are 6[sup]420[/sub] possibilities (no sorting or counting, just a list of 420 digits).
To get exactly 70 ones, this is the number of combinations of 70 items taken from the 420:
Code:
420! / (70! * (420-70)!) = 420! / (70! * 350!)
This tells you which of each roll was a one.
To get exactly 70 twos, this is the number of combinations of 70 items taken from the remaining 350: 350! / (70! * 280!). This tells you which of each roll that was not a one was a two.
Threes: 280! / (70! * 210!)
Fours: 210! / (70! * 140!)
Fives: 140! / (70! * 70!)
Sixes: 70! / (70! * 0!) [yes there is only one way]
So, the probability of getting exactly 70 of each is the product of the combinations of getting exactly 70 of each number, divided by
all the combinations. Noting that the two 350!'s cancel, as do the 280!s, etc., this leaves:
Code:
420! / (70!6 * 6420)
Plugging all this into MS Calculator, I get 5.99 x 10
-7, or one in a bit under 1,668,000.
I expected it to be long, but not quite that long. Let's try that for smaller numbers of throws (multiplied by a milllion for simplicity):
6: 15,432
12: 3,438
18: 1,351
60: 74.6
120: 13.5
The numbers make sense and seem to follow a reasonable progression.