Probability: Roll a die 10x, get 5 one s and 5 two s.?

Stan did 2) and 3) correctly. I won't repeat the , work.

The first question is the toughest.

5 1's and 5 2's?

There are 10!/5!5! = 252 ways to rolls 5 1's and 5 2's

There are 6^10 ways to rolls the dice 10 times.

So the probability is:

252 / 6^10 = 7/ 6^8 = .0000041676

I ran a simulation of that half a billion times, and got a result about 14% higher than that, but the calculation and the simulation both look correct. I'm getting a probability of .00000474. That's about 6+ standard deviations, so it's just not consistent. Any comments from any stat experts reading this would be appreciated. Something just doesn't seem right.

The Excel models I build for things like this usually work pretty well. This problem has such a low probability, that I think the Excel random number generator does not do a good enough job. If I take every random number, I get results, for 10^8 trials, in the range of 472-476 (7 runs). Very consistent. But if I take every other random number, it's in the range of 373-376 (3 runs). Every 3rd gives me 385-386 (3 runs). If I use "randomize" at the start of every trial, that uses the system clock to reset the random number seed, and the results are pure garbage: 87, 302, 259, 438. I knew the random number generator wasn't perfect -- none is -- but this is the first time I've had such inconsistent results. I chalk that up to the extremely low probability involved. I need to get 10 random numbers in a row that lie between 0 and 1/3 in order to have a successful set of rolls. That's asking a lot of the random number generator.

So I conclude the model is OK, but fails to do a good enough job on this problem because of the generator's limitations.

Again, any insights from knowledgeable people on this would be appreciated.

2) Prob of no ones or twos: (4/6)^10 = (2/3)^10 = 0.0173415299

3) Prob of only 1,2 & 3 being rolled: = (3/6)^10 = (1/2)^10 = 0.0009765

Per Freond1 on problem 1, I believe using the multinomial distribution is the correct way to go:

P = (10! / (5!5!)) (1/6)^5 (1/6)^5 = 0.0000041676