This math problem is a cumulative binomial distribution.
Let's look at binomial distribution.
Each die toss is an event with two mutually exclusive possible outcomes: either you roll a 6 (success), or you roll 1 -5 (failure). If we have N events, then the binomial distribution can be used to determine the probability of getting exactly r successes in N outcomes. The formula is:
P(r) = (N! / (r!(N - r)!)) ( p^r) (1 - p)^(N-r)
where N = number of trials,
P(r) = probability of getting exactly r successes in the N outcomes
p = probability of success on any one trial.
So, in our situation, N = 1000 dice rolls, p = 1/6.
David got 182 successes out of 1000 trials and Claude got 326 out of 1000.
So, the probability of David getting exactly 182 out of 1000, by using the above formula, is
P(182) = (1000! / (182!(1000-182)!))((1/6)^182)(1-1/6)^(1000-182)
And the probability of Claude getting 326 out of 100 is:
P(326) = (1000! / (326!(1000-326)!))((1/6)^326)(1-1/6)^(1000-326)
But so far we are talking about the probability of getting exactly the number we got. I think we're more interested in the cumulative case, where we ask, what is the probability of getting at least 182 successes out of 1000, or at least 326 out of 1000?
Well, then we have to add them up, i.e.
P(182 or more) = P(182) + P(183) + P(184) + .... + P(1000)
and
P(326 or more) = P(326) + P(327) + P(328) + ... + P(1000)
I thought, no problem, I'll do this on Excel, but I immediately found that Excel can't handle 1000! because it's too big a number.
