In a luck race, 100 people are gonna throw 5 dice each. If a person throws 3 or more "6", they win. Out of 100 people, what is the probability that 6 people will win?

Answer 1

What is the probability of any particular person winning? It's the probability of throwing 3 or more 6's in 5 fair 6 sided dice (I'm assuming they're fair and 6 sided).

Let's call the number of 6's someone throws X. X follows a binomial(n=5, p=1/6) distribution, which is a discrete distribution, so:

P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5) = (5 choose 3)(1/6)³(5/6)^5-3 + (5 choose 4)(1/6)⁴(5/6)^5-4 + (5 choose 5)(1/6)⁵(5/6)^5-5 ≈ 0.03549383.

But we're not done. Now we need to figure out the probability that exactly 6 people win out of 100, if the probability that each person wins is 0.03549383, and one person winning doesn't effect the probability that someone else wins (i.e. they're all independent). Let's call the number of people who win Y. This also follows a binomial distribution, but this time n = 100 and p = 0.03549383.

P(Y = 6) = (100 choose 6)(0.03549383)⁶(1-0.03549383)^100-6 ≈ 0.0798.

Answer 2

First we need to calculate the probability of an individual person winning.

Whether they win or not is a binomial variable. The probability of an individual roll of the die being six is 1/6. So the probability of throwing 3 or more sixes in 5 tries is the binomial probability of rolling 3, 4 or 5 sixes.

For this problem, C(5,x) * x^0.167 * (5-x)^0.833 is the probability of rolling a particular value x on the die, so we need to compute that probability for x=3,4 and 5 and add together and you should get 0.036.

To compute the probability 6 people win out of 100 we again use the binomial formula, where the probability of each person winning is 0.036, so we compute the probability as C(100,6)*6^0.036*94^0.964, which should get you 0.083.