Question

Suppose you and your three friends have two dice each. 

1. When you roll your dice, what are the chances you get a pair? (example (1:1), (2:2))
2. When 2 dice are thrown, what is the probability of the sum being 9? (example (4:5)
3. When everyone rolls, what are the chances that there is at least one 6?

Answer 1

n - the number of dice, s - the number of a individual die faces, p - the probability of rolling any value from a die, and P - the overall probability for the problem, r - sum of dices values.

p = 1/s

1. P = spⁿ = s (1/s)ⁿ = $({\frac{1}{6}})^2 \times6 = 0.16667$

2. P(n, r, s) = $\frac{1}{S^n}\sum_{k=0}^{\frac{r-n}{s}}(-1)^k(^n_k)(^{r-s\times k-1}_{n-1})$

The formula is quite complicated. However, we can also try to evaluate this problem by hand.The number of permutations with repetitions in this set is 36. All the possibilities to obtain 9 are : 3+6, 4+5, 5+4, 3+6, so the total amount is 4. $\frac{4}{36} = 0.11111$

3. Consider the complement problem, there is a 5/6 probability of not rolling a six for any given die, and since the eight dice are independent, the probability of not rolling a six is (5/6)⁸ = 5⁸ /6⁸ = 390625/1679616. The probability of rolling at least one six is therefore 1 − 390625/1679616 ≈ 0.7674.