Question

In a certain board game, a player rolls two fair six-sided dice until the player rolls doubles (where the value on each die is the same). The probability of rolling doubles with one roll of two fair six-sided dice is 1/6. What is the probability that it takes three rolls until the player rolls doubles?

Answer 1

Answer:

The probability is 25/216. Approximately 0.116

Step-by-step explanation:

You need

- The first throw to fail (probability 1-1/6 = 5/6)

- The second throw to fail (probability 5/5)

- The third throw to be a success (probability 1/6)

Since each throw is independent of the others, we have to multiply all probabilities to obtain the total probability of the event. Thus, the probability of requiring 3 rolls until getting doubles is

5/6 * 5/6 * 1/6 = 25/216 = 0.115741

This problem can also be solved with sophisticated theory;

the random variable which counts the number of tries until the first success is a geometric distribution. The only parameter of the distribution is the probability of success p. If X is geometric  with parameter p, then the probability of X being equal to k (in other words, requiring k tries for a success) is

P(X = k) = (1-p)^(k-1) * p

If p = 1/6, then

P(X = 2) = (1-1/6)^2*(1/6) = (5/6)²*1/6 = 25/216

Answer 2

Answer:

(1/6)(5:6)^2

Step-by-step explanation: