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?

Question

Jamiya Knox

Mathematics

7 May, 11:50

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?

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Answers (2)

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Darian Preston · Answer 1

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

Gallagher · Answer 2

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