Two people, P and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6

Q. Two people, P and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equi-probable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is .

Ans: 0.023

Sol:

Given there are two identical dices, each having 6 faces.
Favorable events for tie = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Since two dice are thrown, total sample space will be 6 x 6 = 36

Therefore, P(tie) = 6/36 = 1/6
and Probability of not tie = (1 – 1/6)

To one of them win on third trial, previous two trials should be tie.

= 1/6 * 1/6 * (1 – 1/6)
= 1/36 * 5/6
= 5/216

= 0.023