Question

Three persons, A, B, and C, take turns in throwing a die. They throw in the order A, B, C, A, B, C, A, B, etc., until someone wins. A wins by throwing a "one". B wins by throwing a "one" or a "two". C wins by throwing a "one", a "two", or a "three". Find the probability that each of the players is the winner.

The answers are 3/13, 5/13, 5/13

Please show your work

Answer 1

Let A denote the event that A rolls a "one" and A* denote the event that A does not roll a "one"

Clearly, P(A) = 1/6 and P(A*) = 5/6

Similarly, let B denote the event that B rolls a "one" or "two" and B* denote the event that B does not roll a "one" or "two"

Clearly, P(B) = 2/6 and P(B*) = 4/6

Similarly, let C denote the event that C rolls a "one", "two" or "three" and C* denote the event that C does not roll a "one", "two" or "three".

Clearly, P(C) = 3/6 and P(B*) = 3/6

Now, the probability that A wins is given by:

Now, the probability that B wins is given by:

Now, Probability that C wins is given by:
P(C wins) = 1 - P(A wins) -P(B wins)

= 1 - (3/13) - (5/13)

= 5/13 [Answer]

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