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since the player A only wins when he gets ‘6’ and player B must not get 6.

so the probability(P(a)) of player A getting 6 is 1/6.

the probability(P(b)) of player B not getting 6 is 5/6.

The probability(1-P(a)) of player A not getting 6 is 5/6.

the final Probability(P) is :

P=P(a)+(1-P(a)).P(b).P(a)+……… (because for every alternative Player A must-win such that P(a) is taken)

by solving we get P=6/11.

so the probability(P(a)) of player A getting 6 is 1/6.

the probability(P(b)) of player B not getting 6 is 5/6.

The probability(1-P(a)) of player A not getting 6 is 5/6.

the final Probability(P) is :

P=P(a)+(1-P(a)).P(b).P(a)+……… (because for every alternative Player A must-win such that P(a) is taken)

by solving we get P=6/11.