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Two players, $A$ and $B$, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player $A$ starts the game, the probability that $A$ wins the game is

  1. $5/11$
  2. $1/2$
  3. $7/13$
  4. $6/11$
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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.
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