if two players A and B are alternatively throwing a coin and a die together. A player who first throws head and 6 wins the game. If A starts the game, then the probability that B wins the game is
Question
if two players A and B are alternatively throwing a coin and a die together. A player who first throws head and 6 wins the game. If A starts the game, then the probability that B wins the game is
Solution
The game is won when a player throws a head on the coin and a 6 on the die. The probability of this happening in a single turn is (1/2) * (1/6) = 1/12, because the coin and the die are independent events.
Let's denote:
P(A) = Probability that A wins P(B) = Probability that B wins
Since A starts the game, there are two possibilities:
- A wins on the first turn. This happens with probability 1/12.
- A doesn't win on the first turn and the game continues. This happens with probability 1 - 1/12 = 11/12.
In the second case, after A's first turn, the game is effectively starting over, but with B now going first. So in this case, the probability that B wins is just 1 - P(A).
Therefore, we can write the total probability that B wins as:
P(B) = 0 + (11/12) * (1 - P(A))
But we also know that P(A) and P(B) must add up to 1, because someone has to win the game. So we can write:
P(A) + P(B) = 1
Substituting the expression for P(B) from above, we get:
P(A) + 0 + (11/12) * (1 - P(A)) = 1
Solving this equation for P(A), we get:
P(A) = 12/23
And therefore:
P(B) = 1 - P(A) = 1 - 12/23 = 11/23
So the probability that B wins the game is 11/23.
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