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A, B, C, D cut a pack of 52 well shuffled playing cards successively in the same order. If the person who cuts a spade first, wins the game and the game continues until this happens, then the probability that A wins the game is
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The correct answer is:
$\frac{64}{175}$
Let probability of spade be $P$ and that of not spade be $P^{\prime}$. Required probability
$$
\begin{aligned}
& =P(A)+P^{\prime}(A) \cdot P^{\prime}(B) \cdot P^{\prime}(C) \cdot P^{\prime}(D) \cdot P(A) \\
& \quad+\left\{P^{\prime}(A) \cdot P^{\prime}(B) \cdot P^{\prime}(C) \cdot P^{\prime}(D)\right\}^2 P(A)+\ldots \ldots \infty \\
& =\frac{P(A)}{1-P^{\prime}(A) P^{\prime}(B) P^{\prime}(C) P^{\prime}(D)} \\
& =\frac{\frac{1}{4}}{1-\frac{81}{256}}=\frac{1}{4} \times \frac{256}{175}=\frac{04}{175} .
\end{aligned}
$$
$$
\begin{aligned}
& =P(A)+P^{\prime}(A) \cdot P^{\prime}(B) \cdot P^{\prime}(C) \cdot P^{\prime}(D) \cdot P(A) \\
& \quad+\left\{P^{\prime}(A) \cdot P^{\prime}(B) \cdot P^{\prime}(C) \cdot P^{\prime}(D)\right\}^2 P(A)+\ldots \ldots \infty \\
& =\frac{P(A)}{1-P^{\prime}(A) P^{\prime}(B) P^{\prime}(C) P^{\prime}(D)} \\
& =\frac{\frac{1}{4}}{1-\frac{81}{256}}=\frac{1}{4} \times \frac{256}{175}=\frac{04}{175} .
\end{aligned}
$$
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