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Step-by-Step Solution
Step 1: Identify the Relevant Probabilities
• The probability that a single throw of two fair dice sums to 6 is
$P(6) = \frac{5}{36}$.
(The possible outcomes summing to 6 are: (1,5), (2,4), (3,3), (4,2), (5,1), totaling 5 favorable cases out of 36.)
• The probability that a single throw of two fair dice sums to 7 is
$P(7) = \frac{6}{36} = \frac{1}{6}$.
(The possible outcomes summing to 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), totaling 6 favorable cases out of 36.)
Step 2: Understand the Game’s Winning Conditions
• Player A wins if he gets a sum of 6 before Player B gets a sum of 7.
• Player B wins if he gets a sum of 7 before Player A gets a sum of 6.
• Players throw in turns, starting with Player A. As soon as either player achieves their target sum (6 for A or 7 for B), the game ends.
Step 3: Derive the Probability of A Winning
Let $P(A)$ be the probability that Player A eventually wins.
• On A’s first throw, if A gets 6 immediately (with probability $5/36$), A wins outright.
• If A does not get 6 (probability $1 - 5/36 = 31/36$) and B also fails to get 7 on his throw (probability $1 - 6/36 = 30/36$), the game essentially “resets” to the same situation for A’s next throw.
Hence we can set up the infinite series for A’s winning probability as follows:
$P(A) = \frac{5}{36} + \left( \frac{31}{36} \times \frac{30}{36} \right) \frac{5}{36} + \left( \frac{31}{36} \times \frac{30}{36} \right)^2 \frac{5}{36} + \dots$
Step 4: Sum the Infinite Geometric Series
This is an infinite geometric series of the form:
$P(A) = \frac{5}{36}\Bigl[1 + \left(\frac{31}{36} \times \frac{30}{36}\right) + \left(\frac{31}{36} \times \frac{30}{36}\right)^2 + \ldots \Bigr].$
The common ratio $r$ of this geometric series is $\left(\frac{31}{36} \times \frac{30}{36}\right)$. For an infinite geometric series $a + ar + ar^2 + \dots$, the sum is
$ \frac{a}{1 - r}, $
where $a$ is the first term and $r$ is the common ratio with $|r| < 1$.
Step 5: Final Calculation
Here, $a = \frac{5}{36}$ and $r = \frac{31}{36} \times \frac{30}{36}$. Thus,
$P(A)
= \frac{\frac{5}{36}}{1 - \left(\frac{31}{36} \times \frac{30}{36}\right)}
= \frac{5/36}{1 - \frac{31 \times 30}{36 \times 36}}
= \frac{5/36}{\frac{36 \times 36 - 31 \times 30}{36 \times 36}}
= \frac{5/36}{\frac{1296 - 930}{1296}}
= \frac{5/36}{\frac{366}{1296}}
= \frac{5 \times 1296}{36 \times 366}
= \frac{30}{61}.
Answer: $ \frac{30}{61} $
