In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws total a of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is :

Probability PYQ — JEE Main 2020 · 4 Sep · Shift 2 · Q26 with Solution

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Correct answer: D

For a pair of fair dice:

Total $6$ can occur in $(1,5),(2,4),(3,3),(4,2),(5,1)$ so $P(\text{sum }6)=\frac{5}{36}.$
Total $7$ can occur in $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ so $P(\text{sum }7)=\frac{6}{36}=\frac{1}{6}.$

Thus:

On A's turn, probability of success (getting $6$ ) is $p=\frac{5}{36}.$
On B's turn, probability of success (getting $7$ ) is $q=\frac{1}{6}=\frac{6}{36}.$

A starts first.

A wins immediately on his first throw with probability $p$ .

If A fails and B also fails, then the situation resets to the original state.

Probability that both fail in one full round is

=\frac{31}{36}\cdot\frac{5}{6}= rac{155}{216}. $$ 3. **Form the infinite series** A can win: - in round 1: with probability $p$ - after both fail once, then A succeeds: with probability $(1-p)(1-q)p$ - after both fail twice, then A succeeds: with probability $[(1-p)(1-q)]^2p$ - and so on. So, $$P(A)=p\left[1+r+r^2+\cdots\right],$$ where $$r=(1-p)(1-q)=\frac{155}{216}.$$ Using geometric series, $$P(A)=\frac{p}{1-r}.$$ 4. **Compute** $$P(A)=\frac{\frac{5}{36}}{1-\frac{155}{216}} =\frac{\frac{5}{36}}{\frac{61}{216}}.

Now, $\frac{5}{36}\cdot\frac{216}{61}=\frac{5\cdot 6}{61}=\frac{30}{61}.$

Hence, $\boxed{P(A)=\frac{30}{61}}.$

The correct option is: $\boxed{\text{D: } \frac{30}{61}}.$