A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8,  and B wins if he throws a sum of 8 before A throws a sum of 5.  The probability that A wins if A makes the first throw, is:

When a pair of dice is thrown, the ways in which we can get a sum of $$5$$ are: $$(1,4)$$, $$(4,1)$$, $$(2,3)$$ and $$(3,2)$$, giving the total probability among all the $$6\times 6$$ possibilities as $$\dfrac{4}{36}$$

When a pair of dice is thrown, the ways in which we can get a sum of $$8$$ are: $$(2,6)$$, $$(6,2)$$, $$(5,3)$$, $$(3,5)$$, and $$(4,4)$$, giving the total probability among all the $$6\times 6$$ possibilities as $$\dfrac{5}{36}$$

The probability that A wins in the very first throw is $$\dfrac{4}{36}$$

The probability that A wins in the second time he throws is $$\dfrac{36-4}{36}\times \dfrac{36-5}{36} \times \dfrac{4}{36}$$

The probability that A wins in the third time he throws is $$\dfrac{36-4}{36}\times \dfrac{36-5}{36} \times \dfrac{36-4}{36}\times \dfrac{36-5}{36} \times \dfrac{4}{36}$$

And so on,

The cases form a GP with the first term $$\dfrac{4}{36}$$ and the common ratio of $$\dfrac{36-4}{36}\times \dfrac{36-5}{36}$$

Thus, the combined probability will be;

$$\dfrac{4/36}{1- \left(\dfrac{36-4}{36}\times \dfrac{36-5}{36}\right)} = \dfrac{144}{304} = \dfrac{36}{76} = \dfrac{9}{19}$$

Option B is the correct answer.