$A$ 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

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $P(	ext{sum }5)=\frac4{36}=\frac19$, $P(	ext{sum }8)=\frac5{36}$. $P(A	ext{ wins})=\frac{a}{1-(1-a)(1-b)}$ with $a=\frac19$, $b=\frac5{36}$: $=\frac{1/9}{1-\frac89\cdot\frac{31}{36}}=\frac{1/9}{76/324}=\frac{36}{76}=\frac9{19}$. Answer (2).

Q. $A$ 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

Explanation

Reviewed by Dr. Abubacker

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