Two old gamblers alternate at rolling a fair 6-sided die until the first 6 comes up. Compute the probability of the first player winning.

Question

Two old gamblers alternate at rolling a fair 6-sided die until the first 6 comes up. Compute the probability of the first player winning.

Let X be the number of trials the first one can one. Then P(X=1)=$\frac {1}{6}$

P(X=3)= $(\frac {5}{6})^2$* $\frac {1}{6}$

P(X=2n+1)=$(\frac {5}{6})^{2n}$* $\frac {1}{6}$

I am not sure how to conclude from this.

Answer 1

Work out the geometric series for the first player. If he wins on roll $2n+1$ then the first $2n$ rolls must all be not $6$. Thus we have the sum

$(1/6)+(5/6)^2(1/6)+(5/6)^4(1/6)+...$

$=(1/6)×(1+(5/6)^2)+(5/6)^4+...)$

$=(1/6)×(1/(1-(5/6)^2))=\color{blue}{6/11}$