Question

$X$ and $Y$ throw a die alternatively till one of them get a $4$. find the probability of winning of $Y$ if $X$ starts first.

• A. $\frac{5}{11}$
• B. $\frac{1}{11}$
• C. $\frac{2}{5}$
• D. $\frac{4}{5}$

Answer 1

Answer: (A)

$\frac{5}{11}$

Winning the game is getting a 4 on the dice. Hence total number of outcomes = 6

P(getting a 4) = $\frac{1}{6}$

P(not getting a 4)=$1-\frac{1}{6}=\frac{5}{6}$

First throw by X: X gets a 4, P(X wins)=$\frac{1}{6}$

Second throw by Y: X does not get 4, Y gets 4. So, P(Y wins)=$\frac{5}{6}\times \frac{1}{6}$

Third throw by X: X doesn't get 4, Y does not get 4, X gets 4. P(X wins) = $\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}$

Fourth throw by Y: X doesn't get 4, Y does not get 4, X does not get 4, Y gets 4. P(Y wins) = $\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}$ and so on

So the probability that Y wins is

$\frac{5}{6}\times \frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+...⟹(\frac{5}{6}\times \frac{1}{6})+{\left(\frac{5}{6}\right)}^3\left(\frac{1}{6}\right)+{\left(\frac{5}{6}\right)}^5\left(\frac{1}{6}\right)+...$

This forms a GP and sum of GP is $\frac{a}{1-r}$, here, $a=\frac{5}{36},r={\left(\frac{5}{6}\right)}^2$

Hence P(Y wins) = $\frac{\frac{5}{36}}{1-{\left(\frac{5}{6}\right)}^2}$

$=\frac{5}{36}\times \frac{36}{36-25}=\frac{5}{11}$