Mission IIT Screening Test

Question

Raju and Rani are playing betting game. Raju rolls a dice and if 5 comes on face then they stop the game, else Raju needs to pay 1 rupee ro Rani and continue rolling. Expected no of rupees Rani gets in this game?

Answer 1

Probability that game stops in $1^{st}$ roll = $\dfrac{1}{6}$. Rani gets Rs. $0$.

Probability that game stops in $2^{nd}$ roll = $\dfrac{5}{6} \times \dfrac{1}{6}$. Rani gets Rs. $1$.

Probability that game stops in $3^{rd}$ roll = $\dfrac{5}{6} \times \dfrac{5}{6} \times  \dfrac{1}{6}$. Rani gets Rs. $2$.

Probability that game stops in $4^{th}$ roll = $\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}$. Rani gets Rs. $3$.

and so on .

Thus, expected amount that Rani gets = $0 \times \left (\dfrac{1}{6} \right )+ 1 \times \left (\dfrac{5}{6} \times \dfrac{1}{6}\right ) + 2 \times \left (\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} \right)+ 3 \times \left(\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}\right)$ and so on.

i.e. $0 \times \left (\dfrac{1}{6} \right )+ 1 \times \left (\dfrac{5}{6} \times \dfrac{1}{6}\right ) + 2 \times \left (\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} \right)+ 3 \times \left(\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}\right)+ ... \infty$

$= 1 \times \left(\dfrac{5}{6} \times \dfrac{1}{6}\right) + 2 \times \left(\left(\dfrac{5}{6}\right)^2 \times \dfrac{1}{6}\right) + 3 \times \left(\left(\dfrac{5}{6}\right)^3 \times \dfrac{1}{6}\right)+ ... \infty$

$=\dfrac{5}{6} \times \dfrac{1}{6} \left(1+2 \times \dfrac{5}{6}+3 \times \left(\dfrac{5}{6}\right)^2+4 \times \left(\dfrac{5}{6}\right)^3 +... \infty\right) $

$= \dfrac{5}{6} \times \dfrac{1}{6} \left(1-\dfrac{5}{6}\right)^{-2}$ $\qquad \color{maroon}{ \because \left [ (1-x)^{-2}= 1+2x+3x^2+4x^3+...\infty\right]}$

$= \dfrac{5}{6} \times \dfrac{1}{6} \times \left( 1- \dfrac{5}{6} \right )^{-2}$

$= \dfrac{5}{6} \times \dfrac{1}{6} \times \left( \dfrac{1}{6} \right )^{-2} $

$= \dfrac{5}{6} \times \dfrac{1}{6} \times 36$

$=5$

Answer 2

let random variable x represents the amount Rani getting from Raju

p(x=1) = p(getting other than 5) * p(getting 5 as outcome) = $\frac{5}{6}$*$\frac{1}{6}$

p(x=2) = ($\frac{5}{6}$)² * $\frac{1}{6}$

p(x=3) = ($\frac{5}{6}$)³ * $\frac{1}{6}$

so on

so expected number of rupees rani gets in the game = 1 * $\frac{5}{6}$*$\frac{1}{6}$  +  2 * ($\frac{5}{6}$)² * $\frac{1}{6}$  +  3 * ($\frac{5}{6}$)³ * $\frac{1}{6}$  +.......$\infty$

                                                                                               = 5