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  • In a dice game, a player pays a stake of Re1 for each throw of a die. She receives Rs 5 if the die shows a 3, Rs 2 if the die shows a 1 or 6, and nothing otherwise. What is the player’s expected profit per throw over a long series of throws?

In a dice game, a player pays a stake of Re1 for each throw of a die. She receives Rs 5 if the die shows a 3, Rs 2 if the die shows a 1 or 6, and nothing otherwise. What is the player’s expected profit per throw over a long series of throws?

Answers (1)

Solution

Let X = the random variable of profit per throw

Probability of getting any number on dice is \frac{1}{6} .

Since, she loses Rs 1 on getting any of 2, 4 or 5.

Therefore, at X= -1,

P(X) = P (2) +P(4) +P(5)

\\P(X)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6} \\=\frac{3}{6} \\=\frac{1}{2}
In the same way,  =1  if dice shows of either 1 or 6 .
\mathrm{P}(\mathrm{X})=\mathrm{P}(1)+\mathrm{P}(6) \\P(X)=\frac{1}{6}+\frac{1}{6} \\=\frac{1}{3}
and at X=4  if die shows a 3
\mathrm{P}(\mathrm{X})=\mathrm{P}(3) \\\mathrm{P}(\mathrm{X})=\frac{1}{6}

\begin{aligned} &\begin{array}{|l|c|l|l|} \hline \mathrm{X} & -1 & 1 & 4 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\ \hline \end{array}\\ &\therefore \text { Player's expected profit }=E(X)=X P(X)\\ &=-1 \times \frac{1}{2}+1 \times \frac{1}{3}+4 \times \frac{1}{6}\\ &=\frac{-3+2+4}{6}\\ &=\frac{3}{6}\\ &=\frac{1}{2}=\text { Rs } 0.50 \end{aligned}

The player’s expected profit per throw over a long series of throws is Rs. 0.50

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