In a game a man wins $₹ 40$ if he gets 5 or 6 on a throw of a fair die and loses ₹ 20 for getting any other number on the die. If he decides to throw the die either till he gets a five or six or to a maximum of three throws, then his expected gain/loss (in rupees) is

Let $S$ denote success (5 or 6) and $F$ denote failure (1, 2, 3, or 4). The probabilities are $P(S) = \frac{1}{3}$ and $P(F) = \frac{2}{3}$, with gains of ₹40 and –₹20 respectively.

The process terminates at the first success or after three throws. The four possible outcomes are:

Success on first throw: $P = \frac{1}{3}$, net gain = ₹40

Failure then success: $P = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}$, net gain = –20 + 40 = ₹20

Two failures then success: $P = \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{1}{3} = \frac{4}{27}$, net gain = –20 – 20 + 40 = ₹0

Three failures: $P = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$, net gain = –20 – 20 – 20 = –₹60

The expected gain is:

$E = 40 \cdot \frac{1}{3} + 20 \cdot \frac{2}{9} + 0 \cdot \frac{4}{27} - 60 \cdot \frac{8}{27}$

Using a common denominator of 27, this becomes:

$E = \frac{360 + 120 - 480}{27} = \frac{0}{27} = 0$

The expected gain is therefore ₹0.