How to Find the Average Win in a Game of Chance Before You Play

A stall at a school fair charges ₹10 to roll a die and pays out according to the number shown. Is the game worth playing? Turning each outcome into a number, listing the chance of each number and averaging gives the answer before a single rupee is spent.

This lesson covers random variables and their probability distributions, the conditions every distribution must satisfy, and the mean of a random variable.

**A random variable is a rule that assigns a real number to each outcome of a random experiment, and its probability distribution lists each possible value $x_i$ with its probability $p_i$.

Discrete random variables take separate values that can be listed, such as the number of heads or the number of defective items.

Worked example (two coins).** Let X be the number of heads when two coins are tossed. The outcomes HH, HT, TH and TT give

- $P(X=0)=\frac{1}{4}$, from TT
- $P(X=1)=\frac{2}{4}=\frac{1}{2}$, from HT and TH
- $P(X=2)=\frac{1}{4}$, from HH

Worked example (drawing without replacement). Two balls are drawn from a bag of 4 red and 2 black balls. Let X be the number of black balls. There are ${}^6{C}_2=15$ possible pairs.

- $P(X=0)=\frac{{}^4{C}_2}{15}=\frac{6}{15}$
- $P(X=1)=\frac{4\times 2}{15}=\frac{8}{15}$
- $P(X=2)=\frac{1}{15}$

An everyday example. The number of customers who walk into a chemist's shop in a given ten minutes is a random variable that the owner can describe with a distribution.

The substance. Many outcomes can share one value — HT and TH are different outcomes but give the same value $X=1$, so their probabilities add.

**A function $P(X=x_i)=p_i$ is a valid probability distribution only if every $p_i$ lies between 0 and 1 and all the $p_i$ add up to exactly 1.

Worked example (checking).** Is $P(X=x)=\frac{x}{10}$ for $x=1,2,3,4$ valid? Each value lies between 0 and 1, and $\frac{1+2+3+4}{10}=1$, so yes.

Worked example (finding k). X takes the values 1, 2, 3 and 4 with probabilities k, 2k, 3k and 4k. Then $10k=1$, so $k=0.1$ and $P(X\le 2)=0.1+0.2=0.3$.

Worked example (rejecting a root). If $P(X=0)=k$, $P(X=1)=2k$ and $P(X=2)=3k^2$, then $3k^2+3k-1=0$, so

$$k=\frac{-3\pm \sqrt{9+12}}{6}=\frac{-3\pm \sqrt{21}}{6}$$

The negative root would make $P(X=0)$ negative, so it is rejected, leaving $k\approx 0.264$.

An everyday example. A forecast giving a 50% chance of sun, a 30% chance of cloud and a 30% chance of rain as the only possibilities cannot be right, because the chances add up to 110%.

The substance. A value of k that makes any probability negative must be rejected even though it satisfies the sum condition.

**The mean, or expected value, of a random variable is $E(X)=\mu =\sum x_i{p}_i$ — each value multiplied by its probability, then added.

Worked example (two coins).** $E(X)=0\times \frac{1}{4}+1\times \frac{1}{2}+2\times \frac{1}{4}=1$ head.

Worked example (a die). $E(X)=\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=3.5$

Worked example (the fair stall). The stall charges ₹10. It pays ₹20 for a six, ₹10 for a four or a five, and nothing otherwise. The expected payout in rupees is

$$E=20\times \frac{1}{6}+10\times \frac{2}{6}+0\times \frac{3}{6}=\frac{40}{6}\approx 6.67$$

so the expected gain per game is about $6.67-10=-3.33$ rupees — a loss.

Worked example (balls). For the black balls above, $E(X)=0\times \frac{6}{15}+1\times \frac{8}{15}+2\times \frac{1}{15}=\frac{10}{15}=\frac{2}{3}$.

An everyday example. A vegetable seller estimating how many kilograms of tomatoes sell on an average day is using the mean of a distribution built from past sales.

The substance. The mean need not be a possible value — no single roll of a die shows 3.5, yet 3.5 is the long-run average.