Expected Value & Discrete Probability Distributions

Q: What makes a game fair?

An expected net gain of $0$. Find the expected winnings, subtract the cost to play, and if the result is $0$ the game breaks even over the long run. A negative result means it loses money over time.

A discrete random variable takes separate, countable values, such as the number of heads in three coin flips or the payout of a raffle ticket. A probability distribution is the table that pairs each of those values with the probability of getting it. Once you have that table, you can answer the question every game and every business decision comes down to: what happens on average if this is repeated many times?

That average is called expected value, written $E(X) = \sum x \cdot P(x)$ . You multiply each value by its probability, then add. The result is a weighted average, so outcomes that are more likely pull it harder, and it is perfectly normal for the answer to be a number the variable can never actually equal.

What makes a valid distribution

A discrete probability distribution is usually shown as two columns: the values $x$ on one side, the probabilities $P(x)$ on the other. Two rules have to hold. Every probability is between $0$ and $1$ , so a negative entry is an immediate disqualification, and all the probabilities together add to exactly $1$ , because one of the outcomes has to happen.

That second rule is also a tool. If a distribution lists three of its four probabilities as $0.2$ , $0.3$ , and $0.4$ , the missing one is $1 - 0.9 = 0.1$ . Check the total before doing anything else, because an expected value computed from a table that does not add to $1$ is meaningless.

Expected value is a weighted average

To find $E(X)$ , go row by row: multiply the value by its probability, then add the products. For a distribution with values $2$ , $4$ , and $6$ carrying probabilities $0.5$ , $0.3$ , and $0.2$ , the expected value is $2(0.5) + 4(0.3) + 6(0.2) = 1 + 1.2 + 1.2 = 3.4$ . Notice that is not the plain average of $2$ , $4$ , and $6$ , which would be $4$ . The probabilities do the weighting.

Expected value is a long run average, not a prediction of the next trial and not the most likely outcome. The expected number of heads in one flip is $0.5$ , and no flip has ever produced half a head. What it means is that over thousands of flips the running average settles near $0.5$ .

Fair games and comparing decisions

For a game with a cost, work out the expected winnings first, then subtract what you paid to get the expected net gain. A game is called fair when that net gain is $0$ , meaning that in the long run you break even. If the net gain is negative, the game loses money over time no matter how a single round turns out.

The same computation compares two options that are not games at all. Give each outcome its value, with losses written as negative numbers, multiply by the probabilities, and add. The plan with the larger expected value is the better long run choice, and it is often not the one with the flashiest single outcome.

Worked examples

Example 1: expected value from a table

A distribution has values $0$ , $1$ , $2$ , $3$ with probabilities $0.1$ , $0.3$ , $0.4$ , $0.2$ . Find $E(X)$ .

Check that the probabilities add to

1

0.1 + 0.3 + 0.4 + 0.2 = 1

Multiply each value by its probability

0(0.1) + 1(0.3) + 2(0.4) + 3(0.2)

Add the products

0 + 0.3 + 0.8 + 0.6 = 1.7

Answer: $E(X) = 1.7$

Example 2: finding a missing probability

Three of a distribution's four probabilities are $0.2$ , $0.3$ , and $0.4$ . What is the fourth?

Add the three that are given

0.2 + 0.3 + 0.4 = 0.9

The total has to be

1

1 - 0.9 = 0.1

Answer: $0.1$

Example 3: is the game fair?

A game costs $3 to play. You win $10 with probability $0.25$ and nothing otherwise. Find the expected net gain.

Expected winnings

10(0.25) + 0(0.75) = 2.5

Subtract the cost to play

2.5 - 3 = -0.5

Compare to

0

-0.5 < 0

Answer: An expected net gain of $-0.5$ , so the game is not fair. It loses about $0.50 per play in the long run.

Example 4: comparing two plans

Plan A earns $500 with probability $0.6$ and loses $200 with probability $0.4$ . Plan B earns $1000 with probability $0.3$ and loses $100 with probability $0.7$ . Which has the better expected profit?

Expected value of Plan A

500(0.6) + (-200)(0.4)

Compute

300 - 80 = 220

Expected value of Plan B

1000(0.3) + (-100)(0.7)

Compute

300 - 70 = 230

Answer: Plan B, with an expected profit of $230 against $220 for Plan A.

Try one yourself

Common questions

Can the expected value be a number the variable never takes?

Yes, and it usually is. The expected number of heads in one flip is $0.5$ , which no single flip can produce. Expected value describes the average over many repetitions, not any one result.

Why not just average the values?

A plain average treats every outcome as equally likely. Expected value weights each value by how often it actually happens, which is why $2$ , $4$ , $6$ with probabilities $0.5$ , $0.3$ , $0.2$ gives $3.4$ rather than $4$ .

What makes a game fair?

An expected net gain of $0$ . Find the expected winnings, subtract the cost to play, and if the result is $0$ the game breaks even over the long run. A negative result means it loses money over time.

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