Expected Value

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Expected Value: meaning, method and solved example (Precalculus) – Maths PDF

In short: The expected value of a random outcome is its long-run average, found by multiplying each possible value by its probability and adding the products: E(X) = Σ x · P(x).

PrecalculusProbabilityS-MD.A.2

What is Expected Value?

The expected value answers the question “what happens on average if I play many times?” It need not be a value you can actually win on one play; it is the balance point of the whole probability distribution.

For a random variable \(X\) taking values \(x_1, x_2, \ldots, x_n\) with probabilities \(p_1, p_2, \ldots, p_n\), \(E(X) = x_1p_1 + x_2p_2 + \cdots + x_np_n\). In a game, subtract the cost to play to get the average net result. A game with a negative net average loses money over time, however lucky a single round may be.

How to spot it

  • A game, raffle or insurance plan lists payouts together with the chance of each one.
  • The question asks what you should expect to win or lose per play over many plays.
  • A table shows values of a random variable with their probabilities, and a mean is wanted.
  • The word fair appears, which means the average net result equals zero.

The method

  1. List every possible value \(x\) and its probability \(P(x)\), and check that the probabilities total 1.
  2. Multiply each value by its probability.
  3. Add all the products to get \(E(X)\).
  4. For a game, subtract the entry cost; a positive net favors the player and a negative net favors the house.
P = 0.5P = 0.3P = 0.2$0$2$9balance: $2.40E = 0(0.5) + 2(0.3) + 9(0.2)= 0 + 0.6 + 1.8 = 2.4
The balance point of the payoffs is $2.40, the expected value.

Expected Value: solved example

Problem

At a school fair, a spinner pays \(\$0\) with probability 0.5, \(\$2\) with probability 0.3 and \(\$9\) with probability 0.2. One spin costs \(\$3\). Find the expected payout and the expected net result per spin.

Solution

Payout: \(0(0.5) + 2(0.3) + 9(0.2) = 0 + 0.6 + 1.8 = 2.4\), so \(\$2.40\) per spin on average.

Net of the \(\$3\) cost: \(2.40 – 3 = -0.60\). The player loses \(\$0.60\) per spin on average.

Try it

Question 1

A variable takes the values 1, 2, 3, 4 with probabilities 0.1, 0.2, 0.3, 0.4. Find its expected value.

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\(0.1 + 0.4 + 0.9 + 1.6 = 3.0\).

Question 2

A raffle sells 500 tickets at \(\$2\) each, with one prize of \(\$400\). What is the expected net result for one ticket?

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\(\frac{400}{500} – 2 = 0.80 – 2 = -1.20\), a loss of \(\$1.20\) per ticket.

Expected Value vs. Mean, Median and Mode

The mean of a data set averages numbers you already have. The expected value predicts the average of an outcome that has not happened yet, using probabilities as weights. They coincide when you treat the data as a perfect picture of the chances, but one looks back and the other looks ahead.

Mean, Median and Mode

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