Prob and StatsRandom VariablesS-MD.B.5

Decisions with Expected Value

Compare games, insurance and strategies by expected payoff.

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Insurance companies, game designers and anyone deciding whether to buy a warranty are really comparing expected values. The expected value tells you what happens on average over many repetitions, which is exactly what a business that sells thousands of policies cares about.

Key idea

To compare choices that involve chance, compute the expected value of each, counting costs as negative and payoffs as positive. The choice with the higher expected value does best in the long run. A fair game has expected net gain 0. Expected value says nothing about risk, which the standard deviation measures.

Why it works

Over many repetitions, the average result settles near the expected value (the law of large numbers). An insurance company with 100,000 policies will pay out very close to the expected amount, so it can set prices above that and earn a steady profit. One customer is in a different position: a single rare loss could be ruinous, so paying a little more than the expected loss to remove that risk can be a sensible choice.
One policy: the company pays a 5,000-dollar claim with probability 0.02.

Worked example: Pricing insurance

A company sells a policy for \$150. With probability $0.02$ it pays a \$5,000 claim. Find its expected profit per policy.
  1. Expected payout: $0.02 \times 5{,}000 = 100$ dollars.
  2. Expected profit: $150 - 100 = 50$ dollars per policy.

Answer: \$50 per policy, on average.

Worked example: Two games

Both games cost the same. Game A pays \$200 with probability $\frac{1}{20}$. Game B pays \$20 with probability $\frac{1}{4}$. Which is better in the long run?
  1. A: $\frac{1}{20} \times 200 = 10$ dollars on average.
  2. B: $\frac{1}{4} \times 20 = 5$ dollars on average.
  3. A pays twice as much on average, though it wins less often.

Answer: Game A.

Worked example: A warranty

A \$80 warranty covers a repair that costs \$400 and is needed with probability $0.15$. Is it worth it on average?
  1. Expected repair cost without it: $0.15 \times 400 = 60$ dollars.
  2. The warranty costs \$80, more than \$60.
  3. On average it costs \$20 more than it saves. (Someone who could not afford a \$400 surprise might still choose it.)

Answer: Not on average.

Common mistake

Comparing only the prizes or only the chances of winning. A big prize with a tiny chance can be worth less than a small, likely one. Check: multiply each payoff by its probability, subtract the costs, and compare the totals.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Builds on

Leads to

This is one of the top rungs of Probability and Statistics.

Sources and further reading

  • OpenStax, Introductory Statistics 2e, section 4.2 Mean or Expected Value and Standard Deviation
  • Common Core State Standards for Mathematics, S-MD.B.5 (Weigh the possible outcomes of a decision by assigning probabilities and finding expected values)
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