Probability as Expectation Math Example 2

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Example 2

medium
A game has three outcomes: win \10 (prob 0.2), break even \0 (prob 0.5), lose \$5 (prob 0.3). Calculate the expected value and interpret what it means for 1000 games.

Solution

  1. 1
    Expected value per game: E = 10(0.2) + 0(0.5) + (-5)(0.3) = 2 + 0 - 1.5 = \0.50$
  2. 2
    The game has positive expected value: on average, gain \$0.50 per game
  3. 3
    Over 1000 games: expected total gain = 1000 \times 0.50 = \500$
  4. 4
    Interpretation: in the long run, play this game โ€” expected profit is \$500 over 1000 games

Answer

E = \0.50 per game; expected profit over 1000 games = \500.
Expected value is the probability-weighted average of all outcomes. Positive EV means you profit in the long run; negative EV means you lose. Expected value is additive across independent games, making long-run prediction straightforward.

About Probability as Expectation

Probability can be interpreted as the long-run relative frequency of an event over infinitely many identical trials of a random experiment.

Learn more about Probability as Expectation โ†’

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