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How should you interpret the expected value of -$0.50 per ticket in the context of a single play?

The expected value of -$0.50 should be interpreted as a long-run average loss per ticket, not as a guaranteed result for a single play. In any individual game, the player will either win $499 or lose $1; the -$0.50 figure represents the mean outcome if the game were played many times.

Conditions

  • The expected value has been calculated as -$0.50.
  • The context involves understanding the meaning of expectation in probability.

Reasoning, step by step

  1. Recall the definition of expected value as a probability-weighted average.
  2. Identify the possible outcomes for a single play: win $499 or lose $1.
  3. Note that neither outcome is -$0.50.
  4. Understand that the expected value is the center of mass of the probability distribution.
  5. Conclude that -$0.50 predicts the average result over a large number of trials, not the specific result of one trial.

Example

The script states: 'The expectation is 499/1000−999/1000=−0.50499/1000-999/1000=-0.50 dollars per ticket. It is a long-run mean, not a guaranteed result for one play.'

Common misconceptions

  • Believing that a single play will result in a loss of exactly $0.50.
  • Thinking that the expected value is a possible outcome of the random variable.
  • Confusing the mean of the distribution with the median or mode.

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