Uniform Distribution 4: Expectation Game
Suppose you generate a uniformly random number in the interval (0,1). You have the option to either keep this number or generate a new number once more. Your final payout will be the last number generated. What is your expected payout under a rational strategy?
Answer
Let \( X \) be the first number generated, uniformly distributed on \([0,1]\).
Under a rational strategy:
\[
E = P(X < \tfrac{1}{2}) \times \tfrac{1}{2} + P(X \geq \tfrac{1}{2}) \times \tfrac{3}{4} = \tfrac{1}{2} \times \tfrac{1}{2} + \tfrac{1}{2} \times \tfrac{3}{4} = \tfrac{5}{8}
\]
Therefore, the expected payout is \( \tfrac{5}{8} \).
Notes and comments
Comment 1: This solution follows a rational strategy by conditioning on whether the first number is below or above \( \tfrac{1}{2} \). If it's below, a new number is generated with an expected value of \( \tfrac{1}{2} \). If it's above, the number is kept with an expected value of \( \tfrac{3}{4} \).