Asked by Alex
Suppose someone gives you 8 to 2 odds that you can not roll two even numbers with the roll of two fair dice. This means you win $8 if you succeed and you lose $2 if you fail. What is the expected value of this game to you? Should you expect to win or lose the expected value in the first game? What can you expect if you play 200 times? Explain.
Answers
Answered by
MathMate
Probability of rolling two even numbers
= (1/2)(1/2)=1/4
Expected value of the game, E(x) is the sum of the following:
outcome X, probability P(X), X*P(X)
2 evens=+$8, 1/4, +$2
at least one odd=-$2, 3/4, -$3/2
The sum of X*P(X) = $0.50
The expected value is thus $0.50 for each game.
Since odds of losing is 3 to 1, it would be reasonable to expect to lose the first game.
On the other hand, if we play 200 times, the law of large numbers would take over and there should be a winning of around $200*0.5=$100.
= (1/2)(1/2)=1/4
Expected value of the game, E(x) is the sum of the following:
outcome X, probability P(X), X*P(X)
2 evens=+$8, 1/4, +$2
at least one odd=-$2, 3/4, -$3/2
The sum of X*P(X) = $0.50
The expected value is thus $0.50 for each game.
Since odds of losing is 3 to 1, it would be reasonable to expect to lose the first game.
On the other hand, if we play 200 times, the law of large numbers would take over and there should be a winning of around $200*0.5=$100.
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