Doubled up
A gambler invites you to play a game. He has two dice with him: a fair 6-sided die with faces numbered 1, 2, 3, 4, 5, and 6, and a fair 8-sided die with faces numbered 3, 4, 5, 6, 7, 8, 9 and 10.
You roll both dice simultaneously. If you get a double, he pays you the sum of the two dice in dollars (i.e., if both dice show a 4 you receive 4 + 4 = 8 dollars.) If you don't get a double, you win nothing.
The price to play is 1 dollar. What is your mathematical expectation in cents? (100 cents = 1 dollar.)
The answer is -25.
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1 solution
Moderator note:
Clear simple explanation :)
There are 48 possible rolls, all of which have an equal chance. Let us say you play 48 times, thus paying 48 dollars, and every possibility occurs once. You will win on 4 occasions (3-3, 4-4, 5-5, and 6-6) and win a total of 3 + 3 + 4 + 4 + 5 + 5 + 6 + 6 = 36 dollars.
So your net loss in 48 plays will be 12 dollars. So per play, your net is -12/48 dollars = -1/4 of a dollar = -1/4 * 100 = -25 cents.