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I propose to you a game. You roll 2 dice. If the sum of the numbers...

I propose to you a game. You roll 2 dice. If the sum of the numbers showing is either 6, or 7, or 8, I win. If it is 2, 3, 4, 5, 9, 10, 11, 12, you win. Since you have lots more possible winning combinations than I do, the rules are that you pay me $2.00 when I win and I pay you $1.00 when you win. If we play this game 30 times, how much do you think you will win or lose? Explain why. You may wish to draw a histogram or create a table as an illustration in addition to showing your calculations.

math   Statistics-And-Probability   
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Answer #1

P(the sum of the numbers in a roll of 2 dice is either 6,7 or 8)

= 5/36 + 6/36 + 5/36 = 4/9

P(the sum is either 2,3,4,5,9,10,11,12) = 1 - 4/9 = 5/9

Expected winning on playing this game once

= $1*5/9 + (-$2)*4/9

= -$1/3

Thus, expected winning on playing this game 30 times

= -$1/3 * 30 = -$10

Thus, you would lose an expected $10 on playing this game 30 times

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