document.write( "Question 806336: If a player rolls doubles when she uses two dice, she wins $5. If the person rolls a 3 or 12, she wins $15. The cost to play the game is $3. Find the expectation of the game.\r
\n" ); document.write( "\n" ); document.write( "2(5)+3+12(15)*3= \n" ); document.write( "

Algebra.Com's Answer #485802 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "Refer to the below table of dice results for 2 six-sided dice:\r
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Sum	Ways	# of Ways
2	1,1	1
3	1,2; 2,1	2
4	1,3; 2,2; 3,1	3
5	1,4; 2,3; 3,2; 4,1	4
6	1,5; 2,4; 3,3; 4,2; 5,1	5
7	1,6; 2,5; 3,4; 4,3; 5,2; 6,1	6
8	2,6; 3,5; 4,4; 5,3; 6,2	5
9	3,6; 4,5; 5,4; 6,3	4
10	4,6; 5,5; 6,4	3
11	5,6; 6,5	2
12	6,6	1

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\n" ); document.write( "\n" ); document.write( "Note that there are 36 different possible results, 6 of which are doubles, 2 of which are 3, and one of which is 12. However, the 12 result is also doubles. Because you don't specify what happens when a 12 is rolled (score 15 for the 12 which overrides the doubles or score 5 for the doubles which overrides the 12, or score 20 for meeting both criteria), I'm going to go with a literal interpretation of the given rules: The player gets 5 for rolling doubles AND 15 for rolling 12.\r
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\n" ); document.write( "\n" ); document.write( "So the probability of rolling doubles OTHER than 2 sixes = 12 is

, the probability of rolling 12 is

, the probability of rolling 3 is

, and the probability of rolling anything else, which we must presume represents a loss of the initial $3 wager, is

(28 ways to lose out of 36 outcomes).\r
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\n" ); document.write( "\n" ); document.write( "And then the expected payout is:\r
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\n" ); document.write( "\n" ); document.write( "You can do your own arithmetic, but in the long run you lose a quarter every time you play.\r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "Egw to Beta kai to Sigma
\n" ); document.write( "My calculator said it, I believe it, that settles it
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$\"The$

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