The expectation of this game is -$7.33
The probability of an event is defined as the possibility of an event occurring against sample space.
Let us tackle the problem.
If a gambler rolls two dice and gets the sum of 7 , there are 6 ways to get the sum of 7 , i.e : ( 1,6 ) , ( 2,5 ) , ( 3,4 ) , ( 4,3 ) , ( 5,2 ) , ( 6,1 ) from total 36 sample spaces, then the probability is :
[tex]P(7) = \frac{6}{36}[/tex]
[tex]\large {\boxed {P(7) = \frac{1}{6} } }[/tex]
If a gambler rolls two dice and gets the sum of 4 , there are 3 ways to get the sum of 4 , i.e : ( 1,3 ) , ( 2,2 ) , ( 3,1 ) from total 36 sample spaces, then the probability is :
[tex]P(4) = \frac{3}{36}[/tex]
[tex]\large {\boxed {P(4) = \frac{1}{12}} }[/tex]
Because the gambler wins $20 when he gets the sum of 7 , and wins $40 when he gets a sum of 4 , then :
[tex]Expectation ~ = ~ Winning ~ - ~ Cost[/tex]
[tex]Expectation ~ = ~ ( \$20 \times \frac{1}{6} + \$40 \times \frac{1}{12}) ~ - ~ \$14[/tex]
[tex]Expectation ~ = ~ ( \$20 \div 3 ) - \$14[/tex]
[tex]Expectation ~ = ~ ( \$6.67 ) - \$14[/tex]
[tex]\large {\boxed {Expectation ~ = ~ -\$7.33} }[/tex]
Grade: High School
Subject: Mathematics
Chapter: Probability
Keywords: Probability , Sample , Space , Six , Dice , Die
