Question #5138

A contestant pays $5 to play the follwing carnival game. They roll two 6-sided dice and win an amount of dollars equal to the sum of the two numbers shown.(i.e. if they roll a 3 and a 6 they win $9)Calculate the expected value of the game. If a contestant plays 10 times how much can they expect to win/lose?(You must state if these are winnings or loses.)

Expert's answer

a-number on 1st dice

b- number on second

then

P(a+b=2) =1/36 1+1

P(a+b=3)=2/36 1+2 2+1

P(a+b=4)=3/36 1+3 2+2 3+1

P(a+b=5)=4/36 1+4 2+3 3+2 4+1

P(a+b=6)=5/36 1+5 2+4 3+3 4+2 5+1

P(a+b=7)=6/36 1+6 2+5 3+4 4+3 5+2 6+1

P(a+b=8)=5/36 2+6 3+5 4+4 5+3 6+2

P(a+b=9)=4/36

P(a+b=10)=3/36

P(a+b=11)=2/36 5+6 6+5

P(a+b=12)=1/36 6+6

So we have

Ex=1/36*(2*1+3*2+4*3+5*4+6*5+7*6+8*5+9*4+10*3+11*2+12*1)=7

so he in average wins $7

to make game fair he has to bet $7 instead of $5

b- number on second

then

P(a+b=2) =1/36 1+1

P(a+b=3)=2/36 1+2 2+1

P(a+b=4)=3/36 1+3 2+2 3+1

P(a+b=5)=4/36 1+4 2+3 3+2 4+1

P(a+b=6)=5/36 1+5 2+4 3+3 4+2 5+1

P(a+b=7)=6/36 1+6 2+5 3+4 4+3 5+2 6+1

P(a+b=8)=5/36 2+6 3+5 4+4 5+3 6+2

P(a+b=9)=4/36

P(a+b=10)=3/36

P(a+b=11)=2/36 5+6 6+5

P(a+b=12)=1/36 6+6

So we have

Ex=1/36*(2*1+3*2+4*3+5*4+6*5+7*6+8*5+9*4+10*3+11*2+12*1)=7

so he in average wins $7

to make game fair he has to bet $7 instead of $5

## Comments

## Leave a comment