Question

A pair of fair dice is rolled once. Suppose that you lose $10 if the dice sum to 4 and win $11 if the dice sum to 3 or 2. How much should you win or lose if any other number turns up in order for the game to be fair?

Answer 1

you need the probabilities of sum being 2,3,or4

there are 36 total possible combinations
only 1 gives sum of 2 ( 1,1) --> probability = 1/36

two give sum of 3 (1,2) or (2,1) --> probability = 2/36

3 give sum of 4 (1,3) or (3,1) or (2,2) --> probability = 3/36

probability of something else = (36-3-2-1)/36 = 30/36

multiply probabilities by gain or loss, then set equal to 0 to make game fair
\[(\frac{1}{36}+\frac{2}{36})*11 + (\frac{3}{36})*(-10) + (\frac{30}{36})*X = 0\]
solve for X

Answer 2

There are 3/36 ways to roll a 4, so the expected value is -$10*3/36
There is 1/36 ways to roll a 2, so the expected value is $11*1/36
There are 2/36 ways to roll a 2, so the expected value is $11*2/36
So with the defined outcomes, the expected value is ($11-$10)*3/36 = $1/12
There are 6 outcomes with defined payouts, leaving 36-6=30 others. $x*30/36 = $1/12
Solve for x to get your win or loss for the other numbers.

Answer 3

it appears great minds think alike :-)

Answer 4

ahh yes :)

Answer 5

and we got the same answer, too! :-)

Answer 6

side note: games in real life tend not to be fair :(

Answer 7

and the scary looking dude urging you to hurry up and roll the dice probably won't let you break out pencil and paper to check the odds :-)

Answer 8

agreed

Answer 9

hi guys would it be a loss then?

Answer 10

and what would be the answer be the the nearest cents?
it cant be in fractions

Answer 11

\[x*\frac{30}{36} = \frac{$1}{12}\]\[x=\frac{$1}{12}*\frac{36}{30}= \]

Answer 12

And yes, you would pay \(x\) if you rolled any number other than 2,3 or 4.

Answer 13

what is the value of x than?