A dice game is played with two distinct 12 sided dice. It costs $3 to roll the p

Question

A dice game is played with two distinct 12 sided dice. It costs $3 to roll the pair of dice one time. The payout scheme is as follows 1. Sum of 13 pays $10 Sum of 11 or 15 pays $6 Sum of 7, 9, 17, or 19 pays $3 Any other roll doesn't pay. What is the expected gain/loss after playing the game one time? A "fair" game is one in which the expected gain/loss after playing once is $0. Is this game fair? If it is not fair, who does the game favor, the player or the "house?" adjust ONE of the payouts so that the game is fair. 2. A dice game is played with two distinct 10 sided dice. It costs $10 to roll the pair of dice one time. A player "wins" when the product of the dice is odd and "loses" when the product is even. Provide payouts for each outcome so that the game is fair. Also provide payouts so that a player would expect to gain $5 for playing the game one time.

Explanation / Answer

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1.

Sum of 13 is got with probability ?

{12,1}*2
{11,2}*2
{10,3}*2
{9,4}*2
{8,5}*2
{7,6}*2 = 12 ways

{11 or 15}
11 - {10,1}*2
{9,2}*2
{8,3}*2
{7,4}*2
{6,5}*2
=10 ways

15 - {12,3}*2
{11,4}*2
{10,5}*2
{9,6}*2
{8,7}*2
=10 ways

Sum of 7
{6,1}*2
{5,2}*2
{4,3}*2
=6 ways

Sum of 9
{6,3}*2
{7,2}*2
{8,1}*2
{4,5}*2
=8 ways

Sum of 17
{12,5}*2
{11,6}*2
{10,7}*2
{9,8}*2 = 8 ways

Sum of 19
{12,7}*2
{11,8}*2
{10,9}*2
= 6 ways

= 6+8+8+6 = 28 ways

Total kind of dice plays =12*12 = 144

Expected value = (12/144)*10 + (10+10)*6/144 + (28*3/144) - $3
= -$0.75

So, expected value is a loss of 75 cents or a loss of $.75

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