What is the house percentage if the house pays the player 1½ and 2 dollars per dollar staked for a main point win and a chance point win, respectively?

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In the popular English game of Hazard, a player must first determine which of the five numbers from 5,…, 9 will be the “main” point. The player does this by rolling two dice until such time as the point sum equals one of these five numbers. The player then rolls again. He/she wins if the point sum of this roll corresponds with the “main” point as follows: main 5 corresponds with a point sum of 5, main 6 corresponds with a point sum of 6 or 7, main 7 corresponds with sum 7 or 11, main 8 corresponds with sum 8 or 12, and main 9 corresponds with sum 9. The player loses if, having taken on a main point of 5 or 9, he/she then rolls a sum of 11 or 12, or by rolling a sum of 11 against a main of 6 or 8, or by rolling a sum of 12 against a main of 7. In every other situation the sum thrown becomes the player’s “chance” point. From here on, the player rolls two dice until either the “chance” point (player wins) or the “main” point (player loses) reappears. Verify that the probability of the player winning is equal to 0.5228, where the main and the chance points contribute 0.1910 and 0.3318, respectively, to the probability of winning. What is the house percentage if the house pays the player 1½ and 2 dollars per dollar staked for a main point win and a chance point win, respectively?

 

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