Suppose that the rules were changed so that the don’t pass bet would be the exact opposite of the pass line bet.
What would be the probability to win a don’t pass bet if rolling a 12 on the come out roll were a win instead of a push ?
| Let p | = | P( win ) | |
| Let A | = | the event of rolling a 2, 3, or 12 on the come out | |
| p | = | P( A ) + 2 * [ P( 4 ) * P( 7 before 4 ) + P( 5 ) * P( 7 before 5 ) + P( 6 ) * P( 7 before 6 ) ] | |
| = | ( 4 / 36 ) + 2 [ ( 3 / 36 ) * ( 6 / 9 ) + ( 4 / 36 ) * ( 6 / 10 ) + ( 5 / 36 ) * ( 6 / 11 ) ] | ||
| = | ( 1 / 9 ) + 2 * ( 1 / 36 ) * [ 2 + ( 24 / 10 ) + ( 30 / 11 ) ] | ||
| = | 251 / 495 = 0.5070707.... | ||
Since p > 0.5, this game would be biased in the player's favor.