I have a YouTube channel on throwing and playing craps. I like to play the hard way bets knowing there is a large house advantage when playing them. I had a comment on my channel that playing those bets is giving the house money. My strategy is treating it like a bonus when I have money on the 6 and/or 8 and money on the corresponding hard way. What I am trying to determine is what happens to the house advantage when you play both bets together.
Is it as simple as taking the expected outcome (house advantage) of throwing a 6 multiplied by the expected outcome of throwing a hard 6?
I am trying to calculate the house advantage with multiple bets, a 6 dollar place bet on the 6 and a 1 dollar bet on the hard 6 and understand the following to be true:
The True Odds of throwing a Hard 6 is 10:1 and pays 9:1 for a house advantage of 9.09%
(1/11)×9 + (10/11)×(-1) = -1/11 ≈ -9.091%
The True Odds of throwing a Place Bet 6 is 6:5 and pays 7:6 for a house advantage of 1.52%
[(5/11)×7 + (6/11)×(-6)]/6 = (-1/11)/6 = -1/66 ≈ -1.515%
I am trying to calculate the house advantage and I am coming up with the following formula:
((5/11)*7)+((1/11)*9) + ((6/11)*-6) + ((10/11)*-1)/6 = 0.575758
Am I calculating this correctly? The expected outcome of winning on any 6 is (5/11) x $7 The expected outcome of winning on a hard 6 is (1/11) x $9 The expected outcome of losing to a 7 (6/11) * -$6 The expected outcome of losing the hard way to a 7 or easy 6 (10/11) * -$1. I am getting a house advantage of 0.575758. Is this anywhere close?
UPDATE:
Is this as simple as multiplying the two expected outcomes together?
((1/11)×9 + (10/11)×(-1)) x ([(5/11)×7 + (6/11)×(-6)]/6 = (-1/11)/6) = 0.1377%
Does that look correct?
I got the calculations from WizardOfOdds.com
Best Answer
If you bet $6$ on a bet that is losing $1.52\%$ and $1$ on a bet that is losing $9.09\%$ you are losing $6 \cdot 1.52\% + 1\cdot 9.09\%=0.0912+0.0909=0.1821$ on a bet of $7$. The percentage loss is then $\frac {0.1821}7\approx 2.6\%$