the Meaning of Craps HE Taking the Odds

Kind of bad to have to ask this question, but all of a sudden it is bugging me.

Playing right, when the HE is figured for 'Craps with Free Odds' at the varying allowed levels, as the WoO has tables for, does what is shown take into consideration the times that the bet is resolved on the Line? or if you see the table that has 5X4X3X at .273% HE, do you need to figure that 1/3 of the time the HE is 1.41% and thus the HE for your average bet is roughly averaged at .70% (or something similar) because you count the pass line bets into the mix?

I'm pretty sure that is already taken into account.

Yes, I agree that the pass line with odds figures all the possible outcomes to arrive at a certain HE.
remember in the short run where all craps players are, variance dominates the HE, especially taking odds.

thanks for the input, I thought so too, but could not find a confirmation and doing the math myself would kick my ass.

For 3/4/5 odds, the HE is .374%:

((8/36 - 4/36 + 6/36(7 x 1/3 - 4 x 2/3) + 8/36 (7 x 4/10 - 5 x 6/10) + 10/36 (7 x 5/11 - 6 x 6/11)) / (1 x 12/36 + 4 x 6/36 + 5 x 8/36 + 6 x 10/36) = -.01414 / 3.77777778 = -.00374.

The HE is 0.374% but the EV is still 1.414%. (That is, on a $10 bet you will lose, on average 14.14 cents but the average bet is $37.78).

Quote: boymimbo

For 3/4/5 odds, the HE is .374%

thanks for the correction, I was gazing at the wrong table when composing the original *sigh*.

Gazing at all that number crunching, I can't discern anything that would tell me it answers the question. I'm thinking it is in the denominator , but 1 times something plus 4 times something plus 5 times something plus 6 times something doesnt "click" for me.

One way to understand the HE calculation (for simple cases where the max odds are the same for each point) is to divide the house edge on the passline by the average wager. The house edge on the passline bet is 1.41% of your pass line wager. What is the average wager? Because there are 12 ways out of the 36 rolls for the decision to be determined immediately, 1/3 of the time your wager is just the pass line wager. The remainder of the time the wager is larger by the odds bet. Therefore on average you wager (1 + 2/3*oddsMultiple)*passBet. For example, if you bet $5 on the pass line and always take double odds, your average wager is (1+2/3*2)*5 ~ $11.6 dollars. So the HE = .0141*$5/($11.6)~0.61. Since the wager cancels out, you can just write this as: 1.41%/(1+2/3*oddsMultiple).

Quote: DrEntropy

One way to understand the HE calculation (for simple cases where the max odds are the same for each point) is to divide the house edge on the passline by the average wager. The house edge on the passline bet is 1.41% of your pass line wager. What is the average wager? Because there are 12 ways out of the 36 rolls for the decision to be determined immediately, 1/3 of the time your wager is just the pass line wager. The remainder of the time the wager is larger by the odds bet. Therefore on average you wager (1 + 2/3*oddsMultiple)*passBet. For example, if you bet $5 on the pass line and always take double odds, your average wager is (1+2/3*2)*5 ~ $11.6 dollars. So the HE = .0141*$5/($11.6)~0.61. Since the wager cancels out, you can just write this as: 1.41%/(1+2/3*oddsMultiple).

very good, certainly anybody can just plug in those numbers and it does answer definitively, thanks.