House edge in laying odds after don't pass bet in Craps

Can someone help me understand how laying the odds after placing a don't pass bet in Craps results in a lowered house edge? I'm having trouble wrapping my head around it, since you get paid less than 1:1 on the laid odds bets. Intuitively it seems like you'd be better off sticking with just the don't pass bet, since once a point is set you earn a greater expected return on just that bet rather than that bet combined with laid odds.

I'm sure the math on the site is right, but I'm just trying to understand the intuition.

We've discusssed it before, but the edge is on the first bet, where you have to survive the Come Out Roll. It's true you have a positive expectation after you avoid a seven or eleven on the Come Out, but you don't make it there 2/9s of the time. But that would be like saying you have a positive expectation at 3CP after you looked at your cards and saw 2 Aces. Because there is an edge on the whole thing, by adding Free Odds after you've gotten through the bad part still lowers the edge.

I'll try to answer this without using math...

Quote: bgriffin

....since once a point is set....

That's the key to the problem.


You're more likely to lose than to win on a come-out. I.E. 8 combinations of 7 or 11 vs. 3 combinations of 2 or 3.

But once a point is established, you're more likely to win (seven-out) than to hit the point.

So, once you get over that hurdle, you can add to your expected win by adding the odds.

Therefore, whether you are playing the pass or don't pass, the logical thing to do is to keep the line bet low, and the odds high.

Does that help it make sense?

Quote: Wizard

The odds are simply an additional wager that the point will be rolled before a 7. Because seven is the most likely total roll, you win more than even money when you win. Specifically, if the point is a 6 or 8 the odds pay 6 to 5, if the point is a 5 or 9 the odds pay 3 to 2, and if the point is a 4 or 10 the odds pay 2 to 1. These are exactly statistically fair payouts......

It is worth noting that while taking the odds lowers the combined house edge, the expected loss remains the same. That is because you are applying a lower average house edge to a higher average bet. Readers often write to me, making this point, as an argument against betting the odds. However, if it is you goal to minimize expected loss as much as possible, then don't play at all! If you are going to gamble anyway, and your goal is to minimize the ratio of losses to amount bet, then you should bet as much as you can on the odds.

The info is all on the Wizards main site and explains it better than I could.

EDIT - looks like some people beat me to the answer. Hope this helps anyway.

House edge = Expected Value / Total Average Bet.

Let's look at 3/4/5 odds and don't pass:

2-3: Win one unit Odds 3/36
7-11: Lose one unit Odds 8/36
12: Push Odds 1/36
4/10: Odds bet 6 units, win 4 units 6/9 times, lose 7 units 3/9 times, happens 6/36 times
5/9: Odds bet 6 units, win 5 units 6/10 times, lose 7 units 4/10 times, happens 8/36 times.
6/8: Odds bet 6 units, win 6 units 6/11 times, lost 7 units 5/11 times, happens 10/36 times.

EV: +1 * 3/36 - 1 * 8/36 + 6/36 * (4 * 6/9 - 7 * 3/9) + 8/36 * (5 * 6/10 - 7 * 4/10) + 10/36 * (6 * 6/11 - 7 * 5/11) = -.01364 units

HA = -.01364 / Average Bet = -.01364 / (1 * 12/36 + 7* 24/36)= -.01364 / 5 = -.00273 = -0.273% (which agrees to the Wizard's calculation).

Check this post: was reading wiz's craps faq explain me this

The question you want to ask yourself is, what else could I do with the $xx instead of using it to lay odds? The odds bet has a 0% house advantage. The alternative is to use it for any other bet in the casino, all of which have > 0% house advatange.