My 2-Bet Strategy w/Odds

A Strategy that takes the center-bets out of the picture and concentrates on the Come-Out roll and a few points. The object is no "Pass/Don't Pass" betting.
For Example here lets start with $200 placing $2 bets.

1.) When a bet wins, it comes down with the win. Bets are not repeated until resolved.
2.) The two bets are Pass-Line and Field. The Field bet MUST pay 3 to 1 when a 12 is rolled, and 2 to 1 for a 2.
3.) Strategy by Come-Out Roll
...a.) 2, 11, or 12: Both bets resolve, replace bets (Win $2, $4, or $4)
...b.) 3 or 7: Both bets resolve as a PUSH, replace bets
...c.) 4, 9, or 10: The Field wins, take bet down, ($4) and put $2 as Free Odds behind the Pass-Line.
......The 4 or 10 Win: The Pass Line resolves as a $6 win for a total of $8 gained (You get the winning Field bet as Odds returned).
......The 9 Winner: The Pass-Line resolves as a $5 win for a total of $7 gained (as above).
......Lose the 4-9-10: The Pass-Line loses. Since you placed winnings from the Field as the Odds, the loss is $2.
...d.) 5, 6, or 8: No Odds, the Pass Line bet remains.
......5-6-8 Winner: The Pass-Line wins, covering the loss upon the Field bet for a PUSH. Replace the bets.
......5-6-8 Loser: Both bets resolve as a Loss of $4. Replace the bets.

I guess it's not a bad little hedge, if you hate losing on the come-out roll. Not sure why the field bet must pay three to one - you still get a net win on 2 or 12. The edge is steeper, naturally, but if you really care so much about edge, why not just bet line and odds?

Also, I'm inclined to recommend you take double odds in the case of a 4, 9, or 10.

I kinda figured it as not losing the Field bet for a net loss of 1 bet, as opposed to a loss of 2 bets if the 4-9-10 loses.
The Pass-Line - 1.41% The Triple-Odds Field bet - 2.78%... if 2&12 pay 2 to 1 Field bet is - 5.56%.
I have not done the math, but it is a nice hedge using 2 simple bets.

My craps strategy (and it's free for all of you to use!):

(1) Make minimum pass line bet.
(2) Take max odds.
(3) Make minimum come bet.
(4) Take max odds.
(5) Lose badly.

Okay, so it needs some tweaking on the back end...

Quote: 98Clubs

I kinda figured it as not losing the Field bet for a net loss of 1 bet, as opposed to a loss of 2 bets if the 4-9-10 loses.
The Pass-Line - 1.41% The Triple-Odds Field bet - 2.78%... if 2&12 pay 2 to 1 Field bet is - 5.56%.
I have not done the math, but it is a nice hedge using 2 simple bets.

But you're losing two bets on the other points anyway, so why not make your wins a little bigger?

The 2 to 1 field bet does give a steeper edge, it's true, but... so what? You're sacrificing a fair amount of edge either way, and you're usually not playing such a table by choice.

As for the math, I can tell you very quickly that edge per bet resolved is 2.10%, or 3.48% if the field bet pays 2 to 1. Since they're put up together, and odds bets are fair, it's just the mean.

Doing it your way, I'm getting:
5.56% - Win four units.
4.44% - Win 3.5 units.
8.33% - Win two units.
2.78% - Win one unit.
39.29% - Push
17.78% - Lose one unit.
21.82% - Lose both units.

For a σ of 1.69 units, or 0.84 total bets (i.e., $3.37). On every 990 come-out rolls, you can expect to win on 209, and lose on 392.

If you instead take odds with your entire field bet, you'll get a σ of 2.25 units - the win/loss percentages will only change in that the top two will be bigger wins, and the two losing scenarios will merge to 39.60%. Of course, all this assumes a 3:1 field bet.

Quote: teddys

My craps strategy (and it's free for all of you to use!):

(1) Make minimum pass line bet.
(2) Take max odds.
(3) Make minimum come bet.
(4) Take max odds.
(5) Lose badly.

Okay, so it needs some tweaking on the back end...

My strategy:

(1) Make minimum don't pass bet.
(2) Lay max odds.
(3) Make minimum don't come bet.
(4) Lay max odds.
(5) Lose badly.

If you're focused enough on edge to play the darkside exclusively, you shouldn't be gambling against the house.

deleted

I think the number of bets avaialble in craps lead people to think they can find some bet combination that would insure a win, no matter what the dice do. If that were the case, though, there'd be no craps tables anywhere.

But I wonder if a team could find a combination, a very complicated one, involving three or four or more players each making different bets, which would insure a small but consistent win. I tend to doubt it. My "gut feeling" is that at best there may be a combination that insures only a small, consistent loss.

Every bet has a negative (or zero) expected value per roll. Expected values always add. A few times, you can reduce the effective edge by manipulating when a bet resolves, but you can never get it below zero when every roll has an edge.

And don't say "Parrando's Paradox." Parrando's Paradox is no paradox at all, since one of the bets is only -EV if the last digit of one's bankroll is random; all it shows is that it isn't.

Quote: Nareed

I think the number of bets avaialble in craps lead people to think they can find some bet combination that would insure a win, no matter what the dice do. If that were the case, though, there'd be no craps tables anywhere.

But I wonder if a team could find a combination, a very complicated one, involving three or four or more players each making different bets, which would insure a small but consistent win. I tend to doubt it. My "gut feeling" is that at best there may be a combination that insures only a small, consistent loss.

Your gut feeling is correct.