I havent seen this strategy anywhere but I am sure its out there and studied. I have been forming a way to try and play craps low variance and get comps along the way. It shows to do OK but I am never going to win a lot and I risk losing a little.

This is what I have been doing:

$350 DONT
Assume the point is 6
$300 PLACE 6 and $50 hard 6
$350 DONT COME
Assume 9 comes
$350 PLACE 9
$350 DONT COME
Assume 4
$300 BUY 4 and $25 hard 4


Now if we just stop here, I can only push on the 6 but freeroll a $50 hard 5 that would pay $450. I can only push the 9 OR I can win $140 if a 9 comes. I can WIN $10 on the 4 if a 7 comes (total) OR I can win $600 on a 4 OR $600+175 on a hard 4.

Obviously I lose on 7 comes outs and I lose $350 when a 7 comes because I keep firing the DONT COME. I lose on 11's but win on 2 and 3 so I ignore the 11 losing. I win a DONT flat bet on 3/36 and lose on 8/36. I just hope to make up for that by winning on the place bets. You make the most when you can hit hardways or you can just save that money and make an extra $25-50 per number by not playing (I have tried both, we know hardways are a bad bet though. But I am basically freerolling them so I play).

Winning strategy? No. But it does give you the lowest variance if you accept you are losing $350 or whatever your DONT bet is on a 7.

I thought it would be interesting for someone to walk up and play a table max $10,000 DONT bet. Avoid 7/11 and point travels to a 9. Now place that for $10,000 and you can win $4,000 with no risk. You can even hedge and place it for $9,000 and win on each side.

Can someone tell me where this has be studied and what the outcome is? I want to see how bad it actually is.




What usually happens is something like this:

$350 DONT
8
$300 8 and $50 hard 8
$350 DONT
4
$300 4 and $25 hard 4 (have to pay $15 to buy)
$350 DONT
5
$350 5
$350 DONT
5
5 now pays $490
$350 DONT
10
$350 10 and $25 hard 10
$350 DONT
4
4 pays $600
$350 DONT
7
Lose $350 and break even on rest of bets.

Lose $350 and Win $740 on that roll. NET +$390


Now if we would have played 3-4-5 odds and $50 pass line bets and it came 8-4-5-5-10-4-7 with come bets we would have won $750 and lost $1100 (if my math is right on this). NET -$350

"As long as you avoid the 7/11"

"No Risk"

Just bet $10k on the hard 8, as long as you avoid anyone rolling a 7 or easy 8 there's no risk

You must not understand what I am saying with such a simple response. A 7 is OK bc we are hedged against it. An 11 isnt good but we make up for that with a 3 plus we make money on a 2. 7's always come and thats ok. This way we work with it and expect it to come and it doesnt hurt us except for the last DONT bet out there.

And you are quoting me from something I didnt say and out of context.

I'll break down the stats later if I have time. I employ a somewhat similar method myself. The cost is higher than the direct comps I receive (duh), however the cost is actually lower than the overall comp value I receive when factoring in the backend comps, so it's a minor AP play for me.

Thanks. I know for sure its not a winning play but its an AP I have to assume. You do have the upside of a bunch of 4-10 or 5-9 coming and helping out. All that upside with such a small downside seems good. I just dont know how to calculate it.

It's not an AP, trust me. Your expected loss is worse than mine, and your casino's comp system isn't as good as mine and doesn't have the same exploitative value. I care not to talk about it more out here. Don't be fooled by the low variance, the math and EV don't lie.

what happens to the dont come bets on a roll of 12?

push

your strategy is worthless i advise u stop gambling and take up knitting

How do you get a hard 5 ?

but what happens to your other bets?

Nothing. It's only a 7 that forces them to come down if they're off.

Don't bother with this strategy. You'll get picked off on winner 7's on the come out and you won't have a chance.


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You also aren't figuring the vig on the 4/10 resulting in a lower hedge.


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