The One Trick Pony Play

[heavy]

A strategy I've been playing a bit with of late focuses on one number of random rollers. Essentially, I might start out with something like $66 inside. On the first hit on any number I press it or power press it. The five or nine get pressed to $35. The six or eight move up to $42. At that point I regress or take down all of my bets except on that one I have pressed up. Should the shooter toss one more hit on the "one trick pony" we get paid $50 for $1 and (assuming you took down your other action) you have $25 net profit in the rack. Once in a profit position I'll take the next hit, then full press the one after that. My six or eight would go to $90 with $1 change to me. The five or nine would go to $75 and you'd lock up $9. To date I've gotten my six or eight up to $420 on this play several times, but I've never gotten the $500 for $10 pay day. This has me thinking in terms of slowing down the press once I hit the $180 level = or to simply regress back down to a $60 or $90 bet.

Anyone else play a one-number strategy on other shooters? Inquiring minds and all that.

[wild child]

H

The sure thing lock up $ has pure more sure value than any WOULD-A ,COULD-A, SHOULD-A
plus the obligatory "I knew it ,I just knew it "after statement


Just me saying

W C

[gargoil]

How about wait for a box number after the point is made (avoid the PSO) then place that number pressed (35 on the 5/9 or 42 on the 6/8). One hit then down. $50.00 in the rack
Or keep the bet out there for another hit then regress to 52/54 across.

Another play would be wait for the point then make a $10.00 bet on the "COME". A PSO will pay you back and a box number will give you the same results with odds behind your bet.
The downside is you can remove your odds anytime but not the $10.00 bet.

[freak]

Over time I have drifted towards a "hero" number strategy. I used to want to have a lot of box numbers covered, especially if it seemed a hand was developing. But I realized over time that when the long roll comes it usually hits many different box numbers. If I only have one number up it usually gets hit on a good roll. But if I have many numbers up they all get taken on a short roll. One of our most profitable rolls in Vegas actually started on the don't. When the shooter hit his point we put $10 on the passline hoping for double-play. He rolled nine (not seven as we'd hoped) and so we had a nine and were behind the four. He went on to roll a 37 hand, but the only box number hitting in a streak was the six. We made the six our "hero" and kept it up even though it didn't hit in the last 10 rolls or so. He also made the six as the point once and the eight as the point twice. When he finally sevened out the eight was the point again. We only lost the point and our six to the seven. We lost the no-4 to a 4 mid hand. He made seven passes in all. We missed the first PL but got paid on the other six PL winners. It's tempting to get in on every number on a roll like that but when the occasional trash would come I'd get reminded not to over-invest. If I put up two more numbers I need two hits just to be even. The one number he seemed to be hitting most was the point, so we stayed on that. I'm learning to keep my need for a win on every roll in check. Take the push as a win and be satisfied with getting paid on my one number once every 5 rolls or so.

[Drab]

Last weekend, I was placing the sister number at my max ($25/30) for a roll or two and down--more out of desperation than anything else--while I waited for my turn to roll.

[heavy]

Yep. I've seen the sister number play be very profitable - especially when played for one hit and down.

[dork]

A strategy I've been playing a bit with of late focuses on one number of random rollers. Essentially, I might start out with something like $66 inside. On the first hit on any number I press it or power press it. The five or nine get pressed to $35. The six or eight move up to $42. At that point I regress or take down all of my bets except on that one I have pressed up. Should the shooter toss one more hit on the "one trick pony" we get paid $50 for $1 and (assuming you took down your other action) you have $25 net profit in the rack. Once in a profit position I'll take the next hit, then full press the one after that. My six or eight would go to $90 with $1 change to me. The five or nine would go to $75 and you'd lock up $9. To date I've gotten my six or eight up to $420 on this play several times, but I've never gotten the $500 for $10 pay day. This has me thinking in terms of slowing down the press once I hit the $180 level = or to simply regress back down to a $60 or $90 bet.

Anyone else play a one-number strategy on other shooters? Inquiring minds and all that.

I kinda have a one-number strategy that I use... I'm playing on a $5 table (the only table) at a quiet casino nowadays. When the table seems 'good', on RR's I'll go $32/34 inside when the point's an inside number. When I've made a $32/34 inside bet, the underlying strategy is to let the dice chose a "trick pony" to parlay-parlay-parlay that number and take profits on the 3x winner.

Let's just say the point is '6'...I'll have a $5 PL w/$20 odds and $32 inside. (When I'm the shooter, I'll regress to this system from a $5 PL w/$30 odds and $90 inside bet that's paid me twice in any manner.) I'll parlay the very first box number to $24, reducing the others to single units, with an eye on parlaying that $24 bet to either $60 (for the 5/9) or $54 (for the 8). When it hits again, I'll regress to a single unit, clearing either $139 (5/9) or $111 (8) .

If the point (6) is made before I can collect on a 3x parlay--let's say my place-8 sits at $54 from previous play--and a new point (arbitrarily we'll assign a '9') is made. I'll regress the '8' to a $6 place bet and add $2 to the regressed "profit" $48, and convert it into at $50 odds bet on the new point (9). Now all my bets are "casino money" (not really, but close enough). In this situation, all the place bets will sit at minimums, so the first new place number to hit is parlayed to a 3rd winner. Then it's reduced from it's 3x occurance of $54 or $60 to one unit. When a hand's bet is 'casino money', AND I've made one 3x profit, I'll 'take a chance' and attempt one more parlay (to the 4x) on the trick pony place bet hoping to regress then to one unit. Some other place number will hopefully have at least one "up one unit" bet at that point, and that will become the new 'trick pony' number until the 5th winner; then I'd saddle up again on the place numbers to find the new horse. But I've never cashed in on a 5th repeater. I've cleared $300 on $32/34 inside lots of times though. The best part is that when the 7 comes, I don't hardly ever have $250 in lost bets on the boxes having made only $100. Ifn I wanna win bigger though, I think I gotta bet bigger. That's tough; I'm not ready for that yet.

The main strategy is to have minimal betting amounts on all but one "trick pony" somewhere--I'm attempting to concentrate most of the money on a single number... sometimes I don't get single-number re-re-repeaters, and instead, the dice will hit each number in turn--i.e., "randomly scattered results"--let's say the shooter rolls a 5, 9, 8, leaving me with three first presses of $24 (5), $10 (9), $12 (8). In this case, I'm all over the board, waiting for two more parlays to take a profit. So instead, I'll reduce them all to one unit each, and use the resulting regression monies- $19+5+6 = $30 as a supplement to my odds bet, making the odds $50. For the single unit place bets that are left behind, I'll follow an "up-one-unit" strategy except for a dice-chosen new "1st"-number 'trick pony' in the place boxes. In this case, since my 'trick pony-5' won't repeat again-and-again quickly, I'll skim the payoffs and chose my own 'pony'--the best paying odds available--PL odds.

[oChRoNiCo]

Hello all new member and my 1st post. I have been playing craps for about 12 years since I was of legal age to do so. I have been playing a 1 number strategy for the last 2+ years. I call it 4 down progression I pick the 6 or 8 just one of them and start with a $36 bet for this type of play I should mention my buy in is $360 so I have 10 bets of $36. I have also done this system with 5 or 9 and 4 or 10 as well but I have had the best results with 6 or 8. So I place $36 on 6 or 8 after charting the table to see which is more popular and of course I use my own set for whichever number I choose when my roll comes up. So after picking the number I wait until after a point has been established and place the 6 or 8 for $36 if they roll the number I take everything down original $36 wager + $42 winnings and wait for a new point. Sometimes I will let it ride depending on the shooter something I have tweaked over time. So once I decide to play another wager now I bet $78 on the 6 or the 8 depending on which number we selected. If that bet wins we bring everything down so we have $78 wager and $91 winnings. Now we cannot lose any money on this roll our next bet is place for $132. you had $169 total after your last bet won so with only $132 in action you have pocketed $37 so we are up $1 after 2 hits. Our next bet will be $132 on the 6 or 8 once again sticking with the original number you choose if you win this bet you will get paid $154 once again bring everything down this gives you a total of $286 pocket $10 and your 4th and final bet will be a place for $276 which if you hit will pay $322. Capping this all if you can win 4 bets in a row you will have won a total of $609 and replenished your original $36 as well. Lets hear everyone's thoughts on this thank you in advance.

[oChRoNiCo]

I should note that in my home casino I'm allowed to do put bets which gives me a little better payout once I start my 3rd bet I actually do a $12 flat with $120 odds to make $156 instead of $154 on place bet for $132, and my last bet is $25 flat with $250 odds which pays $325 instead of the $276 to make $322 but I know this isn't allowed in all cations so I listed the alternate format instead.