
June 23rd, 2009, 05:05 PM


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New shoe strategy
So at the beginning of the shoe, depending on the game, the house has a typical .5% advantage. I've been using a strategy mainly for cover, but after reading around it seemed to be somewhat on par with opposition betting (referring to the wild bets at .5% to .5% HA) and I was trying to make sense of the math in my head and wondered if I was correct.
Basically off the top of the shoe, instead of betting 1 unit, I'll bet 2 or 3. If I win, I let it ride at that level (not letting it ALL ride, just the same 2 or 3 units) until I lose and then go to 1 unit until the count says otherwise. If I lose the very first hand, then the second had Im back to the minimum 1 unit.
Now I assume that the deck has an equal chance of going positive as it does going negative and therefore, in the long run any continued wins would be cancelled out by losses correct? So ultimately, in the long run, I'm giving up .5% on a single bet. Is this right?
And then theres part of me that says a little more than 1/2 the time on the 1st bet I will lose and be out lets say 1 unit, if I bet 2 initially. So just on the outcome of the 1st had I will be just about equal in the long run, down .5%. But then since the 2nd hand, has a chance greater than 0% of winning, and so forth, it seems that Im not losing anything by keeping the same bet on the table and only increase my chance of being profitable. I know this cant be right, but if someone could help me with where my logic is flawed. Thanks

June 23rd, 2009, 05:16 PM

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sounds like my play style. I figured that .5 against me, basic strat evens it out to an even game for the most part from the start. play 3 hands and best 2 out of 3 and you win

June 23rd, 2009, 05:35 PM


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ummm
Chance of winning or pushing First Hand roughly 47.9% (edited this one too.)
Chance of that hand being a BJ roughly .0475
Chance of winning 2 in row is .2255 (I think.. I'm doing this on my iphone) (I edited this number it was wrong)
Last edited by daddybo; June 23rd, 2009 at 06:32 PM.

June 23rd, 2009, 05:47 PM


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Quote:
Originally Posted by daddybo
Chance of winning or pushing First Hand roughly 47.5%
Chance of that hand being a BJ roughly .0475
Chance of winning 2 in row is .0242 (I think.. I'm doing this on my iphone)

Wouldn't the 47.5% give the house a 2.5% advantage? or is that where the natural comes in?

June 23rd, 2009, 06:13 PM


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ooops
I screwed up some of the numbers.. I edited them in my previous post.
I probably need to go back and look at the numbers.. I had some of those numbers in my head... I think they are for 6D S17 games (which might not be .5% HA )
I've done these same calculations for myself before... Let me get home on the computer and I'll look it up. (on the phone now)
BTW.. sometime I'll do a max bet off the top and go from there... your playing at a disadvantage but it looks good to the pit and it's nice when the first one's BJ or double. ;).
Last edited by daddybo; June 24th, 2009 at 05:48 PM.

June 23rd, 2009, 06:51 PM


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ok maybe this will make more sense
LOL... I had to go look this up at the wizard of odds .... and these numbers are approximate.
Chance of win = 43.31
Chance of push = 8.8
Chance of losing = 47.89
Chance of BJ = .0475 (4.75 per 100 hands)
That probably makes more sense. it should be pretty close to .5%HA

June 23rd, 2009, 07:12 PM

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i still think doing 3 hands hold the most value. a 25 50 100 bet with the 100 on the last hand would mean you can play simple quick basic on the first and as you see those cards you can adjust and use the exact count on the third hand. lots of variation here. plus like he said it looks good to lose the 25 and win the 100, id figure they aren't paying enough attention to notice 100 and 25.

June 24th, 2009, 12:35 AM


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Quote:
Originally Posted by daddybo
LOL... I had to go look this up at the wizard of odds .... and these numbers are approximate.
Chance of win = 43.31
Chance of push = 8.8
Chance of losing = 47.89
Chance of BJ = .0475 (4.75 per 100 hands)
That probably makes more sense. it should be pretty close to .5%HA

Ok, so if my math is correct with EV, then on a $10 bet it would be:
.4331($10) + .088($0) + .4789($10) + .0475($15)
$4.331 + $0 + $4.789 + $.007125 = $.4509 (approx.)
And therefore the chance of winning X number of hands in a row would basically be .4331 ^ X right? Hence giving you:
2 Hands = .1876
3 Hands = .0821
4 Hands = .0352
5 Hands = .0124
And from there forth the probability is less than 1%. So out of 100 shoes played, in the long run, one would expect to win the 1st 5 hands in a row about 1 time. My assumption here is that there is an equal chance for the deck to go positive and negative and that in the long run these would balance out leaving the same basic EV for each hand, correct?
Finally, to figure the EV for every possible outcome of the 1st 5 hands where $10 was the initial bet to ride the wins, and a $5 bet was placed for the remainder after a loss, would require calculating 1024 hands (4^5)? I might just attempt the 1st 3 hands, which would be a little more doable.
EDIT: Those probabilities, do they assume your playing perfect BS and do they factor in doubling down and splitting do you know?
Last edited by airborneinf82; June 24th, 2009 at 12:38 AM.

June 24th, 2009, 09:44 AM


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Quote:
Originally Posted by airborneinf82
Ok, so if my math is correct with EV, then on a $10 bet it would be:
.4331($10) + .088($0) + .4789($10) + .0475($15)
$4.331 + $0 + $4.789 + $.007125 = $.4509 (approx.)
And therefore the chance of winning X number of hands in a row would basically be .4331 ^ X right? Hence giving you:
2 Hands = .1876
3 Hands = .0821
4 Hands = .0352
5 Hands = .0124
And from there forth the probability is less than 1%. So out of 100 shoes played, in the long run, one would expect to win the 1st 5 hands in a row about 1 time. My assumption here is that there is an equal chance for the deck to go positive and negative and that in the long run these would balance out leaving the same basic EV for each hand, correct?
Finally, to figure the EV for every possible outcome of the 1st 5 hands where $10 was the initial bet to ride the wins, and a $5 bet was placed for the remainder after a loss, would require calculating 1024 hands (4^5)? I might just attempt the 1st 3 hands, which would be a little more doable.
EDIT: Those probabilities, do they assume your playing perfect BS and do they factor in doubling down and splitting do you know?

You're headed in the right direction...
Actually the simple calculation for chances of winning x hands in a row, ignoring pushes, would be 0.4749^x. The house edge and the win/loss computation are separate animals (though related). The win loss calculations would count splits as two hands and would not consider double downs. (that would be part of the house edge computation.) BJs would merely be considered a win for the win/loss calculation.
Go to the link below... the wizard has done all the calcs you need..
http://wizardofodds.com/askthewizard/blackjackfaq.html

June 24th, 2009, 03:22 PM


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Quote:
Originally Posted by daddybo
You're headed in the right direction...
Actually the simple calculation for chances of winning x hands in a row, ignoring pushes, would be 0.4749^x. The house edge and the win/loss computation are separate animals (though related). The win loss calculations would count splits as two hands and would not consider double downs. (that would be part of the house edge computation.) BJs would merely be considered a win for the win/loss calculation.
Go to the link below... the wizard has done all the calcs you need..
http://wizardofodds.com/askthewizard/blackjackfaq.html

I went there and I see where your getting those numbers from, but I don't understand how he went from .4331 for a win and then just ignoring the ties makes the probability for a win .4749
I guess Im just trying to get the probabilities for each individual situation. Like If I wanted to calculate my EV for getting a 3 hand "combination" of for example, WinLossBlackjack or BlackjackWinPush, and utilizing the individual probability for each to calculate the EV based on the betting mentioned already.
Also wouldn't at least double downs have to be factored in to the probability considering that you are forced to only take one more card and not additional cards available to take?
Last edited by airborneinf82; June 24th, 2009 at 04:03 PM.

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