New shoe strategy

[airborneinf82]

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

 

[Colde86]

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

 

[daddybo]

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)

 

[airborneinf82]

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?

 

[daddybo]

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. ;-).

 

[daddybo]

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

 

[Colde86]

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.

 

[airborneinf82]

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?

 

[daddybo]

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/blackjack-faq.html

 

[airborneinf82]

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/blackjack-faq.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, Win-Loss-Blackjack or Blackjack-Win-Push, 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?

 

[daddybo]

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 don't know the answer to that... I just took him for his word.

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, Win-Loss-Blackjack or Blackjack-Win-Push, and utilizing the individual probability for each to calculate the EV based on the betting mentioned already.

Actually that part is pretty easy.. assuming HA=.5% 3 hands $10 bet -- 3x$10x0.005 = 0.15 .... You'll loose $0.15 per 3 hands (over the long haul)

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?

Double downs have more to do with the total house edge. I believe the average hand in BJ is 2.7 cards... so using that info; it shouldn't make a significant difference on winning or losing future hands.

wow.. you've got me doing math I haven't done in a while!

 

[KenSmith]

OK, I admit I haven't actually read most of this thread, so if I'm off on a tangent, let me know.

However, I thought I would comment on two points:

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.)

Actually, the 4.75% chance of getting a blackjack is already on the 43.31% chance of winning a hand. (Add them all up and you get 1.0475).

As others have noted, you can't figure the EV this way, because some of the wins are one bet, some 1.5 bets, some 2 bets, etc. Likewise, some of the losses are 1 bet, 2 bets or more.

Second point...

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

0.4331 / (0.4331 + 0.4789) = approx 0.4749.

 

[daddybo]

OK, I admit I haven't actually read most of this thread, so if I'm off on a tangent, let me know.

:laugh::laugh::laugh: No Ken... I think you hit the nail on the head. We're just taking something simple and making it complicated. Thanks

That's what I get for trying to do that on my phone while driving. :whip:

 

[airborneinf82]

Actually that part is pretty easy.. assuming HA=.5% 3 hands $10 bet -- 3x$10x0.005 = 0.15 .... You'll loose $0.15 per 3 hands (over the long haul)

Well this is where my "need" to try and figure out the math comes in. What you put up shows the EV for flat betting $10, and my curiosity arises from, for example and sake of simplicity, initial $10 bet, reduced to the $5 minimum only after a loss.

Thus Blackjack-Blackjack-Blackjack (prob .0475^3) would result in $45 won. Now let say you Win-Win-Win those 3, then the payout is $30, where as if you Lose-Lose-Lose then you lose $20. Now granted the prob of winning that 3rd hand in a row is reduced from that of the 1st and the 2nd, but there is the realistic probability, and Im wondering if that probability times the expected win, is enough to reduce your expeted loss from flat betting $5 or even, given the right amout might eliminate the expected loss.

The only way I can see to calculate this is to know the exact probability of each hand, from a blackjack, to just a win (not including bj), a push, and a loss. Then calculating the exact probability of each 3-hand combo times the expected win $ (as mentioned above). This is where my headache begins lol!

wow.. you've got me doing math I haven't done in a while!

haha sorry... im totally one of those people that once I get an idea in my head, I like to prove or disprove it, and in this case it requires tapping in to some math abilities. Thanks for the help tho!

 

[airborneinf82]

OK, I admit I haven't actually read most of this thread, so if I'm off on a tangent, let me know.

However, I thought I would comment on two points:

Actually, the 4.75% chance of getting a blackjack is already on the 43.31% chance of winning a hand. (Add them all up and you get 1.0475).

As others have noted, you can't figure the EV this way, because some of the wins are one bet, some 1.5 bets, some 2 bets, etc. Likewise, some of the losses are 1 bet, 2 bets or more.

Second point...


0.4331 / (0.4331 + 0.4789) = approx 0.4749.

On the issue of the BJ probability incorporated in to the win %, then would the probability of just a win (no bj) just be .3856? So a BJ (.0475), Win (.3856), Loss (.4789) and Push (.088) then all totals to 1?

 

[Sonny]

What you put up shows the EV for flat betting $10, and my curiosity arises from, for example and sake of simplicity, initial $10 bet, reduced to the $5 minimum only after a loss.

If we know the house edge then we can figure this out. Let’s assume it is 0.5% for this example. Your EV for the first hand is always -$0.05. The EV for your second hand depends on what happened on your first hand. There is a 52.11% chance that you won (or tied) the first hand and you will continue to bet $10 on the second hand with the same EV of -$0.05. There is a 47.89% chance that you will lose the first hand and bet $5 on the second hand for an EV of -$0.025. Your total EV for that strategy is:

EV = -$0.05 + .5211(-$0.05) + .4789(-$0.025)
EV = -$0.088

Another (simpler) way to figure this out is to calculate your average bet for both hands and multiply it by the house edge, which is constant for this example. Your average bet will be $10 + .5211($10) + .4789($5) = $17.6055. Since the house edge for both hands is 0.5% your EV is $17.6055 * -0.005 = -$0.088.

Thus Blackjack-Blackjack-Blackjack (prob .0475^3) would result in $45 won.

Just because you got a BJ doesn't mean that you won the hand. Keep in mind that a BJ will occasionally be a tie.

Now let say you Win-Win-Win those 3, then the payout is $30, where as if you Lose-Lose-Lose then you lose $20. Now granted the prob of winning that 3rd hand in a row is reduced from that of the 1st and the 2nd, but there is the realistic probability, and Im wondering if that probability times the expected win, is enough to reduce your expeted loss from flat betting $5 or even, given the right amout might eliminate the expected loss.

Why are you less likely to win the 3rd hand? If you have already won 2 hands in a row then the probability of winning the next hand has not changed significantly. Just like flipping a coin and getting 2 heads in a row does not mean that tails is more likely to come up next.

-Sonny-

 

[airborneinf82]

If we know the house edge then we can figure this out. Let’s assume it is 0.5% for this example. Your EV for the first hand is always -$0.05. The EV for your second hand depends on what happened on your first hand. There is a 52.11% chance that you won (or tied) the first hand and you will continue to bet $10 on the second hand with the same EV of -$0.05. There is a 47.89% chance that you will lose the first hand and bet $5 on the second hand for an EV of -$0.025. Your total EV for that strategy is:

EV = -$0.05 + .5211(-$0.05) + .4789(-$0.025)
EV = -$0.088

Another (simpler) way to figure this out is to calculate your average bet for both hands and multiply it by the house edge, which is constant for this example. Your average bet will be $10 + .5211($10) + .4789($5) = $17.6055. Since the house edge for both hands is 0.5% your EV is $17.6055 * -0.005 = -$0.088.

Ok that makes total sense and I think I see the flaw in logic, which was a simple fact that was overlooked. Thanks for the breakdown. So by that example, this betting strategy compared to flat betting $5 (EV of -$.05 after 2 hands) and $10 (EV of $.10 after 2 hands) really wouldnt make "that" much of a difference.

Just because you got a BJ doesn't mean that you won the hand. Keep in mind that a BJ will occasionally be a tie.

I realized in my calculations and logic I wasnt taking that in to account or the EV from double downs paying off and not paying off.

Why are you less likely to win the 3rd hand? If you have already won 2 hands in a row then the probability of winning the next hand has not changed significantly. Just like flipping a coin and getting 2 heads in a row does not mean that tails is more likely to come up next.

-Sonny-

The debate in any stats class is alive. Does it REALLY have an equal chance? I guess you could really bring in quantum physics if you wanted to make the other case!