Expected...
1. Does anyone know if a statistical table exists that shows the expected win ratio of every Blackjack hand permutation?
For example;
if the dealer shows a 2 and the players initial cards are AA then, if basic strategy is followed, the player is expected to win x % of the time
if the dealer shows a 2 and the players initial cards are A2 then, if basic strategy is followed, the player is expected to win x % of the time
if the dealer shows a 2 and the players initial cards are A3 then, if basic strategy is followed, the player is expected to win x % of the time
all the way through to;
if the dealer shows an A and the players initial cards are KK then, if basic strategy is followed, the player is expected to win x % of the time
2. Also, if such a table exists, would there be a difference in the figures between American and European Blackjack in terms of a hole-card being present or not?
https://youtu.be/jCF-Btu5ZCk
Yes. Some slight differences. If you can follow that workbook, you can modify it.
Quote: si2. Also, if such a table exists, would there be a difference in the figures between American and European Blackjack in terms of a hole-card being present or not?
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Yes, as game rules change, basic strategy & probable outcomes get adjusted.
If memory serves, the adjustments are covered under peek vs no-peek and lose all vs OBBO (original & busted bets only).
What I'm finding tough to grasp is how I'm to extrapolate the data in that spreadsheet into the figures that I need.
For example, with a Hard 4 v the Dealer's 2, the EV of;
Standing = -0.292783727
Hitting = -0.114913328
Doubling = -0.585567454
So it's clear that the correct option is to hit.
Is it therefore possible for me to extrapolate that to produce an expected win % ? So that, in the event of a Hard 4 and the dealer has a 2, IF I follow the Basic Strategy, I can expect the hand to win x% of the time.
This isn't necessarily about EV and how much I can expect to win, but the % chance of the hand successfully beating the dealer ?
I'm missing something but I can't see the woods for the trees...
Quote: siThanks for pointing me in that direction. It's a fascinating video and I managed to follow the process (not so much the more complex of the spreadsheet formulas).
What I'm finding tough to grasp is how I'm to extrapolate the data in that spreadsheet into the figures that I need.
For example, with a Hard 4 v the Dealer's 2, the EV of;
Standing = -0.292783727
Hitting = -0.114913328
Doubling = -0.585567454
So it's clear that the correct option is to hit.
Is it therefore possible for me to extrapolate that to produce an expected win % ? So that, in the event of a Hard 4 and the dealer has a 2, IF I follow the Basic Strategy, I can expect the hand to win x% of the time.
This isn't necessarily about EV and how much I can expect to win, but the % chance of the hand successfully beating the dealer ?
I'm missing something but I can't see the woods for the trees...
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I don't think you can, as (a) you don't know how often you would push, and (b) assuming this is allowed, you have to allow for the possibility that, if you hit, your next card is a 7, so you would double on your 11.
You also didn't specify how pushes should be handled in your percentage (e.g. treat them as losses, since you "didn't beat the dealer," or just ignore them, and calculate wins divided by (wins + losses)).
If there were no pushes, or pushes counted as 1/2 win and 1/2 loss, and you assume that you can't double with three or more cards, then you could do it; an EV of -0.1149 has a win chance of (50 - 0.1149 / 2) %, or 49.94255%
As pushes are a factor in the normal course of BJ, any idea how this could be taken into account?
Quote: siIf I understand your post correctly, you're saying that if pushes weren't a factor, then we can extrapolate the -0.1149 into an 'expected win' of 49.94255%, from that Hard 4 v 2 scenario?
As pushes are a factor in the normal course of BJ, any idea how this could be taken into account?
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Only if you know the probability of a push in that situation - in which case, you should also already know the probability of a win and of a loss.
Assume you know the probability of a push and the ER, but not the probability of a win or loss (and also assume that you can't double on three or more cards).
Let P be the probability of a push: the probability of a win or a loss is 1 - P.
Let W be the probability of a win, and L the probability of a loss: EV = W - L, and W + L = 1 - P.
Add them up: EV + 1 - P = (W - L) + (W + L) = 2W, so W = (EV + 1 - P) / 2. Note that, in this case, EV is negative.
Question for those more familiar with casino blackjack: how common is being allowed to double on three (or more) cards?
Never seen online, nor in UK B&MQuote: ThatDonGuy
Question for those more familiar with casino blackjack: how common is being allowed to double on three (or more) cards?
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We usually have Double after split except on aces. Unlimited splits in UK B&M
Nice analysis, thanks.
Quote: ThatDonGuyQuote: siIf I understand your post correctly, you're saying that if pushes weren't a factor, then we can extrapolate the -0.1149 into an 'expected win' of 49.94255%, from that Hard 4 v 2 scenario?
As pushes are a factor in the normal course of BJ, any idea how this could be taken into account?
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Only if you know the probability of a push in that situation - in which case, you should also already know the probability of a win and of a loss.
Assume you know the probability of a push and the ER, but not the probability of a win or loss (and also assume that you can't double on three or more cards).
Let P be the probability of a push: the probability of a win or a loss is 1 - P.
Let W be the probability of a win, and L the probability of a loss: EV = W - L, and W + L = 1 - P.
Add them up: EV + 1 - P = (W - L) + (W + L) = 2W, so W = (EV + 1 - P) / 2. Note that, in this case, EV is negative.
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It sounds like we've come full circle. The probability of a win from a specific starting hand is precisely what I'm trying to ascertain.
I'm just struggling to find out whether such information is available. Or even possible to calculate.
Quote: ThatDonGuy
Question for those more familiar with casino blackjack: how common is being allowed to double on three (or more) cards?
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It may be available in a few carnival pit 21 games, but I can't recall seeing it for Blackjack.
edit: I vaguely recall some bizarre stories from a while back about somewhere in South America where it might have been offered.
Quote: siQuote: ThatDonGuyQuote: siIf I understand your post correctly, you're saying that if pushes weren't a factor, then we can extrapolate the -0.1149 into an 'expected win' of 49.94255%, from that Hard 4 v 2 scenario?
As pushes are a factor in the normal course of BJ, any idea how this could be taken into account?
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Only if you know the probability of a push in that situation - in which case, you should also already know the probability of a win and of a loss.
Assume you know the probability of a push and the ER, but not the probability of a win or loss (and also assume that you can't double on three or more cards).
Let P be the probability of a push: the probability of a win or a loss is 1 - P.
Let W be the probability of a win, and L the probability of a loss: EV = W - L, and W + L = 1 - P.
Add them up: EV + 1 - P = (W - L) + (W + L) = 2W, so W = (EV + 1 - P) / 2. Note that, in this case, EV is negative.
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It sounds like we've come full circle. The probability of a win from a specific starting hand is precisely what I'm trying to ascertain.
I'm just struggling to find out whether such information is available. Or even possible to calculate.
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W = (EV + 1 - P) / 2 requires you to know both EV and P.
You cannot calculate the winning percentage from just the EV; you need to know the probability of a push as well.
Quote: ThatDonGuyQuote: siQuote: ThatDonGuyQuote: siIf I understand your post correctly, you're saying that if pushes weren't a factor, then we can extrapolate the -0.1149 into an 'expected win' of 49.94255%, from that Hard 4 v 2 scenario?
As pushes are a factor in the normal course of BJ, any idea how this could be taken into account?
link to original post
Only if you know the probability of a push in that situation - in which case, you should also already know the probability of a win and of a loss.
Assume you know the probability of a push and the ER, but not the probability of a win or loss (and also assume that you can't double on three or more cards).
Let P be the probability of a push: the probability of a win or a loss is 1 - P.
Let W be the probability of a win, and L the probability of a loss: EV = W - L, and W + L = 1 - P.
Add them up: EV + 1 - P = (W - L) + (W + L) = 2W, so W = (EV + 1 - P) / 2. Note that, in this case, EV is negative.
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It sounds like we've come full circle. The probability of a win from a specific starting hand is precisely what I'm trying to ascertain.
I'm just struggling to find out whether such information is available. Or even possible to calculate.
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W = (EV + 1 - P) / 2 requires you to know both EV and P.
You cannot calculate the winning percentage from just the EV; you need to know the probability of a push as well.
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For simplicity, can we just set P=8% for every blackjack decisions? If we have this number correct, then we will have the win percentage easily ready from your formula,
W = (EV + 1 - P) / 2.
On another thought, this formula does not apply to the player blackjack situation.
Quote: ThatDonGuy
W = (EV + 1 - P) / 2 requires you to know both EV and P.
You cannot calculate the winning percentage from just the EV; you need to know the probability of a push as well.
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Understood, however is it even possible to calculate accurately the probability of a push from every starting permutation ?
It seems to me that there are too many variables...?
If you want a combinational analysis, then you have to do the above for all possible hands.
For infinite deck is relatively easy since you can treat (say) every hard 8 vs 10 as the same and add up (soft19,10,11,12,13,14,15,16,17,18) with probabilities of {1/13,1/13...4/13}. With combination it's laborious, as the probabilities aren't easy and 5,3 is different than 6,2 4,4 2,3,3 etc, but it's possible.
Quote: aceside<snip>For simplicity, can we just set P=8% for every blackjack decisions?<snip>link to original post
aceside,
I hope you see that the probability of a push is much higher for 9,8 vs. 7 than for 10,10 vs. 7.
Furthermore, any time the player stands on a stiff total, the push probability is 0. For example, 10,5 vs. 6 has a push probability of 0.
Thus, your idea is incorrect.
Hope this helps!
Dog Hand
That's why you hit hard 17 vs 7 on your free money hand: your chances of beating the dealer on 17 are (infinite deck) the same as being on 16 but if you hit 17 and do get 18-21 your chances of winning are that much greater because you beat any dealer 17. (I think similar logic applies to 17v8 and 17v9 but it's not so obvious.)Quote: DogHand...I hope you see that the probability of a push is much higher for 9,8 vs. 7 than for 10,10 vs. 7....link to original post
Quote: DogHandQuote: aceside<snip>For simplicity, can we just set P=8% for every blackjack decisions?<snip>link to original post
aceside,
I hope you see that the probability of a push is much higher for 9,8 vs. 7 than for 10,10 vs. 7.
Furthermore, any time the player stands on a stiff total, the push probability is 0. For example, 10,5 vs. 6 has a push probability of 0.
Thus, your idea is incorrect.
Hope this helps!
Dog Hand
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You explained this very well. Another example is the surrender decision in blackjack. Let us use this formula,
W = (EV + 1 - P) / 2,
where EV=-0.5, P=0.
This gives W=25%, the possibility of a win.
Although we don’t know the exact value of P, we know its range as 0<=P<=30.8%. If we plug P=30.8% in the above formula, we have
W=(-0.5+1-0.308)/2=9.6%, the possibility of a win.
Quote: siUnderstood, however is it even possible to calculate accurately the probability of a push from every starting permutation ?
It seems to me that there are too many variables...?
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Well, whoever calcualted the EV values for hit, stand, and double was able to calculate the win, push, and loss probabilities for each of those, but as far as I know, this is a brute force problem.
Yes I once wrote a program to calculate all the possible player hands and then work through all hands the dealer could make given the cards left in the pack. That does give the total times for wins, ties and pushes if you stand on various hands; but one only really needs the EVs for correct strategy decisions, so you don't carry over win percentages. In theory one also has to do it based on cards already gone from previous splits. You then work through the hands, starting from those with most cards, and compare standing with the combined totals of hitting (or doubling). When down to two cards you can look at splitting decisions as well.Quote: ThatDonGuy...Well, whoever calcualted the EV values for hit, stand, and double was able to calculate the win, push, and loss probabilities for each of those, but as far as I know, this is a brute force problem...
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That is why, to get an approximation which tends to be good enough to get an idea of the correct strategy, it's easier to use infinite decks.
Quote: charliepatrickYes I once wrote a program to calculate all the possible player hands and then work through all hands the dealer could make given the cards left in the pack. That does give the total times for wins, ties and pushes if you stand on various hands; but one only really needs the EVs for correct strategy decisions, so you don't carry over win percentages. In theory one also has to do it based on cards already gone from previous splits. You then work through the hands, starting from those with most cards, and compare standing with the combined totals of hitting (or doubling). When down to two cards you can look at splitting decisions as well.Quote: ThatDonGuy...Well, whoever calcualted the EV values for hit, stand, and double was able to calculate the win, push, and loss probabilities for each of those, but as far as I know, this is a brute force problem...
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That is why, to get an approximation which tends to be good enough to get an idea of the correct strategy, it's easier to use infinite decks.
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I actually did it the other way around: First, for each of the ten possible dealer up cards, I calculated the probabilities of the dealer having 17, 18, 19, 20, 21, and busting, taking the number of each value card in the deck/shoe into account (remembering not to include two-card 21s). Then, for each up card, starting with 21 and working down, calculate the EV of standing (easy enough: add the values of the dealer results where the player wins, and subtract the values where the dealer wins) and hitting (add up the EVs of (hand + 1) through (hand + 10), multiplying the (hand + 10) value by 4), and use the one that is higher as the strategy. Note that the player calculations assume an infinite deck (i.e. when hitting, a 2 will come up 1/13 of the time, and a 10 will come up 4/13 of the time).
It has to be done from 21 down as, for example, to calculate the hit value for 17, you have to know the EVs for best play of 18 through 21.
Note that, once you get down to 11, you then have to calculate soft 21 down to 12, because, for example, to calculate hard 5, you need to know the play for soft 16 (if you draw an ace).
I totally agree this method for infinite decks: I use a spreadsheet with the first sheet working out the dealer's probabilities, then ten sheets for the player's strategy for each dealer up-card.Quote: ThatDonGuy...I actually did it the other way around: First, for each of the ten possible dealer up cards, I calculated the probabilities of the dealer having 17, 18, 19, 20, 21, and busting...Then, for each up card, starting with 21 and working down, calculate the EV of standing...It has to be done from 21 down as, for example, to calculate the hit value for 17, you have to know the EVs for best play of 18 through 21...down to 11, you then have to calculate soft 21 down to 12...
For combinational analysis you need to know what to do with (say) 12-card hands before looking at 11-card hands (which by hitting can turn into some of the 12-card hands), so start with 21 Aces and work down.
It appears that it IS computable (albeit extremely complex with all of the steps/variables throughout the hand) to calculate the precise amount of possible permutations that any hand could possibly go, from 12 cards all the way down to two versus all variables of the dealers hands. Once that is done, it is possble to calculate the % of times that the hand wins/loses/pushes, when the play follows basic strategy ??
Is this right ?
I doubt anyone has bothered to look at win percentages since, using the specific rules for the BJ, players try to maximize EV. Technically in the UK, playing infinite deck strategy, you could land up with a 14-card hand (as you don't split AA vs A and hit Soft 18).Quote: siAm I to assume then, that no one knows of such an existing analysis ?...Is this right ?
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Personally I think it's easier just to go all the way up to the 21-card hand, that way you don't need to worry about the exact rules for the game. Keeping the code simple one might use 22-card hands, i.e. create all the ways to get a non-bust hand and then include all the ones where they take a card and bust. This makes it easy for the code to see that hitting 21 is wrong as all the new hands are bust with an EV=-1, thus confirming you stand on 21-Aces! (The fact that you can never get to that hand for normal rules is immaterial, as you work back it allows the code to work this out for itself.)
Quote: charliepatrickI doubt anyone has bothered to look at win percentages since, using the specific rules for the BJ, players try to maximize EV. Technically in the UK, playing infinite deck strategy, you could land up with a 14-card hand (as you don't split AA vs A and hit Soft 18).Quote: siAm I to assume then, that no one knows of such an existing analysis ?...Is this right ?
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Personally I think it's easier just to go all the way up to the 21-card hand, that way you don't need to worry about the exact rules for the game. Keeping the code simple one might use 22-card hands, i.e. create all the ways to get a non-bust hand and then include all the ones where they take a card and bust. This makes it easy for the code to see that hitting 21 is wrong as all the new hands are bust with an EV=-1, thus confirming you stand on 21-Aces! (The fact that you can never get to that hand for normal rules is immaterial, as you work back it allows the code to work this out for itself.)
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On my second thought, Blackjack is a little more complicated than a sport game. There are actually four outcomes for every hand, win, loss, push, and surrender (half loss).
Let me calculate the win percentage for the hand of 8,9 vs. 7. Assuming infinite deck, we have P>=4/13=30.8% because there are multiple-card 17s and the basic strategy of standing EV=-0.108, so the win percentage is
W=(-0.108+1-0.308)/2<=29.2%
So the loss percentage is 1-29.2%-30.8%<=40%
We definitely can calculate these numbers, but this is not easy.
Quote: charliepatrickWith 17 versus the dealer it is easy to see how often you can win since you can look up the EVs for 16 (since both 16 and 17 only "win" if the dealer busts). In this case you get about 26%, and this matches up with how often the dealer busts starting from 7 (e.g. see http://www.edgevegas.com/blackjack-dealer-bust-rates/ which presumably has a figure for finite decks). Obviously for other values (only for infinite decks) you could do similar calculations knowing that the EV(16) is only when the dealer busts, EV(17) is win when the dealer busts plus the effect of pushing against 17, etc.
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You are right. For the hand of 8,9 vs. 7, we can easily find the win percentage,
W=26%.
From W=(EV+1-P)/2, we find P=37.2%.
So loss percentage L=36.8%.
Quote: charliepatrickI doubt anyone has bothered to look at win percentages since, using the specific rules for the BJ, players try to maximize EV.
I agree that it may not be a direction that has been previously explored but using basic strategy to maximize EV and the win/loss/push % from every starting position are different things, I think.
The BS table doesn't specifically account for the natural process of the hand. Reference must be made back to the BS table at each decision point.
The table I am trying to source is the end result of all of the possible processes/variables during the normal course of blackjack, so that from each starting point the user can determine that (if he follows basic strategy) he is x% to win/lose/push etc.
I'm not sure that the table highlighting EV can do that ?
Si,Quote: siAm I to assume then, that no one knows of such an existing analysis ?
It appears that it IS computable (albeit extremely complex with all of the steps/variables throughout the hand) to calculate the precise amount of possible permutations that any hand could possibly go, from 12 cards all the way down to two versus all variables of the dealers hands. Once that is done, it is possble to calculate the % of times that the hand wins/loses/pushes, when the play follows basic strategy ??
Is this right ?
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Seriously, why are you interested in probability of particular hands winning. It's not as if you can use that knowledge after a hand is dealt, nor indeed before.
Revisit Wizards infinite deck derivation. Actually do and understand every step and you will soon figure out how to modify it for finite decks and deck constitution analysis. Straightforward but a bit tedious, as you need some long formulae in cells rather than x/13.
As CharlieP has said, you could also derive the probabilities simply enough by programatically running through every combination of hands dealt.
But seriously.... Why? Are you maybe looking to prove the already known: That betting systems don't work.
Quote: OnceDearSi,
Seriously, why are you interested in probability of particular hands winning. It's not as if you can use that knowledge after a hand is dealt, nor indeed before.
Revisit Wizards infinite deck derivation. Actually do and understand every step and you will soon figure out how to modify it for finite decks and deck constitution analysis. Straightforward but a bit tedious, as you need some long formulae in cells rather than x/13.
As CharlieP has said, you could also derive the probabilities simply enough by programatically running through every combination of hands dealt.
But seriously.... Why? Are you maybe looking to prove the already known: That betting systems don't work.
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I guarantee you I'm not a 'betting systems' guy. Having worked in the Casino Industry since 1998, I can testify to that.
I understand that this must appear a strange request, and that there's other figures out there that can give me more 'valuable' information, but gathering these figures is the first step into a new project I am developing.
Without knowing 'why' I want the information, I totally appreciate that the request seems pointless.
The problem with that game is to weigh up the EV for the base bet and the EV for the bonus bet, which only takes effect on a win. Thus the strategy is somewhere between regular and just maximising the chances of winning. Thus it might be like some poker games, raise with XXX, fold with YYY, best of luck between.
EDIT: If you're looking at a single bonus that pays on a win (for arguments sake on the first hand on splits and equal to the original wager) then be prepared for players changing strategy. For instance you may be able to work out that using regular strategy the player wins X%; however if you made it valuable enough, they should change strategy and be able to win more often - examples are not doubling or making unusual split decisions.
Good observation by CP. Could well be design considerations in a game variant. If so, the gamekeeper is seeking advice from the poachers 😀or maybe Si is looking to AP the variant that already exists. The plot thickens. Maybe Si should put this analysis out for tender. Sounds like it might be valuable after all.Quote: si
I understand that this must appear a strange request, and that there's other figures out there that can give me more 'valuable' information, but gathering these figures is the first step into a new project I am developing.
Without knowing 'why' I want the information, I totally appreciate that the request seems pointless.
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Quote: OnceDearGood observation by CP. Could well be design considerations in a game variant. If so, the gamekeeper is seeking advice from the poachers 😀or maybe Si is looking to AP the variant that already exists. The plot thickens. Maybe Si should put this analysis out for tender. Sounds like it might be valuable after all.
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I've said too much.... :D
Quote: siQuote: OnceDearGood observation by CP. Could well be design considerations in a game variant. If so, the gamekeeper is seeking advice from the poachers 😀or maybe Si is looking to AP the variant that already exists. The plot thickens. Maybe Si should put this analysis out for tender. Sounds like it might be valuable after all.
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I've said too much.... :D
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Some of the win percentages are very easy to calculate, while others take some effort to get them.
Lol.you asked for analysis. You got more than you bargained for. Still. We are not hostile either way. Some who have showed assistance might be amenable to Personal messagesQuote: siQuote: OnceDearGood observation by CP. Could well be design considerations in a game variant. If so, the gamekeeper is seeking advice from the poachers 😀or maybe Si is looking to AP the variant that already exists. The plot thickens. Maybe Si should put this analysis out for tender. Sounds like it might be valuable after all.
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I've said too much.... :D
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Quote: OnceDearLol.you asked for analysis. You got more than you bargained for. Still. We are not hostile either way. Some who have showed assistance might be amenable to Personal messages
To be fair, it's not quite the 'gamekeepers & poachers' analogy you mentioned before. It's more an 'amateur poacher & poachers' scenario, which you'll agree is somewhat more palatable.
My OP was essentially asking you guys whether you know of any existing analysis that contains the figures requested. The objective, once that analysis was sourced, was merely to use it to garner further analysis myself...
Not as much 'cloak & dagger' as you might think...
My suggestion would be to create a spreadsheet (or program) that works out the best play based on infinite decks. You can easily work out the chances of the dealer's hand using logic such as from 16 there's 1 in 13 chances of being 17 thru 21 and the rest bust. From 15 there's 1 in 13 chance of being 16 (and you've just worked out those) 1 in13 of 17 thru 21 and the rest bust. Repeat and rinse to get the probabilities of getting to 21,20...17 from 11-16. Now you need to consider soft hands, as 10 plus an Ace is soft 21 (soft 17-21, depending on rules, will stand; soft 16-12 will hit). Use similar logic down to Soft 1. For 10s and Aces you then need to cater for the dealer getting BJ.Quote: si...My OP was...to use it to garner further analysis myself...
It's all possible, just a little (or a lot of) thought required. You can also check your values against the ones wizard has posted (remember EV = Pr(Win)-Pr(Lose).)
Now you can work out the chances of winning, pushing and losing for any player's hand, and then whether it's better to stand, hit, double, split etc. From this you can work out whatever else you need to.
