Question on simple "winning the hand" percentages...

Fun_at_21

Well-Known Member
k_c said:
An even money pay EV can be converted to win percent by this formula:
win percent = (1+EV)/2 and this assumes a push = 50% win, 50% loss.

In blackjack there is the possiblilty of blackjack, doubling, and splitting incorporated into the EV, so that needs to be considered.

If dealer has a ten or ace as up card, EV can be computed 2 ways:
conditional EV: assumes dealer has checked for BJ and doesn't have it
unconditional EV: player non-BJ loses to dealer BJ

Most EV listings for ten and ace up use conditional EV. However, in order to calculate a winning percent unconditional EV needs to be used.

To convert conditional EV to unconditional EV:
pDBJ=prob(dealer blackjack); prob(dealer BJ)=0 if up card is not ten or ace
uev = cev - pDBJ(cev+1)
If up card is not ten or ace, uev and cev are the same thing.

Once uev is determined, a win percent can be calculated from the EV that counts pushes as 50% win and 50% loss.

Stand and hit:
win percent = (1+uev)/2

Double
win percent = (2+uev+pDBJ)/4

Split (1 split)
win percent = (2+uev+pDBJ)/4

Split (multi splits)
win percent = (2+uev+pDBJ(expected hands - 1))/4
expected hands can be computed, but this normally won't be known.

The above is for full peek. If ENHC (no peek) is in effect, just eliminate pDBJ from the above because in that cause player's non-BJ always loses to dealer BJ even for splits and doubles.

If this is done, the percentage of pushes still won't be known so if that's important to you this won't be of much help.
Thanks a lot! I was wondering if there was just a relatively simple "formula" for just figuring percentage odds from the EV tables. Now I can do these (hopefully) myself and even see where the factors are coming from. Of course, when I say I'm after simple winning the hand odds in "5th grade" terms I'm pretty much being serious! lol. So we'll see how I do with this math here. But it doesn't look difficult. No squares, piis, brackets anywhere so hopefully I can survive it... ;)

Now, just to verify that I'm doing NON-Ten or Ace dealer upcard calculations correct....

In Sanford Wong's Professional Blackjack book, for practice purposes, I'm trying to find what the odds of winning a 2,2 versus a 2 hand is if I HIT it. I'm under the 6 deck, S17 tables, pg 320. It gives an EV of -.115. So using the first formula in the post you wrote for non dealer blackjack possibility upcards, I can just take...

(-.115 + 1) / 2

and I get...

(0.885) / 2 = 0.4425

So my average odds of winning a 2,2 versus a 2 hand by hitting it would be roughly 44% correct?

Figuring out the unconditional EV's for dealer Ten & Ace cards looks slightly more complicated but I'm sure I can do those as well when I make it up to those two upcards...

But I really appreciate these! Thanks a lot! This should work...
 
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callipygian

Well-Known Member
Fun_at_21 said:
1) In your original tables you sent in your first post, the pair charts were assumed that you win this many hands out of a 100 if you used the correct basic strategy decision with that pair. To use the 2,2 versus 2 example that I always use as my default example, that first chart says I should win, on average, 42 out of a 100 hands if I SPLIT the 2,2 versus the 2 (as this is what the 42% odds are resting on - that I'll play the 2,2 the correct way). However, in the recent chart you gave for "winning percentages", the "split" table has 2,2 vs a 2 as winning, on average, 0.52 (52%) out of a 100 hands? This is a 10% difference of what I thought, if I'm reading the two tables correctly, was the same action against the 2?
There actually is a problem with the split charts which I'll go into later, but there's an additional problem with your analysis here. The first set of charts was for the hands to be split - that is, for 2,2 vs. 2, you should win a certain percentage of the time. The second set of charts is for the hands after they're split - that is, a single card 2 vs. 2. Because splitting 2,2 vs. 2 involves additional decision-making, I thought I'd be giving you something closer to what you wanted by giving the raw percentages.

If you calculate P(1)*Soft(13)+P(2)*Split(2)+P(3)*Hard(5)+P(4)*Hard(6) ... +P(10)*Hard(12), you should get 42%. (In this case Split(2) = 52%)

Fun_at_21 said:
2) Also, in the recent tables you provided, the winning percentage one has the 0.52 (52%) listed for splitting 2,2 versus a 2 while the losing percentages chart shows that splitting a 2,2 versus a 2 will lose 0.56 (56%) of the time. These don't match up evenly to 100%. Does this have something to do with pushes that are throwing me off again?
No, this is an oversight on my part. I shouldn't have posted the split charts, as they contain a small probability that a single card 2 gets dealt a 7,8,9 and then doubled. The doubling is mixed in with the win percentages so all the split charts are going to be off.

Fun_at_21 said:
3) In the case of the 2,2 vs 2 hand, I still don't see which charts would reveal how many times out of a 100 I'm likely to win by drawing, standing and doubling
You'd look up doubling hard 4 to see what the win percentage is doubled, standing hard 4 to see what the win percentage is standing, hit hard 4 to see what win percentage is hitting, and split 2 to see what win percentage is splitting (except that the split charts are wrong).

Fun_at_21 said:
(which I guess would actually be the same odds as winning the hand by hitting just that you'd lose twice the money)
No, and this is why people usually calculate EV. Doubling means you get one card only, whereas hitting allows for more decision making. It's quite possible - and is true for a lot of soft doubling - that your win percentage actually DROPS as a result of doubling.

For example, let's say you have a coin flip which you win 80% of the time and lose 20% of the time. Would you rather play that game, or a coin flip which you'll only win 70% of the time, but for double the stakes?

EV for first game = 0.80*(+1)+0.20*(-1) = +0.60
EV for second game = 0.70*(+2)+0.30*(-2) = +0.80

Win percentages aren't the end-all, they're a stepping stone on the way to calculating EV.
 

Fun_at_21

Well-Known Member
callipygian said:
There actually is a problem with the split charts which I'll go into later, but there's an additional problem with your analysis here. The first set of charts was for the hands to be split - that is, for 2,2 vs. 2, you should win a certain percentage of the time. The second set of charts is for the hands after they're split - that is, a single card 2 vs. 2. Because splitting 2,2 vs. 2 involves additional decision-making, I thought I'd be giving you something closer to what you wanted by giving the raw percentages.

If you calculate P(1)*Soft(13)+P(2)*Split(2)+P(3)*Hard(5)+P(4)*Hard(6) ... +P(10)*Hard(12), you should get 42%. (In this case Split(2) = 52%)
Think I still might be confused. I'm not sure I'm following what the difference is, chart-wise, between "to split a pair of 2's" and "actually drawing to the split 2's". I just assume that if you decide to split them, you're automatically making the decision to draw, at least one, card to each hand. Like when you see an EV table in books, sites etc, aren't the splitting tables assuming that your EV is such and such IF you, in fact, play out the split hands (draw to them according to best strategy). I assumed your first split chart was assuming the split 2's were played out fully and that your odds are, on average, that you'll win 42 times out of a 100 if you do in fact play them out as "correctly" as you can. But actually the second split chart is what reflects that, not the first?


No, this is an oversight on my part. I shouldn't have posted the split charts, as they contain a small probability that a single card 2 gets dealt a 7,8,9 and then doubled. The doubling is mixed in with the win percentages so all the split charts are going to be off.
That's cool. No problem. I still appreciate all the charts. I just wanted to make sure I wasn't going crazy trying to figure why they weren't adding up to 100%...lol



You'd look up doubling hard 4 to see what the win percentage is doubled, standing hard 4 to see what the win percentage is standing, hit hard 4 to see what win percentage is hitting, and split 2 to see what win percentage is splitting (except that the split charts are wrong).
But the problem is I don't see totals of 4 anywhere on the chart (with the exception of the split charts). All of them, when I scroll the arrow over all the way to the far right sides of the table all start at a minimum player's hand total of 5.



No, and this is why people usually calculate EV. Doubling means you get one card only, whereas hitting allows for more decision making. It's quite possible - and is true for a lot of soft doubling - that your win percentage actually DROPS as a result of doubling.

For example, let's say you have a coin flip which you win 80% of the time and lose 20% of the time. Would you rather play that game, or a coin flip which you'll only win 70% of the time, but for double the stakes?

EV for first game = 0.80*(+1)+0.20*(-1) = +0.60
EV for second game = 0.70*(+2)+0.30*(-2) = +0.80
You're right. That was my mistake. See, and I even warned myself in previous posts that I knew that doubling vs hitting situations are the one "slippery slope" with just using a "winning the hand odds" chart! ;) So I was aware of this. I simply wasn't thinking. I forgot that when doubling, ONLY one card can be drawn. This obviously makes all the difference in the world compared to hitting. I think I was messing up the actions of splitting and doubling in my mind. With splitting (except aces) you can still endlessly draw (until you bust anyways). I was incorrectly thinking of that when I was thinking of doubling vs hitting being the same...

Win percentages aren't the end-all, they're a stepping stone on the way to calculating EV.
I'm sure that's true but, like I said, for whatever reason, just straightforward "how many times can I expect to win this hand out of a 100" percentages seem to register with me better in more "practical", relatable terms (again, provided such above exceptions AREN'T skewing the percentages in a misleading fashion). But this is because I already know basic strategy (largely). It's true that if someone didn't already have a chart memorized, they'd be better off to first look at an EV chart than just a winning the hands chart or they might come across some hands where they'd get the wrong idea. But if you already know WHAT to do with a hand then EV tables just kinda seem confusing to me. At that point, after knowing WHAT to do, then I just enjoy knowing more of the general probabilities of that "what to do" - in other words, just how often will I win this hand now that I know "what to do". It just seems more basic, relatable and even curious, in a trivial way. It doesn't mean I try to memorize all these winning the hand odds but I'm just one who's always been fascinated with just the general probabilities, frequencies of the game. In that sense, such things are more interesting, straightforward to me than all the decimal rich integers mixed together on EV charts (even if the EV charts are more "important" to know).

For instance, I don't necessarily need to know (nor can I relate in practical everyday math terms) that such and such a hand has a -11% expectation to stand rather than a -9% if you hit it. This just simply tells me its a little better to hit. That's critical and helpful yes. But, in everyday math terms, I can "envision" it better if I just knew that such and such a hand I'm only likely to win 38% of the time if I stand or 44% if I hit. Now THAT I can relate to better (in general math terms). I don't have to "imagine" what precisely a -11% expectation really means (in money terms) or how much worse is that really compared to the -9 expectation (in money terms). But seeing things in a "black and white" 38% or 44% is easier to everyday sense. I know, without even thinking, what something 44% means, odds-wise. It just means what it looks like - I should win about half the time but isnt quite a winning situation, regardless which one I do. I still know the hitting percentage is the "correct" one to play. Of course it isn't difficult, necessarily, to get this same idea from using the EV tables BUT it doesn't hit the mind in as straight-forward and relatable a way. You have to kinda just "guess" that a -9% expectation isn't TOO bad, it's probably about, oh.... maybe 40 times out of a 100 I'll win this hand here??? And in the minimum of cases where just a "winning the hands odd" chart would lie to you (doubling vs hitting or surrender), I'd think one could easily color code or note that in this particular percentage, don't pay attention to it, just do this instead...). In effect, such a winning the hand chart could STILL be an EV chart (if these corrections/color codes were added). **Note: I used the 38% and 44% in relation to the -9 and -11 EV numbers as pure random illustration purposes. I actually have no idea if 38% and 44% translate from the EV numbers I made up. That's the whole problem!!! lol. But at any rate, this is just all my opinion and babbling. I'm sure most people don't have any problem at all with EV charts, or maybe just don't like/care what winning the hand percentages are, even just for trivial purposes. It probably just comes back to me and my poor math mind again... ;)
 
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k_c

Well-Known Member
Fun_at_21 said:
Thanks a lot! I was wondering if there was just a relatively simple "formula" for just figuring percentage odds from the EV tables. Now I can do these (hopefully) myself and even see where the factors are coming from. Of course, when I say I'm after simple winning the hand odds in "5th grade" terms I'm pretty much being serious! lol. So we'll see how I do with this math here. But it doesn't look difficult. No squares, piis, brackets anywhere so hopefully I can survive it... ;)

Now, just to verify that I'm doing NON-Ten or Ace dealer upcard calculations correct....

In Sanford Wong's Professional Blackjack book, for practice purposes, I'm trying to find what the odds of winning a 2,2 versus a 2 hand is if I HIT it. I'm under the 6 deck, S17 tables, pg 320. It gives an EV of -.115. So using the first formula in the post you wrote for non dealer blackjack possibility upcards, I can just take...

(-.115 + 1) / 2

and I get...

(0.885) / 2 = 0.4425

So my average odds of winning a 2,2 versus a 2 hand by hitting it would be roughly 44% correct?

Figuring out the unconditional EV's for dealer Ten & Ace cards looks slightly more complicated but I'm sure I can do those as well when I make it up to those two upcards...

But I really appreciate these! Thanks a lot! This should work...
That's correct. Hitting 2-2 v 2 wins ~44.3% of the time. :)
Standing wins ~35.4%.
 

Fun_at_21

Well-Known Member
k_c said:
That's correct. Hitting 2-2 v 2 wins ~44.3% of the time. :)
Standing wins ~35.4%.
Cool! Glad to see I'm on the right track, at least for the non-ten/ace dealer upcard calculations...

Maybe I can do a little bit of math after all. Who knew... ;)
 

Fun_at_21

Well-Known Member
Hmmm.. as a sidenote to the 2,2 versus 2 *default* example that I've always been using throughout this thread, does anybody know how I should be interpreting/calculating Stanford Wong's EV split tables in his Professional Blackjack book? For those who have the book, I'm on pg. 320 (the dealer's upcard of 2).

I like his EV tables, in particular, because they don't just give you an average EV when doing the best move but gives an EV for all four possible actions (right or wrong). However, the two rows for splitting has me a bit confused. For the 2,2 versus a 2, scanning all options makes it look like hitting is the correct play (it has the lowest negative expectation at -.115). Unless you look at the very last row under splitting (the dbs labeled row). It, in fact has the lowest negative expectation at -0.83.

Is the first row for splitting on these tables (the one labeled reg) just telling me that my expectation is a -.148 if I split the 2's, draw to them to the highest total I can but fail to not double when I may have landed on a 9, 10 or 11? Whereas, the second row (dbs) is giving me my "true" expectation of -0.83 if I split the 2's and always play them "perfectly" to best strategy, doubling if allowed etc?

Therefore, if doubling is allowed, these last rows (dbs) are what I should always be concerned with (it assumes I'm playing best strategy fully and correctly after the splits)?

Am I reading these particular tables right (for those who know what I'm looking at)?

I wonder if this is also what I was confused by earlier when callipygian was trying to explain his dual splitting charts to me. I may have answered my own questions. If I'm reading the above correctly???
 

k_c

Well-Known Member
Fun_at_21 said:
Hmmm.. as a sidenote to the 2,2 versus 2 *default* example that I've always been using throughout this thread, does anybody know how I should be interpreting/calculating Stanford Wong's EV split tables in his Professional Blackjack book? For those who have the book, I'm on pg. 320 (the dealer's upcard of 2).

I like his EV tables, in particular, because they don't just give you an average EV when doing the best move but gives an EV for all four possible actions (right or wrong). However, the two rows for splitting has me a bit confused. For the 2,2 versus a 2, scanning all options makes it look like hitting is the correct play (it has the lowest negative expectation at -.115). Unless you look at the very last row under splitting (the dbs labeled row). It, in fact has the lowest negative expectation at -0.83.

Is the first row for splitting on these tables (the one labeled reg) just telling me that my expectation is a -.148 if I split the 2's, draw to them to the highest total I can but fail to not double when I may have landed on a 9, 10 or 11? Whereas, the second row (dbs) is giving me my "true" expectation of -0.83 if I split the 2's and always play them "perfectly" to best strategy, doubling if allowed etc?

Therefore, if doubling is allowed, these last rows (dbs) are what I should always be concerned with (it assumes I'm playing best strategy fully and correctly after the splits)?

Am I reading these particular tables right (for those who know what I'm looking at)?

I wonder if this is also what I was confused by earlier when callipygian was trying to explain his dual splitting charts to me. I may have answered my own questions. If I'm reading the above correctly???
Hitting is the correct play for 2-2 v 2 when doubling after splitting is not allowed.
Splitting is the correct play when doubling after splitting is allowed.

A caveat to the method I gave for computing winning percentages is that it won't give a truly valid answer for doubling after splitting. This is another reason to use EV rather than winning percentage to arrive at a strategy.

I think it might work for splitting where doubling after splitting is not allowed but I'm not 100% sure. It should definitely work for stand, hit, and double though. However, as has been noted you can't compare doubling winning percentages to stand/hit winning percentages to validly arrive at the best strategy.
 

Fun_at_21

Well-Known Member
k_c said:
Hitting is the correct play for 2-2 v 2 when doubling after splitting is not allowed.
Splitting is the correct play when doubling after splitting is allowed.
Cool. Well at least I think I understand why the book has the two particular tables for it now. I guess I was just surprised at first when I saw the expectations for each action. I thought that splitting charts, in general, always assumed that you could double and were written with the expectations as such. That's why I was kinda surprised to see different expectations for splitting. Plus, I, personally, always assume that when splitting is called for that any "value" it has as a strategy move is already taking into account that you can double on it. That's why whenever I see 2,2 versus 2, I think split right away and not hit. But then again in some of the casinos I get a chance to play in they typically all have "doubling on splits" allowed so I kinda view it as a "common rule" though I see that not all places allow it. That's what kinda threw me off seeing TWO different split EV charts -I figured most just treated doubling as a given in their expectations.

A caveat to the method I gave for computing winning percentages is that it won't give a truly valid answer for doubling after splitting. This is another reason to use EV rather than winning percentage to arrive at a strategy.
So the formula you gave me for splitting will give a "winning the hand" odds but it just assumes that the split cards are simply drawn to but never enhanced by doubling correct?
 

Fun_at_21

Well-Known Member
mdlbj said:
My thoughts exactly, why not just know the BS for the table.
Well mostly because I already know it! ;) As stated, I'm not looking for these simple "winning the hand" percentages as a tool to learn basic strategy. And I'm the first to state that it really makes no sense for anyone to know how many hands out of a 100 does such and such win if they arent playing the right option to begin with. But my problem is: okay, I know (for the most part) what to do in every general hand vs dealer card but now, just for innocent curiosity, how many times out of a 100 can I expect to win this hand? It isn't going to alter my move playing basic strategy or anything but it's just "interesting" to know...

And I am using that chart that qfit provided. It's helpful and much appreciated. Although, for instance, on the 2,2 versus a 2 (39%), I'm now uncertain, because it doesn't particularly specify, if the percentage is assuming that the 2's are split and doubled upon (which is the best strategy move) or if it's assuming that 2,2 splits are NOT doubled (in which case hitting is the best strategy move and thus the 39% figure is the hitting odds of winning the hand out of a 100???)
 
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k_c

Well-Known Member
Fun_at_21 said:
Cool. Well at least I think I understand why the book has the two particular tables for it now. I guess I was just surprised at first when I saw the expectations for each action. I thought that splitting charts, in general, always assumed that you could double and were written with the expectations as such. That's why I was kinda surprised to see different expectations for splitting. Plus, I, personally, always assume that when splitting is called for that any "value" it has as a strategy move is already taking into account that you can double on it. That's why whenever I see 2,2 versus 2, I think split right away and not hit. But then again in some of the casinos I get a chance to play in they typically all have "doubling on splits" allowed so I kinda view it as a "common rule" though I see that not all places allow it. That's what kinda threw me off seeing TWO different split EV charts -I figured most just treated doubling as a given in their expectations.



So the formula you gave me for splitting will give a "winning the hand" odds but it just assumes that the split cards are simply drawn to but never enhanced by doubling correct?
If all you had to worry about was whether to stand or hit then winning percent is just as valid as EV in determining a strategy. Doubling is different. You are giving up going for the best winning percentage in favor of adding an additional bet at a possibly lower winning percentage but a greater EV.

When no doubling after splitting is allowed all of the split hands are either hit or stand so the formula I gave could work in that case. When doubling after splitting is allowed then some doubled hands would be mixed in with the hit and stand hands. That's perfectly fine when computing EV but the resulting EV will have lost its relationship to winning percent.

In a nutshell the reason EV is used is that it can account for doubling and the additional payout odds for a player blackjack. Winning percentages simply can't do this.
 

Fun_at_21

Well-Known Member
k_c said:
In a nutshell the reason EV is used is that it can account for doubling and the additional payout odds for a player blackjack. Winning percentages simply can't do this.
True. Hopefully I haven't given the impression I think they can (although I fear that's how I'm sounding...) ;) . But see, even with blackjacks, several charts anywhere commonly will list, even though it's not necessary to know, what the general odds are you'll get a blackjack, in "out of 100" terms. Isn't it roughly 7% or something, from the top of any deck? That's really the kind of questions (just about all hands and all decisions for them) that I'm innocently curious about. Nothing more. It still just seems that even with the complexities of splitting, whether doubling afterwards or winning one hand while losing the other etc that if it IS, in fact, the most advantageous move to make for a particular hand as a whole, then one could still determine (somehow) how many times out of a 100 is that advantageous scenario going to occur per hand(s) with that hand, on average? Again, it isn't necessary to know. If I know it's the best move then I should just do it (which I do) and not worry about it. But I still find it curious...

Nonetheless, I do appreciate all the help!
 
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k_c

Well-Known Member
Fun_at_21 said:
True. Hopefully I haven't given the impression I think they can (although I fear that's how I'm sounding...) ;)
Not at all. Basically you want how often a given strategy wins.

Fun_at_21 said:
But see, even with blackjacks, several charts anywhere commonly will list, even though it's not necessary to know, what the general odds are you'll get a blackjack, in "out of 100" terms. Isn't it roughly 7% or something, from the top of any deck? That's really the kind of questions (just about all hands and all decisions for them) that I'm innocently curious about. Nothing more.
It's a simple calculation:
pBJ=(prob of being dealt ace followed by ten)+(prob of being dealt ten followed by ace)
pBJ=(aces/(total cards)*tens/total cards-1)+tens/(total cards)*aces/total cards-1)
pBJ=aces*tens/(total cards)/(total cards-1)*2

Fun_at_21 said:
It still just seems that even with the complexities of splitting, whether doubling afterwards or winning one hand while losing the other etc that if it IS, in fact, the most advantageous move to make for a particular hand as a whole, then one could still determine (somehow) how many times out of a 100 is that advantageous scenario going to occur per hand(s) with that hand, on average? Again, it isn't necessary to know. If I know it's the best move then I should just do it (which I do) and not worry about it. But I still find it curious...
The formula I gave for splitting is a shortcut. If doubling after splitting is allowed the shortcut won't apply as long as least one split hand is doubled. You'd need to go through all of the possibilities - 2-2 v 2 assume single split, (no resplit):

2-ace prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-2 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-3 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-4 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-5 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-6 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-7 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-8 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-9 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-ten prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?

You'd sum the (hand prob for each of the 10 possible hands)*(winning percent) to get the probability of a single split 2 winning versus a 2. This would just show how often it wins and not whether splitting is right or wrong.

Similarly, winning percent for doubling shows how often it wins, not whether it's right or wrong.

In contrast, winning percent for hitting when compared to winning percent for standing can show whether hitting or standing is best.

Fun_at_21 said:
Nonetheless, I do appreciate all the help!
Hope it helps :grin::)
 

Fun_at_21

Well-Known Member
k_c said:
Not at all. Basically you want how often a given strategy wins.
Pretty much yes, for any 2 card hand versus every upcard in "out of 100" terms although I'm not necessarily looking for winning the "hand" out of 100. As stated, this doesnt always give right answers. I think I might need to re-phrase my dependence on saying "winning the hand" all the time, which leads to a little further clarification below (although probably still as confusing) lol...


It's a simple calculation:
pBJ=(prob of being dealt ace followed by ten)+(prob of being dealt ten followed by ace)
pBJ=(aces/(total cards)*tens/total cards-1)+tens/(total cards)*aces/total cards-1)
pBJ=aces*tens/(total cards)/(total cards-1)*2


The formula I gave for splitting is a shortcut. If doubling after splitting is allowed the shortcut won't apply as long as least one split hand is doubled. You'd need to go through all of the possibilities - 2-2 v 2 assume single split, (no resplit):

2-ace prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-2 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-3 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-4 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-5 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-6 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-7 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-8 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-9 prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?
2-ten prob=?, strategy=?, win percent for strategy v 2 with another 2 removed=?

You'd sum the (hand prob for each of the 10 possible hands)*(winning percent) to get the probability of a single split 2 winning versus a 2. This would just show how often it wins and not whether splitting is right or wrong.
Wow, Thanks! I'm not positive how far I'll get trying to calculate these myself but I may give it a try with a couple. We shall see how it goes...
Similarly, winning percent for doubling shows how often it wins, not whether it's right or wrong.
Logically, I know this is true. I think my problem might be in being too insistent on using the "winning the hands" phrase. I think what I'm really wondering (in just the case of doubling and doubling after splitting) is can't it still be stated somehow (in out of 100 terms) how many of those 100 2,2 versus 2 hands will result in us enjoying the benefits of the EV (make more money) by doubling after we split? I think that's more what I'm asking and actually is a very differen't question (i think) than "how many hands will a 2,2 versus a 2 win if you split then double. As we know, the "hands" won may be misleading. So I'm really asking how many hands "win" out of a 100 if we split and double as we should but not LITERALLY how many "hands" themselves win. If that makes any sense? ;) In other words, how many hands out of a 100 will we win more money by doing the right thing (doubling after splitting), not necessarily how many of the 100 "hands" will we win by doubling after splitting...

I think the problem is that I suspect EV deals exclusively with just looking at "how much money is being gained/lost overall and in the long run, rather than doing what I'm doing and trying to look at EV as something that can be evaluated on just one hand of those 100 at a time and then accumulated for an overall yet "correct" expectation out of a 100, 1000, 10,000 etc. Perhaps EV, by definition isn't (or can't really) be calculated in such ways, in which case I may in fact just be asking illogical questions...
 
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k_c

Well-Known Member
Fun_at_21 said:
Pretty much yes, for any 2 card hand versus every upcard in "out of 100" terms although I'm not necessarily looking for winning the "hand" out of 100. As stated, this doesnt always give right answers. I think I might need to re-phrase my dependence on saying "winning the hand" all the time, which leads to a little further clarification below (although probably still as confusing) lol...




Wow, Thanks! I'm not positive how far I'll get trying to calculate these myself but I may give it a try with a couple. We shall see how it goes...


Logically, I know this is true. I think my problem might be in being too insistent on using the "winning the hands" phrase. I think what I'm really wondering (in just the case of doubling and doubling after splitting) is can't it still be stated somehow (in out of 100 terms) how many of those 100 2,2 versus 2 hands will result in us enjoying the benefits of the EV (make more money) by doubling after we split? I think that's more what I'm asking and actually is a very differen't question (i think) than "how many hands will a 2,2 versus a 2 win if you split then double. As we know, the "hands" won may be misleading. So I'm really asking how many hands "win" out of a 100 if we split and double as we should but not LITERALLY how many "hands" themselves win. If that makes any sense? ;) In other words, how many hands out of a 100 will we win more money by doing the right thing (doubling after splitting), not necessarily how many of the 100 "hands" will we win by doubling after splitting...

I think the problem is that I suspect EV deals exclusively with just looking at "how much money is being gained/lost overall and in the long run, rather than doing what I'm doing and trying to look at EV as something that can be evaluated on just one hand of those 100 at a time and then accumulated for an overall yet "correct" expectation out of a 100, 1000, 10,000 etc. Perhaps EV, by definition isn't (or can't really) be calculated in such ways, in which case I may in fact just be asking illogical questions...
The problem is that when additional bets are added after the hand is dealt, money won per 100 units of initial bet and hands won out of 100 measure different things.

An EV of +10% means on average you will win 10 units for every 100 units that are bet before the round begins. An EV of -10% means on average you will lose 10 units for every 100 units that are bet before the round begins.

This includes doubles, splits, blackjacks, and everything else.

If you continue to logically look at it I think you'll see it.
 

Fun_at_21

Well-Known Member
k_c said:
The problem is that when additional bets are added after the hand is dealt, money won per 100 units of initial bet and hands won out of 100 measure different things.

An EV of +10% means on average you will win 10 units for every 100 units that are bet before the round begins. An EV of -10% means on average you will lose 10 units for every 100 units that are bet before the round begins.
Yeah, I suppose it makes sense, money-wise. Still, seeing the above in bet terms for some reason just doesn't register as "nicely" as if we had general probabilities for just how often these EV's would occur out of a number of hands (if we did the same action on them for 100 times). But I guess nobody ever said that blackjack math has to be nice and simple. ;) I know we play blackjack with money but it still, for some reason, seems to confuse me more when I try to think of my "probabilities" in money terms. I don't know why but it just does. Like you said, I'm sure it just has to do with the idea of probabilities and money won being two different things (unfortunately).

Maybe this is what I get for just essentially being a basic strategy blackjack player who just simply enjoys the game, plays it for fun and just enjoys the general probabilitites of it rather than worrying too much about expecting money from it! It leads to all sorts of trouble lol ;)
 
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k_c

Well-Known Member
Fun_at_21 said:
Yeah, I suppose it makes sense, money-wise.
Maybe this is what I get for just essentially being a basic strategy blackjack player who just simply enjoys the game, plays it for fun and just enjoys the general probabilitites of it rather than worrying too much about expecting money from it! It leads to all sorts of trouble lol ;)
In that case just know that for a single unit even money pay bet where no additional units are added to the bet:

EV = 2 * (win percent) - 1 which is the same as saying
win percent = (1 + EV) / 2

This relationship between EV and win percent works 100% of the time for the condition described, assuming pushes count as 50% win and 50% loss.
 

Fun_at_21

Well-Known Member
k_c said:
In that case just know that for a single unit even money pay bet where no additional units are added to the bet:

EV = 2 * (win percent) - 1 which is the same as saying
win percent = (1 + EV) / 2

This relationship between EV and win percent works 100% of the time for the condition described, assuming pushes count as 50% win and 50% loss.
Thanks, that one I can remember pretty well, I think. I'll try putting it to use. At least I know I was able to do one correctly! ;)
 

Cardcounter

Well-Known Member
How often you win depends on how you play the hand if you have 2,2 vs a 2 and stay you will win about 35% of the time when the dealer busts. If you split them and hit them out till you have 13 or moreyou do a little better. If you just hit them out you might make hand giving you a better chance.
 

arnold

New Member
There are several good resources which cover this very topic, one of which is Revere's Playing Blackjack as a Business. The combination of 12 v 2, when hit will yield 4 more winning wagers / 100 than standing; 12 v. 3 will yield 1. The entire chart can be found on p.55. I believe that Blackjack Attack also covers this, as well as several of the more notable classics.
 
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