Blackjack Calculation help

So, my eventual goal is to dynamically calculate Win %, but I don't want to get ahead of myself so I'm starting out with a theoretical hand in order to form a good grasp of the calculations, and any help would be greatly appreciated!

FOR THIS THREAD I SIMPLY WISH TO STICK WITH THE EXAMPLE AND NOT DEVIATE FROM IT

I simply want to correctly calculate a dealer busting while showing a 10. It'll be the first hand of a single deck shoe (no burn card). The player has a 10 and a 6, while the dealer is showing a Jack with "X" as the under card. The dealer does not have blackjack, and the player stays. What are the chances the dealer breaks?

Numbers I have so far:

X = Dealer Stands = %53.06 = 26/49 = (4-7's,4-8's,4-9's,3-10's,3-J's,4-Q's,4-K's)
X = Dealer Hits = %38.78 = 19/49 = (4-2's,4-3's,4-4's,4-5's,3-6's)

Now if the dealer is to hit:

X = 2(12) = %29.17 = 14/48
X = 3(13) = %37.50 = 18/48
X = 4(14) = %45.83 = 22/48
X = 5(15) = %54.17 = 26/48
X = 6(16) = %58.34 = 28/48 (2-6's,4-7's,4-8's,4-9's,3-10's,3-J's,4-Q's,4-K's)

What is the process, or formula, to calculate the dealer bust % in all situations (perhaps multiple hits) for this scenario?!
I'm guessing I'll be using conditional probability somewhere, but I don't know where or how, if I do.

You need an appropriate recursive algorithm that goes through all possible combianations.
There are way too many combinations just to put them in a excel spreadsheet.
It is especially diffifuclt to do it in an excel worksheet if are using fixed number of decks where you have to take into account the depletion of a rank for each hit.
If you are doing infinite deck is a lot easier and you can work backwards.
From 16 the probabilities are
Bust 8/13

For 15 the probs are
16 1/13
Bust 7/13
So it is 7/13 + 1/13 x 8/13 = 99/169

For 14 the probs are
15 1/13
16 1/13
Bust 6/13
So its is 6/13 1/13 x 8/13 + 1/13 x 99/169 =?? and so on

For finite decks you can find the algorithm probably on Wizzzard site or some other BJ site or book.

Quote: AceTwo

There are way too many combinations just to put them in a excel spreadsheet.

It can be boiled down to a few thousands, so technically it could easily fit into an excel spreadsheet.

However getting the dealers probabilities is only the first step towards exact EVs.

Thank you BOTH for replying, and MangoJ, I have looked at your spreadsheet and previous discussion (both briefly) on some of these calculations. I should probably look at them more, but I've always been more of a kinesthetic learner. I do plan on dealing with finite decks and and getting exact, dependent, calculations, but I do not plan on making an Excel sheet. Rather I want to write a program myself (I also noticed that you had posted some C stuff MangoJ, again, briefly), although using VBA with Excel could give me some practice/use with VB.

Anyway, for some reason things aren't clicking for me in the formulation for probabilities, and I just want to get that out of the way before worrying about recursion and dealing with the problems of a finite amount of decks.

For the dealer having 16 then (in my example):
It would simply be 28/48 for a bust.

It is at 15 (and/or calculating another hit) that eluded me, and whether or where to multiply and/or add. Tell me if I am wrong in my understanding at any point in this discussion.
There is a 26/48 chance that I would bust. There is also a 4/48 chance that I would get to 16, and if I did there would then be a 28/47 chance to bust, bringing the calculation for 15 to be:
26/48 + 4/48 x 28/47 = 667/1128?

If this is correct do I take all the probabilities of each possible down card and factor them together somehow? Please forgive my lack of understanding the math, but I do wish to work it out.

Quote: bjdealer

...For the dealer having 16 then (in my example):
It would simply be 28/48 for a bust.

It depends on what the dealer's 16 is made of. Assuming Player has X 6 and Dealer has X 6 then there would be 48 cards left, 28 of which bust. But say the dealer has X 4 2 then the probability is different.
This is why you need recursion.

As it happens the dealer can only have six additional cards (worst case is X 2 A A A A any), so you could work out all the permutations on a spreadsheet.

But in general (and I've never done it) you need to work out all the various states (e.g. X 6 vs X 3 3) and work them backwards to X6 vs X.

Best of luck

Quote: bjdealer

FOR THIS THREAD I SIMPLY WISH TO STICK WITH THE EXAMPLE AND NOT DEVIATE FROM IT

I want to work out the possibilities in one situation first if I could. Recursion is not a problem. Please see above where I say the dealer is showing a Jack, while the player has a 10,6. In your post however charliepatrick, how would I factor the multiple hits in terms of probability?

All in all I think that's all I really need to know...

If 12
x= the probability of breaking
If 13
y= the probability of breaking
if 14
z= the probability of breaking
if 15
j= the probability of breaking
if 16
k= the probability of breaking

What do I do with x,y,z,j,k in relation to the total probability of busting when the dealer is showing a 10?

Right now I'm thinking I'd have to start at twelve and go through each possible situation, but I still don't know what to do with the result of each possible situation. How do I relate each possible outcome to each other in order to get the total probability of busting?

Quote: bjdealer

Tell me if I am wrong in my understanding at any point in this discussion.
There is a 26/48 chance that I would bust. There is also a 4/48 chance that I would get to 16, and if I did there would then be a 28/47 chance to bust, bringing the calculation for 15 to be:
26/48 + 4/48 x 28/47 = 667/1128?

Basically yes, but you are neglecting card removal effects. Meaning that you calculate dealer bust probabilities for an infinite deck.

If you have a dealers 16, there is not always a 28/48 probability he will bust. It depends on the remaining deck distribution, and thus on all cards already removed from the deck.
Meaning a dealers 16, composed of T+6 has a different probabiltiy to bust than a dealers 16 composed of 3+4+5+4. In the first case there are 30 of 50 cards giving the dealer the bust, while for the second example there are 32 of 48 cards giving the bust.


In essence, what you need is the probability to bust a drawing hand with cards XYZ removed, by considering each drawing possibilities and subsequent bust probabilities *with the additional drawn card removed*.
If Bust(N, XYZ) is the probability to bust the dealers hand with total N, and cards XYZ removed, and Draw(i, XYZ) the probability to draw card i with XYZ removed, you would need to (recursively, or by other means) calculate:

Bust(N, XYZ) = Draw(A, XYZ) * Bust(N+A, XYZA) + Draw(2, XYZ) * Bust(N+2, XYZ2) + ... + Draw(T,XYZ) * Bust(N+T, XYZT)

See below some code (basically this is Code from Griffins book) in Basic that calculates Dealer Probabilities for Stand S17 game.
This goes recursively through all possible dealer combinations for all Upcards to calculate the prob of each dealer total (including the bust totals of 22-26)
The code also handles the Soft Hands


10 For NextCard = 1 To 10
20 If Rank(NextCard) = 0 Then GoTo 270
30 ProbNextCard = ProbNextCard * Rank(NextCard) / TotalCards
40 Rank(NextCard) = Rank(NextCard) - 1
50 DealerTotal = DealerTotal + NextCard
60 If DealerTotal > 16 Then GoTo 130
70 If (DealerTotal - 11) * (DealerTotal - 7) > 0 Then GoTo 170
80 If (NextCard - 1) * (DealerUpcard - 1) = 0 Then GoTo 150
90 For L = 1 To DealerNoCards
100 If C(L) = 1 Then GoTo 150
110 Next L
120 GoTo 170
130 ProbDealTot(DealerTotal) = ProbDealTot(DealerTotal) + ProbNextCard
140 GoTo 240
150 ProbDealTot(DealerTotal + 10) = ProbDealTot(DealerTotal + 10) + ProbNextCard
160 GoTo 240
170 TotalCards = TotalCards - 1
180 DealerNoCards = DealerNoCards + 1
190 C(DealerNoCards) = NextCard
200 GoSub 10
210 NextCard = C(DealerNoCards)
220 DealerNoCards = DealerNoCards - 1
230 TotalCards = TotalCards + 1
240 Rank(NextCard) = Rank(NextCard) + 1
250 DealerTotal = DealerTotal - NextCard
260 ProbNextCard = ProbNextCard * TotalCards / Rank(NextCard)
270 Next NextCard
280 Return

And this is an explanation of what this code does which might help you understand how the algoritm to calculate Dealer probs works:

10 NextCard 1 to 10
20 If run out of specific Rank GOTO 270
30 Probability of sequence of cards = Prob x Cards of Rank/Total Cards
40 Deduct 1 card of the specific rank
50 Calculate Dealers Total : by adding next card
60 If Dealer Total is > 16 GOTO 130
70 If Dealer Total is >11 or <than 7 GOTO 170
80 If DealerUpCard is 1 OR NextCard is 1 GOTO 150
90 For All cards 1 to Last Card
100 If any card is 1 GOTO 150
110 Next Card to check for 1
120 GOTO 170
130 Calculate Probability of Dealers Total over 17 : Prob + Prob of Sequence of cards
140 GOTO 240
150 Calculate Probability of Dealers Total over 17 by treating Ace as 11: Prob + Prob of Sequence of cards
160 GOTO 240
170 Deduct 1 card from the Total number of cards
180 Add 1 to Dealers no of cards
190 C(DealerNoCards)= Next Card ie is the Last NextCard before the Sub at 200 loops
200 GOSUB 10
210 Next card is the last next card before Sub at 200 (ie from line 190)
220 Deduct 1 from Dealers No of Cards
230 Add back one to the Total number of cards
240 Add back 1 card of the specific rank
250 Go back to the Previous Dealer Total by deducting the next card used
260 Recalculate probabilities of sequence of next cards before next card : Prob x Total/Cards of Rank
270 NEXT Next card (ie go back to 10)
280 Return to 210 (end of Sub at 200)

Quote: MangoJ

Bust(N, XYZ) = Draw(A, XYZ) * Bust(N+A, XYZA) + Draw(2, XYZ) * Bust(N+2, XYZ2) + ... + Draw(T,XYZ) * Bust(N+T, XYZT)

@MangoJ
This is awesome! Correct me if I am wrong, but it looks as though this only takes into account one hit percentages? How would I step into multiple hits to get the total, that is, if my assumption is correct lol? Just add them?

EDIT: Nevermind. It looks as though multiple hits would be factored in with the probability of busting, thanks.

@AceTwo
Found the "The Theory of Blackjack" book by Griffin you suggested. We actually have a copy at work lol. I'll have to check it out.

Quote: bjdealer

@MangoJ
@AceTwo
Found the "The Theory of Blackjack" book by Griffin you suggested. We actually have a copy at work lol. I'll have to check it out.

Check chapter 11 'Some techniques for BJ Computations'. The code is on page 158.
Griffins book, despite being so old, is one of the best books for anyone interested in finding out the mathematics behind BJ and counting including the formal theory behind the whole thing.

Quote: bjdealer

It looks as though multiple hits would be factored in with the probability of busting, thanks.

Yes in a way. The above formula is actually meant recursive. So to calculate the bust probability of some hand, you need the bust probabilities of all those hands the dealer might draw to, with a higher N. Eventually if N reaches 17 or above, these probabilities are trivial. Bust=0 for N<=21 and Bust=1 for N>21.

The only way the parameter XYZ affects this calculation is that of the draw. You should try to find out by yourself how to calculate drawing probabilities if the cards XYZ are removed from the deck. Putting all together, this gives you a simple mean (by using recursive programming) to calculate bust probabilties.

Same with all other probabilties(i.e. dealer standing on 17).

Note that this is not the most efficient way. In my personal blackjack analyzer all the ways the dealer draws his hand is compiled into a simple table, which is then weighted by the deck distribution to give efficient and exact results for any number of decks.