References
Griffin. Some Techniques for Blackjack Computations. Chapter 11, TOB.
Hawkins. An Application to Resplitting in Six Deck Blackjack.
Farmer. Exact Expected Values for Pair Splitting.
Cacarulo.Expected Value and Hand Probabilities for.
Wong. PBJ, Appendix E.
Resplitting, Myths and BJ Gurus
Computing exact expected values, when the resplitting rule is in effect, has never been an easy task.
As Griffin tell us in the introduction of Chapter 11:
This has been due largely to a concern for playing the subsequently derived hands optimally, depending on the cards used on earlier parts of the split.
Wong gives us his rationale also, why he didn?t bother with the exact calculations in the appendices of his famous book, while skipping the unpleasant work at the same time:
Thorp numbers assumes resplits to four hands, and I assume no resplits. If splitting is the best way to split a pair, the tables will say so no matter whether the calculations assume no resplits, unlimited resplits, or something in between. Therefore, the tables yield good advice on splits and resplits no matter how the calculations on resplits are done.
Given that his logic looks undisputable, most of the card counters seem to be unconcerned for something which knowledge, won?t add EV ($$$$$) to their pockets. Demand and supply are perfectly correlated in our market economy. And besides and more important the stuff is somehow risk averse. Not a lot of candidates to bite on this!
Griffin?s algorithm to compute resplitting expected values
The player expectation from repeated pair splitting is now given by
(1081*2*A)/1176 + (90*3*B)/1176 + (5*4*C)/1176
where A, B, C are conditional expectations, while 2, 3 and 4 are the number of hands obviously.
The question is. Where do the other numbers come from? Out of his magician?s hat?
Let?s take the first fraction 1081/1176 which is the probability of splitting a pair only.
Example:
6-6 vs 2 (for simplicity we will assume the dealer?s up card is a different one).
We don?t want any further pairs to be drawn, right? The elementary calculations of sampling without replacement probabilities shall guide us. Disregard the two remaining sixes here, and you have only 47 cards at your disposal, to accomplish your objective of no more pairs:
(47/49)*(46/48) = 2162/2352 = 1081/ 1176 we got it!
But wait a moment. What about inferring on the search for a general formula?
47/49 = 49/49 ? 2/49 = 1 ? 2/49 the same for the other
46/48 = 48/48 - 2/48 = 1 ? 2/48 now we?re ready for our first derived formula:
Probability that no further pair will be drawn, after the first one
P = (1- (t - 2)/ (n - 3)) * (1 - (t -2)/ (n - 4))
with n = 52 and t = 4 for single deck calculations, obviously. We aren?t ten-splitters, disregard t = 16.
Probability that no further pair will be drawn, after the second one
We?ll need here his derivation N(I) = F(2, I-2) and quickly solving for 3 hands:
N(3) = F(2, 1) = 2 his ?magic number? for 3 hands.
The rest shouldn?t be a problem for the smart reader. As above:
2*2/49*(47/48)*(46/47)*(45/46) or the more academically:
P = 2*(1-(t-2)/ (n-3)) *(1-(t-3)/ (n-4)) *(1-(t-3)/ (n-5)) * (1-(t-3)/ (n-6))
Assuming that we are practical oriented men, let?s take the standard rule of resplits to a maximum of four, like is the case for our single deck player by force. The derivation of P(4) is as easy as:
P(4) = 1 - P(2) - P(3) given that Sum P(S) = 1, we can use it as an accuracy check, too.
Skipping the unlimited respite rule, we?re ready now to construct our practical P(I) table!
Splitting probabilities to a maximum of four
I Single Deck Double Deck Six Deck[/b]
2 .919217687 .884158416 .862459547
3 .076530612 .101702823 .115126809
4 .004251701 .014138761 .022413644
w(I) 2.085034014 2.129980345 2.159954097
where w(I) = mean of I
Now what we need is a weighted conditional expectation for the three cases that our player
can be confronted with, namely 2, 3 or 4 hands against the dealer. Something like this:
EV = (2*E(2)*P(2)) + (3*E(3)*P(3)) + (4 *(E(4)*P(4))
Since we have already computed the P(I) in the above table, all we need is to write a short subroutine to perform the calculations and avoid us the grind of the pocket calculator. Let the computers do the tedious work!
Extracting conditional expectations with the aid of a CA.
a) Two hands with no resplits.
These calculations are employed when they don?t let us resplit, like is the case with aces. Here theory states:
The expectation for each one of the split hands is the same.
The rationale behind this is that for any drawing sequence that we can imagine to complete one of the hands, there is also the same probability for the other hand to obtain the same sequence of cards in a sort of mutual interchange with his partner the other one of the pair.
b) Two hands with resplits.
We will follow Griffin?s recommendation like in A (page155), or in plain English, we?ll exclude the possibility the player drawing any card from the same value like his already split one. That will help us to calculate the conditional expectation of hand two, that we?re going to label it E(2)
c) Three hands with resplits.
Same procedure, same arguments. We can get E(3) now.
d) The last one.
Just playing one hand against the selected dealer?s up card from an initial pack where the other three split cards have been removed. That?s a 49 card packet. We can get E(4).
An example
2-2 vs 5 (sd, h17, das, rsp =4)
Per hand expectation
2 hands .14056
3 hands .158641
4 hands .178192
Using the above mentioned EV formula we get then:
EV = .2978636 as our average expectation for the resplitting advantage. (look at the table below)
EV = .274 is Wong?s figure for the case of no resplits and the same rules. (PBJ, Appendix E)
SD, H17, DOA, DAS, RSP = 4, NRSA
rsp = 4 nrs
A-A vs 2 .565372
A-A vs 3 .612564
A-A vs 4 .668294
A-A vs 5 .73208
A-A vs 6 .755928
A-A vs 7 .540712
A-A vs 8 .406468
A-A vs 9 .289770
A-A vs T .1942514
A-A vs A .215092
9-9 vs 2 .2049048 .198514
9-9 vs 3 .2151048 .208952
9-9 vs 4 .3073799 .296818
9-9 vs 5 .4062425 .391108
9-9 vs 6 .4085991 .39366
9-9 vs 8 .207497 .195408
9-9 vs 9 -.0990434 -.1029334
8-8 vs 2 .1158342 .0833966
8-8 vs 3 .1938323 .1585036
8-8 vs 4 .2258029 .1915008
8-8 vs 5 .3278566 .289176
8-8 vs 6 .3508971 .311458
8-8 vs 7 .3036633 .250712
8-8 vs 8 -.0595246 -.0735092
8-8 vs 9 -.4011449 -.406326
8-8 vs T -.4470056 -.45183
8-8 vs A -.4642993 -.47032
7-7 vs 2 -.0543433 -.0728482
7-7 vs 3 .055979 .031028
7-7 vs 4 .1877603 .1568014
7-7 vs 5 .2238588 .1934716
7-7 vs 6 .2404408 .21009
7-7 vs 7 -.0512847 -.0657474
6-6 vs 2 -.0974663 -.1141838
6-6 vs 3 .0136899 -.0071997
6-6 vs 4 .1465257 .1188104
6-6 vs 5 .2736864 .239808
6-6 vs 6 .2006868 .1870428
4-4 vs 5 .2678592 .259384
4-4 vs 6 .297765 .288536
3-3 vs 2 -.1080983 -.112358
3-3 vs 3 -.0209689 -.0252378
3-3 vs 4 .160898 .1425266
3-3 vs 5 .3057737 .279208
3-3 vs 6 .3385111 .310796
3-3 vs 7 .0138643 -.0572504
2-2 vs 2 -.0362746 -.0395562
2-2 vs 3 .0400116 .029895
2-2 vs 4 .1316659 .1180946
2-2 vs 5 .2978636 .275354
2-2 vs 6 .335156 .310362
2-2 vs 7 .0138643 .0047523
Composition-dependence BS is assumed to play the hands.
Many years have passed since TOB was written.
Cacarulo seems to be the only researcher with the self confidence and courage to extract full set of expectations and publish them. Maybe we won?t be able to approach his degree of accuracy , but the work can be done. Griffin has showed us the path. This article should be understood and read in this context.
A Barry Meadow?s quote:
Once I told him that I couldn?t understand the mathematical symbols in his book. ?Don?t worry,? he jokingly assured me, ?I don?t understand them myself.? Blackjack Autumn
Enjoy!
Sincerely
Z
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