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how to calculate expected number of hands a player can split to in blackjack?

Posted on 2005-03-31
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Last Modified: 2006-11-18
Hey math experts!

here is 10 million iterations, counting the frequency of 1,2,3 splits occuring in a single deck bj game:

Histogram Results
One Split:
splits   :  587517   0.0587517
nonsplits:  9412483   0.9412483
Two Split:
splits   :  47262   0.0402387317588162
nonsplits:  1127278   0.959761268241184
Three Split:
splits   :  969   0.0205027294655326
nonsplits:  46293   0.979497270534467
 total iterations.
1112236146 is the stop time

now, what would the "expected number of hands the player could split to" be? How do I calculate the EV given the frequency/histogram results?


Thanks!
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Question by:sapbucket
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23 Comments
 
LVL 3

Author Comment

by:sapbucket
ID: 13675546
expNumSplits = 1 * .0587517 + 2 * 0.0402387317588162 + 3 * 0.0205027294655326 ?    // times the number of split

= .200737351914

or maybe:

expNumHands = 2 * .0587517 + 3 * 0.0402387317588162 + 4 * 0.0205027294655326 ?     // time the number of hands

=.32023051313858


but for some reason I was expecting it to be > 1.

IF it IS approx. .3202 than we can conjecture that given 4 decks that the expected number of hands we could split would be 1.28 hands. ?

anyway, trying to figure this out :)
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LVL 31

Expert Comment

by:GwynforWeb
ID: 13675564
Your numbers do not look right to me, but assuming they are then if you have a  histogram (where the values for 1 repesents the probablity of one and  no further splits, same for 2 and 3) then the expected numbers of splits per hand is

1*0.0587517 + 2*0.040238 +3* 0.020502
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by:sapbucket
ID: 13675691
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LVL 31

Expert Comment

by:GwynforWeb
ID: 13675763
that is for an nfinite deck which is very different. This is failry easily calculated for a fixed a single palyer playing.,  but there are many variables effecting this, eg the  number of decks and the number of other players.
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Author Comment

by:sapbucket
ID: 13675792
how about for one player, and one deck?
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LVL 27

Expert Comment

by:d-glitch
ID: 13675873
I guess it's okay to do 10,000,000 iterations.  But it pretty easy in this case to calculate the probabilities.

You can split any pair.  So what's the probablity of getting a pair with a single deck?

It doesn't matter what your first card is.  The probability of getting a pair in two cards is 3/51 ==> 0.058824

With four decks the probabiltiy of getting a pair in two cards is 15/207 ==> 0.072464

You definitely don't multiply  0.058824 by 4.
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LVL 3

Author Comment

by:sapbucket
ID: 13675958
if the probability is 0.058824, than how many hands can I expect to split? (single deck, single player)

how do i go from the probability to the expected number of hands that a player can split?
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LVL 27

Expert Comment

by:d-glitch
ID: 13676021
With a single deck the probability of splitting either your first or second split pair but not both is:

               2 * 3/51 * 2/50 * 48/49 ==> 0.004610


With a single deck the probability of splitting both split pair is:

               3/51 * 2/50 * 1/49 ==> 0.000048
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LVL 27

Expert Comment

by:d-glitch
ID: 13676059
Multiply the number of hands by 0.058....

3/51 is exactly 1/17.  So divide the number of hands by 17.
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Author Comment

by:sapbucket
ID: 13676113
"So divide the number of hands by 17."

you mean the average number of hands per single deck with a single player?

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LVL 31

Accepted Solution

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GwynforWeb earned 2000 total points
ID: 13676136

    split one then not split again  =3/51*48/50=0.0564705882 =>1 split

    split both again 3/51*2/50*1/49= 0.0000480192077         => 3 Splits

    split one again then 1 again
                  = 3/51*2/50*48/49*1/48 =0.0000480192077    => 3 splits

    split one again then not  1 again
                  = 3/51*2/50*48/49*47/48 =0.00225690276     => 2 splits

Totals:-

1 split = 0.0564705882

2 splits = 0.00226690276

3 splits= 2*0.0000480192077=0.0000960384154

Expected value = 0.0564705882 + 2*0.00226690276 +3*0.0000960384154

                     =0.061292509

ie you will split about 6% the number of hands dealt you
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Expert Comment

by:GwynforWeb
ID: 13676185
..dividing by 17 assumes that 2 and 3 splits are so infrequent that they need not be counted, which can only be established through further anlaysis.
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LVL 27

Expert Comment

by:d-glitch
ID: 13676191
Using my numbers:

expNumSplits =   ( 0.058824  +  0.004610  +  0.000048) x (Number of hands played)


You don't multiply the 2nd split probabilities by 2 because you have already counted one of them in splitting your original pair.

Same for the 3rd split.

Our numbers agree for the 1st and 2nd splits.  But they are off on the 3rd.  I don't see the problem off hand.
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LVL 31

Expert Comment

by:GwynforWeb
ID: 13676239
I have been carefull not to recount splits, my probablities add up to one.
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Expert Comment

by:GwynforWeb
ID: 13676259
..I mean 3/51
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LVL 27

Expert Comment

by:d-glitch
ID: 13676302
I agree.  Gwyn's analysis is almost certainly correct.

I didn't recount splits, but I missed one of tne of the two ways to get three splits.
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LVL 3

Author Comment

by:sapbucket
ID: 13676446
Histogram Results
One Split:
splits   :  849616   0.0521803623002999
nonsplits:  15325888   0.941260979564673
Two Split:
splits   :  69751   0.00428385582522954
nonsplits:  1766123   0.108468929500966
Three Split:
splits   :  1515   9.3045857051838e-005
nonsplits:  69751   0.00428385582522954
16282294 number of hands dealt.
1120167040 is the stop time

ok, but in this simulation I took care to REMOVE the histogram results from a prior split. for example:

expNumSplits =   (1* 0.0521803623002999  + 2* 0.00428385582522954  + 3* 9.3045857051838e-005) x (Number of hands played) would be correct

and now I go eat some Sushi and think about what you guys have shared with me - thanks!
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LVL 3

Author Comment

by:sapbucket
ID: 13676853
expNumSplits = .0610272115    (*this SHOULD be slightly lower than calculated because I remove a single card from each deal: the dealers upCard)

ok, correct answers! That's what I like!

Looks as if the simulation is very close to Gwyn's calculation:

((0.061292509 - .0610272115) / 0.061292509) * 100 = .4 % diff (or 99.6% similar)

Thanks again!
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LVL 84

Expert Comment

by:ozo
ID: 13678371
split at least one = 3/51  (two hands in play)
split at least two = 3/51*(2/50+48/50*2/49)  (three hands in play)
split three = 3/51*(2/50*(1/49+1/48+1/47)+48/50*2/49*(1/48+1/47+1/46)) = 0.000294268493706456
split exactly two = 3/51*(2/50+48/50*2/49)-0.000294268493706456=0.00436359465155165
split exactly one = 3/51-3/51*(2/50+48/50*2/49)=0.0541656662665066
Expected value = 1*0.0541656662665066+2*0.00436359465155165+3*0.000294268493706456 = 0.0637756610507293
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LVL 84

Expert Comment

by:ozo
ID: 13678405
split three = 3/51*(2/50*(1/49+48/49*1/48+48/49*47/48*1/47)+48/50*2/49*(1/48+47/48*1/47+47/48*46/47*1/46)) = 0.000288115246098439
Expected value = 1*0.0541656662665066+2*0.00436974789915966+1*0.0541656662665066 = 0.0637695078031213


 
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LVL 84

Expert Comment

by:ozo
ID: 13678414
Expected value = 1*0.0541656662665066+2*0.00436974789915966+3*0.000288115246098439 = 0.0637695078031213
0
 
LVL 3

Author Comment

by:sapbucket
ID: 13679157
So, that is the expected value (approx. 6%). But how do I translate that into how many splits I can expect for a single player playing a single deck BJ game?

Wouldn't I have to know the number of hands per deck? And then the calculation becomes:

numSplitHands = expNumSplits * numHandsPerDeck

Or is there a better way? Because numHandsPerDeck isn't such an easy thing to calculate.

Thanks for any advice!
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LVL 84

Expert Comment

by:ozo
ID: 13679604
split at least one = 3/51  (two potentially splitable hands in play)
split at least two = 3/51*2/50  (three potentially splitable hands in play)
split at least two = 3/51*48/50*2/49  (two potentially splitable hands in play)
split three = (3/51*2/50)*(1+1+1)/49 + (3/51*48/50*2/49)*(1+1)/48 = 0.000240096038415366
split exactly two = 3/51*(2/50+48/50*2/49)-0.000240096038415366 = 0.00441776710684274
split exactly one = 3/51-3/51*(2/50+48/50*2/49)=0.0541656662665066

Expected value = 1*0.0541656662665066 + 2*0.00441776710684274 + 3*0.000240096038415366 = 0.0637214885954382
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