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

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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Commented:

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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Author Commented:
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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Commented:
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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Author Commented:
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Commented:
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 Commented:
how about for one player, and one deck?
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Commented:
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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Author Commented:
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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Commented:
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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Commented:
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 Commented:
"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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Commented:
..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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Commented:
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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Commented:
I have been carefull not to recount splits, my probablities add up to one.
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Commented:
..I mean 3/51
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Commented:
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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Author Commented:
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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Author Commented:
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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Commented:
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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Commented:
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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Commented:
Expected value = 1*0.0541656662665066+2*0.00436974789915966+3*0.000288115246098439 = 0.0637695078031213
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Author Commented:
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.

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Commented:
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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