Calculating Split EVs Exactly


Well-Known Member
What happens when we split?

When we play out a split hand - we have the follwing 2 possible types of second card that can be
played onto a split card - either the paircard (P) or a non-pair card (N).

Since the number of hands that can result are limited by the rules, there are only two possible
hands that can be end points of a split card. Either a non-split card was dealt (N) - or the
maximum number of hands has be reached and any card including a paircard can be played (x) as the
second card of the hand. These hands do not need to consist of only 2 cards, however N and x refer
to which second cards are allowed. Note that "N" can refer to either the non-paircard or to the
hand which began with a non-paircard depending on the context.

A round for the player can consist of any of the possible different hands that will be played
until either the maximum number of splits is reached or until no free pair cards are left.

The probability of the hands and therefore the round is determined by the N and P cards played,
while the EV of the hands/round is determined by the type of hands that result, i.e. N and x.

The p(N) is simply 1-p(P). Determining the EV's of N and x hands is the hard part.

In the case of infinite splits, p(N), p(x), EV(N) and EV(x) are not dependent on previous hands
and therefore do not change. An additional notation will be introduced for finite splits to
account for the effects of removal on N and x hands. For reference, the different possible rounds
that can occur as determined by the maximum number of splits allowed is listed below:

Hand EV Prob

xx 2*EV(x) 1

NN 2*EV(N) p(N)*p(N)
Pxxx 3*EV(x) p(P)
NPxx EV(N)+2*EV(x) p(N)*p(P)

NN 2*EV(N) p(N)*p(N)
PNNN 3*EV(N) p(P)*p(N)*p(N)*p(N)
NPNN 3*EV(N) p(N)*p(P)*p(N)*p(N)
PPxxxx 4*EV(x) p(P)*p(P)
PNPxxx EV(N)+3*EV(x) p(P)*p(N)*p(P)
NPPxxx EV(N)+3*EV(x) p(N)*p(P)*p(P)
PNNPxx 2*EV(N)+2*EV(x) p(P)*p(N)*p(N)*p(P)
NPNPxx 2*EV(N)+2*EV(x) p(N)*p(P)*p(N)*p(P)

NN 2*EV(N) p(N)*p(N)
PNNN 3*EV(N) p(P)*p(N)*p(N)*p(N)
NPNN 3*EV(N) p(N)*p(P)*p(N)*p(N)
PPNNNN 4*EV(N) p(P)*p(P)*p(N)*p(N)*p(N)*p(N)
PNPNNN 4*EV(N) p(P)*p(N)*p(P)*p(N)*p(N)*p(N)
PNNPNN 4*EV(N) p(P)*p(N)*p(N)*p(P)*p(N)*p(N)
NPPNNN 4*EV(N) p(N)*p(P)*p(P)*p(N)*p(N)*p(N)
NPNPNN 4*EV(N) p(N)*p(P)*p(N)*p(P)*p(N)*p(N)
PPPxxxxx 5*EV(x) p(P)*p(P)*p(P)
PPNPxxxx EV(N)+4*EV(x) p(P)*p(P)*p(N)*p(P)
PNPPxxxx EV(N)+4*EV(x) p(P)*p(N)*p(P)*p(P)
NPPPxxxx EV(N)+4*EV(x) p(N)*p(P)*p(P)*p(P)
PPNNPxxx 2*EV(N)+3*EV(x) p(P)*p(P)*p(N)*p(N)*p(P)
PNPNPxxx 2*EV(N)+3*EV(x) p(P)*p(N)*p(P)*p(N)*p(P)
PNNPPxxx 2*EV(N)+3*EV(x) p(P)*p(N)*p(N)*p(P)*p(P)
NPPNPxxx 2*EV(N)+3*EV(x) p(N)*p(P)*p(P)*p(N)*p(P)
NPNPPxxx 2*EV(N)+3*EV(x) p(N)*p(P)*p(N)*p(P)*p(P)
PPNNNPxx 3*EV(N)+2*EV(x) p(P)*p(P)*p(N)*p(N)*p(N)*p(P)
PNPNNPxx 3*EV(N)+2*EV(x) p(P)*p(N)*p(P)*p(N)*p(N)*p(P)
PNNPNPxx 3*EV(N)+2*EV(x) p(P)*p(N)*p(N)*p(P)*p(N)*p(P)
NPPNNPxx 3*EV(N)+2*EV(x) p(N)*p(P)*p(P)*p(N)*p(N)*p(P)
NPNPNPxx 3*EV(N)+2*EV(x) p(N)*p(P)*p(N)*p(P)*p(N)*p(P)

SPL5-Infinity - Enjoy


Well-Known Member
Exact EV's of splits for Infinite Decks

Now that we have the various rounds described - we will first decribe the case where 1 Split is
allowed (SPL1).

Here there is only one possible round that can occur:

(PP)xx EV(x)+EV(x) 1

As mentioned previously, x hands are the result of any card being played - so to calculate EV(x) -
we take the weighted avarage of all the hands that can result from playing a second card on a
split card.

Let's look at 10, 10 vs 6 for simplicity since the strategy is to stand on all hands. After
splitting the 10's - we have a hand total of 10 plus the next card played:

EV(x) = p(1)*EV(21) + p(2)*EV(12)...p(10)*EV(20)

If the strategy were to hit, on let's say 16 , then we would use the same prob p(6), but use the
EV of hitting a 16 instead of the EV of standing on 16.

Once you have calculated this number - you simply multiply it by 2 and you have the exact EV for

Now let's look at being allowed to split twice (SPL2)

Hand EV Prob

NN 2*EV(N) p(N)*p(N)
Pxxx 3*EV(x) p(P)
NPxx EV(N)+2*EV(x) p(N)*p(P)

Here we have 3 different resulting playing rounds - either we can be dealt two non-pair cards
first and end the round, be dealt a pair card and then reach our maximum number of hands, or be
dealt a non-pair card first, using up one of our pair cards, and then be dealt a paircard and have
2 more hands to fill in.

The probability of each of these possible rounds is listed in the 3rd column. The sum of the
probabilities of each of the 3 possible rounds add up to 1. Taking the weighted average of the
EV's of each of these rounds will give us SPL2.

First however we need to figure out what the EV of each round is. Again, since we are dealing with
infinite decks - we only need to calculate EV(x) and EV(N) once, and then we add the appropriate
number of EV(N)'s and EV(x)'s for each respective round as listed above in the second column.

EV(x) is the same EV(x) as that described above. If you remember from above - N refers to the fact
that a non-split card was played. Therefore to calculate EV(N) - we need to calculate the weighted
average of all of the possible resulting hands except those that can result from a paircard being
dealt. Note that each probability needs to be adjusted to account for the fact that the paircard
is not included. The adjustment can be made to the weighted sum all at once. Looking at 10, 10
vs 16 again:

EV(N) = ( p(1)*EV(21)+p(2)*EV(12)...+p(9)*EV(19) ) / (1 - p(10))

Again, if the strategy were to hit/double, then we would use the prob of the second card played
and the EV would be the EV of hitting/doubling the hand.

Now that we have both EV(N) and EV(x) we just figure out the weighted EV's of the respective
rounds as describe above.

If further splits are allowed, you just do the same thing using the different possible rounds that
are listed in the previous post.


Well-Known Member
Exact EV's of Splits for Finite Decks

So far we have described the various deifferent kinds of rounds that can result form splitting and
how to use this knowledge to calculate the values of splits for infinite decks.

The situation however becomes a bit more complicated if you are limited by finite decks.

Let's look at SPL1 again:

Hand EV Prob

xx EV(x)+EV(x) 1

Several issues arise even for this simple hand with finite decks.

First of all, even before trying to calculate EV(x), the question arises
- are the two EV(x)'s the same?

In order to facilitate further discussion, I will introduce a new notation. When looking at a
round, e.g. xx, the first x is played off of the initial shoe (see below what the initial shoe
is), while the second x is played after the first x has been removed. a "-..." will represent the
hand played after the "..." has been removed from the shoe. The second hand would have the


Another example is the hand:


Here we have 2 EV(N)'s and 2 EV(x)'s to deal with and they are:
EV(N-P), EV(N-PN), EV(x-PNNP) and EV(x-PNNPx)

So to rephrase the first question:
Does EV(x) = EV(x-x)?

It turns out that for Fixed Strategies the answer is Yes. For optimal play, or varying strategies,
the answer is sometimes.

The fact that EV(x) = EV(x-x) led us to the next question:

Do hands e.g. EV(x-NPP) = EV(x-NPPx) = EV(x-NPPxx)?

Again, the answer for fixed strategies is yes and optimal strategy is sometimes.

The next question that arises is:
Do hands such as EV(NP) = EV(PN)?

And once again the answer is yes for fixed strategy play and sometimes for optimal play. It is
also important to note that only the effects of removal of the initial N card need to be taken
account and not the third, fourth, etc. card in any hand.

These facts make our lives a lot easier because they greatly simplify the calculations for the
various rounds.

For example let's look at the EV's for SPL1 and SPL2.

EV = 2*EV(x)

NN = EV(N) + EV(N-N)
Pxxx = 3*EV(x-P)
NPxx = EV(N) + 2*EV(x-PN)

The probabilities of the respective rounds are calculated the same way as described for the
infinite deck case, but calculated with the shoe in use though rather than with infinite deck

So now we have the probabilities of the rounds and the EV's that need to be calculated. The last
step we need to do is figure out how to calculate the respective probabilities and EV(x)'s and

Interestingly, it turns out that two different sets of shoe states can be used to calculate the
exact EV's. The first was determined by thinking about the effects of removal, the second
represents the actual shoe state as determined by brute force. The probabilities used for the
rounds are the same. The two possible sets of EV's that can be used are:

SPL1 EOR Actual
xx 2*EV(x) 2*EV(x)

SPL2 EOR Actual
NN EV(N) + EV(N-N) 2*EV(N-N)
Pxxx 3*EV(x-P) 3*EV(x-P)
NPxx EV(N) + 2*EV(x-PN) EV(N-P) + 2*EV(x-PN)

SPL3 EOR Actual
NN EV(N) + EV(N-N) 2*EV(N-N)
PPxxxx 4*EV(x-PP) 4*EV(x-PP)
PNPxxx EV(N-P) + 3*EV(x-PPN) EV(N-PP) + 3*EV(x-PPN)
NPPxxx EV(N) + 3*EV(x-PPN) EV(N-PP) + 3*EV(x-PPN)
PNNPxx EV(N-P) + EV(N-PN) + 2*EV(x-PPNN) 2*EV(N-PPN) + 2*EV(x-PPNN)
NPNPxx EV(N) + EV(N-PN) + 2*EV(x-PPNN) 2*EV(N-PPN) + 2*EV(x-PPNN)

Notice how in the "Actual" column, only a single EV(N) and EV(x) are used for any given round.
I'm not exactly sure why this is but I believe it is because the dealer plays his hand out after
the player has played all of his hands. Since the player has equal probability of playing PN or
NP for any given pair of rounds that occur with equal probability - the effects of removal appear
to be negated.


Well-Known Member
Calculating EV(x) and EV(N) for Finite Decks

As we know from our calculations on infinite deck splits, EV(x) and EV(N) are the weighted
averages of the various possible hand that may arise when playing out a single split card, with
EV(x) being the result of any card, and EV(N) being limited to the results of non-pair cards.

To calculate the x hands where only a P is removed, e.g. EV(x-P), EV(x-PP), adjusting the shoe is
easy - simply remove the relevant number P cards from the deck, and calculate the EV's as
previously described.

Before we go any further, it is necessary to quickly review the conditional probability equation:
EV(A|~B) = (EV(A)-EV(A|B)*p(B))/(1-p(B))

One application of this equation that we can use right away, is in the calculation of EV(N) for
any given shoe state where only P cards are removed, if we know EV(x) and the EV of a pair that is
not split for the given shoe state, or EV(pair). EV(N) for these states is simply the conditional
probability of the hand given that the hand does not include a pair:

EV(N) = ( EV(x) - EV(pair)*p(P) ) / (1-p(P))

This relationship obviously saves calculation time since we determine EV(pair) and p(P) while
determining EV(x).

Another example of this relationship would be if we wished to calculate EV(N-P):

EV(N-P) = ( EV(x-P) - EV(pair-P)*p(P-P) ) / (1-p(P-P))

It's also helpful to remember that p(N) = 1 - p(P).

The real question then is how do you remove an N hand?

One could obviously brute-force this, however it's much easier and faster to think in terms of
conditional probaiblities again. It turns out that the EV of a hand given that an N card was
removed is the conditional probability relative to if a paircard had been removed. This is very
important because it means that only hands in which P cards were removed need to be calculated.

For example, if we wish to determine EV(x-N), all we do is first figure out EV(x-Full Shoe),
EV(x-P) and p(P), where "Full Shoe" represents the deck after the original 2 paircards and dealer
upcard are removed, and "x-P" is the shoe after 3 paircards and the dealer upcard are removed.
Then we simply apply the conditional probability equation:

EV(x-N) = (EV(x) - EV(x-P)*p(P))/(1-p(P))

Similarly, to get EV(x-PN) we first need to adjust p(P) to account for the extra paircard removed,
and then we have:

EV(x-PN) = (EV(x-P) - EV(x-PP)*p(P-P))/(1-p(P-P))

In cases where 2 or more N's are removed, we use the conditional probability that a P has been
removed after an N was removed, relative to having just an N removed. So for example, in order to
get EV(x-PNN), we simply use the equation:

EV(x-PNN) = (EV(x-PN) - EV(x-PPN) * p(P-PN)) / (1 - p(P-PN))

The same relationship holds between EV(N) when N hands are removed, however the probability of the
paircard must be adjusted for the N card that was removed. For example for EV(N-N) and EV(N-PNN)
we have:

EV(N-N) = (EV(N) - EV(N-P) * p(P-N) / (1 - p(P-N))

EV(N-PNN) = (EV(N-PN) - EV(N-PPN) * p(P-PNN)) / (1 - p(P-PNN))

The significance of the above examples is that in order to determine the exact EV's of any fixed
strategy splits, the only hands that need to be calculated directly are x hands for shoe states
with only P cards removed. For SPL1 only 1 shoe state needs to be directly calculated, for SPL2
we need 3 shoe states and for SPL3 we need only 5 shoe states.

The shoe states and the conditional relationships for the probabilities and EV's are summarized
below. Rows in which nothing only the state is listed need to be calculated directly:

Shoe State		Abbr	EV(A)	EV(A|B)		p(P) for x		p(P) for N
1   Full Deck		Full         
2   1P Removed		-P         
3   2P's Removed	-PP         
4   1N Removed		-N	Full	-P		Full			-N
5   1P, 1N Removed	-PN	-P	-PP		-P			-PN
6   3P's Removed	-PPP         
7   4P's Removed	-PPPP         
8   2P's, 1N Removed	-PPN	-PP	-PPP		-PP			-PPN
9   3P's, 1N Removed	-PPPN	-PPP	-PPPP		-PPP			-PPPN
10  1P, 2N's Removed	-PNN	-PN	-PPN		-PN			-PNN
11  2P's, 2N's Removed	-PPNN	-PPN	-PPPN		-PPN			-PPNN
The last thing that needs to be addressed is dealing with an Ace or 10 upcard when the dealer has
a hole card. The only difference is that all the probabilities and EV's for the hands in which P
cards are removed are adjusted to account for the conditional probability that the dealer does not
have blackjack. Once those are calculated however, the same relationships hold and the EV's are
calculated in the same way.

And that's all there is to it :)


Well-Known Member
Bringing it all together

Below is a table with the first column being the various rounds for the corresponding SPL's. The second column is how I calculate the EV for a given round, and the third column is where your numbers are coming from.

SPL1	MGP					Eric 
xx	2*EV(x)					2*EV(x) 
SPL2	MGP					Eric 
NN	EV(N) + EV(N-N)				2*EV(N-N) 
Pxxx	3*EV(x-P)				3*EV(x-P) 
NPxx	EV(N) + 2*EV(x-PN)			EV(N-P) + 2*EV(x-PN) 
SPL3	MGP					Eric 
NN	EV(N) + EV(N-N)				2*EV(N-N) 
PPxxxx	4*EV(x-PP)				4*EV(x-PP) 
PNPxxx	EV(N-P) + 3*EV(x-PPN)			EV(N-PP) + 3*EV(x-PPN) 
NPPxxx	EV(N) + 3*EV(x-PPN)			EV(N-PP) + 3*EV(x-PPN) 
PNNPxx	EV(N-P) + EV(N-PN) + 2*EV(x-PPNN)	2*EV(N-PPN) + 2*EV(x-PPNN) 
NPNPxx	EV(N) + EV(N-PN) + 2*EV(x-PPNN)		2*EV(N-PPN) + 2*EV(x-PPNN)
Eric's values are what match actual brute force shoe states so that's currently what I use.


Well-Known Member
Adjusting for BBO/OBBO/Australian OBBO

You need to keep track of the possiblity of busting when the dealer has an upcard of Ace/Ten and then adjust accordingly. I did this a long time ago and it's the comments in my code so I can't explain right now more than what's below:

BBO = Sum(State UCEVs with loss bust) - (1 - Prob(All Bust))*pBJ

OBBO = UCEV(Original Bet with Loss 1) + Sum(UCEV's other hands with Loss Bust)

AustOBBO = Sum(State UCEVs with Loss 1)

ENHC = (1-p(BJ))*(EV|no dBJ) + p(BJ)*((1-sum(p(Res))*(-1) + sum(p(Res)*EV(Res|dBJ))

BBO = (1-p(BJ))*(EV|no dBJ) + p(BJ)*((1-sum(p(Res))*(0) + sum(p(Res)*EV(Res|dBJ))
BBO = ENHC + p(BJ)*(1-sum(p(Res))*(1)

OBBO = (1-p(BJ))*(EV|no dBJ) + p(BJ)*((1-sum(p(Res))*(-1) + sum(p(Res)*EV(Res|dBJ))