MGP's BJCA

[lij45o6]

But it's not different; the above is more work than we need to compute the expected value of splitting without resplitting. Let's use your example of splitting 2-2 vs. dealer 6, in single deck with S17 SPL1, and let's assume nDAS to simplify the "brute force" calculations. We are assuming CDZ- playing strategy, which in this case is to hit any hand drawn to the split.

First, hopefully we can agree that the overall CDZ- expected value of splitting 2-2 vs. 6 is 0.1189818683947885 (as a fraction of initial wager). This can be confirmed by any of the accessible CA's (e.g., mine, MGP's, or k_c's).

We can compute this value as 2E[X;1], where E[X;1] is the expected value of:

1. Starting with the full single deck, removing the dealer's up card, and two pair cards.
2. Randomly draw a card to yield a split hand, e.g., drawing an ace yields the hand 2-A.
3. Play out that single "half" of the split and note the outcome.

It's important to note that this isn't just computational convenience. That is, it's not that we simply "get the same answer" by doing the calculation this way. The expected outcome of the two halves of the split are indeed identical.

And even for resplitting, we can similarly take advantage of that same linearity of expectation by "grouping" the various possible split/resplit outcomes according to the "pattern" of number of pair-cards and non-pair-cards drawn to the split hands (weighted by the number of ways that each such pattern can occur). For example, consider SPL3: among those outcomes where we only split 2-2 once (i.e., we draw a non-deuce to both halves of the split), let X be the outcome of the first half of such a split, and Y be the outcome of the second half of such a split. Then E[X]=E[Y]. See this paper for more details on how we can use this to efficiently (and still exactly) compute split EVs. (This is how my CA does it, for example.)

Eric

Okay, so:

1.) From what I am gathering here is that the method of computing exact split Expectation's for single-split hands is 2 times the overall weighted Expectations for each new hand drawn, conditioned on a pair card removed. That is, even when computing the overall split Expectation for 22 vs 6, we only need to be aware of one single hand conditioned on its pair-rank removed from a full deck. We don't need to be cognizant of the other card nor of the other hand's make-up. So, for splitting 22 vs 6 , we don't need to evaluate {2A, 2A} directly; taking into account the missing deuce and Ace in the second hand for the Expectation of the first. We simply need to do {2x, 2A} and add the new Expectation of 2A conditioned on the missing 2 pair-rank removed to properly find the correct E/Action for 2A. Summing for each new draw rank 1-10, we combine them (by their overall weighted Expectations as E[Split] = P(A) * E(A) + ...P(10) * E(10)) taking a factor of two (for the two hands) and we should properly derive the conditional expectation of splitting 22 vs 6 for 1D, S17, (n)DAS, SPL1.

2.) Further splitting gets more complicated as we most of the time hit a "wall" of non-pair ranks. So, to split 22 vs 6 for SP2: we can have {2x, 2x}, draw a 2 and develop {2x, 2x, 2x}, we then cycle through each rank x, from 1-10, for the right-most 2x, and times it by the number of ways that hand state can be ordered. But, from your paper, this is (not?) the way this is done. I would assume the multinomial coefficient of the given split rank values to determine the overall Expectation of splitting 22 v 6 for SP2. Assume we draw an Ace after splitting to 3 deuces, we have a MC of 4. We then take this MC and times it by the overall weighted expectation for each optimal action (similar to our SP1 example.) However; drawing a third deuce is not guaranteed and so this method is wrong, correct? As per your paper, what I just described is incorrect and there involves some level of detail that I am missing.