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

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