Originally Posted by

**dogman_1234**
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.

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