Re: AA cracked twice in one hand--how bad is this beat?
i just hacked my hand calculator code to allow me to figure out how often a hand comes in last place (instead of first). running A[img]/images/graemlins/club.gif[/img]A[img]/images/graemlins/spade.gif[/img] against K[img]/images/graemlins/heart.gif[/img]K[img]/images/graemlins/diamond.gif[/img] and 8[img]/images/graemlins/club.gif[/img]8[img]/images/graemlins/diamond.gif[/img] i get
.56% a tie for last (which players tie for last isn't specified)
4.78% the aces come in last (this is the situation you had)
19.12% the kings come in last
75.54% the 8s come in last
this doesn't care if the Ks and 8s beat the As by each hitting their set, or (in this case) by catching four diamonds to make the flush, or 9TJQ to make a straight, or anything else i may have overlooked. it just sees who comes in last for every possible board if all three hands stay to the river (as they will when two are all in before the flop). so almost one time in twenty, AA vs KK vs 88 will lose both the main and side pots. assuming i have no bugs in my code.
also, changing 8[img]/images/graemlins/club.gif[/img] to 8[img]/images/graemlins/heart.gif[/img] bumps the losing % up to 5.16%, so suit selection is non-trivial.
if you want to know how often both players will beat you specifically by hitting their set, or something else even more specific (like the K hits on the flop AND the 8 hits on the river), then i concede that combinatorial counting is a better approach, but it's not clear to me that people agree on what it is they should be counting.
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