Quote:
Originally Posted by whosnext
Actually doing all the math correctly in the case you describe can be quite tedious since you need to keep track of each possible runout (or at least categories of runouts).
To give some concreteness, suppose Villain has Kh 9d and Hero has Ac 7c. Flop comes down Kc 9c 2h. Players get it all in on the flop. If the turn and river cards are only dealt once, Villain will win 647 possible runouts and Hero will win 343 runouts of the possible 990 runouts.
If I tallied these correctly, here is how Hero can win the first or only runout:
3 AA
9 A7
3 77
36 two clubs
286 exactly one club
6 A2 (no clubs)
-----
343 TOTAL
Suppose players agree to run it twice. Now, to determine the exact probs of each player winning both runouts and the prob that they chop, you need to keep track of what comes on the first runout (at least in which above category) since that affects the prob of winning the second runout.
You can see immediately that how Hero wins the first runout would affect the prob of winning a second runout. Of course, the converse is also true. How Villain wins the first runout (compare first turn/river of Qc/Kd to 4s/3h) affects the prob of Hero winning the second runout.
Due to "card removal" the prob of either player winning the second runout is not independent of the prob of either player winning the first runout. In theory, this can be handled by a huge "runout tree" or computer program.
Of course, in the simplest case of Hero having exactly one out, the tree is easily constructed and the respective probs can be easily calculated. But things are not so easy in more complicated cases such as getting top two pair vs. ace-high flush draw all-in on the flop.
I ran a computer program last night to do these tallies in this specific case. Recall from above that:
0.34646 = 343/990 = Overall prob of Player 2.
The computer program tallies show:
0.32353 = Prob(Player 2 wins 2nd runout | Player 2 wins 1st runout)
0.35862 = Prob(Player 2 wins 2nd runout | Player 1 wins 1st runout)
So we have the probs of the possible outcomes of running it twice:
0.41916 = Prob(Player 1 wins both runouts)
0.46875 = Prob(Players chop)
0.11209 = Prob(Player 2 wins both runouts)
You can verify that (0.34646)*(0.32353)+(1-0.34646)*(0.35862) = 0.34646 so that Player 2's overall equity is the same whether or not they run it twice.
Note also that the prob of Player 2 winning both runouts (0.11209) is less than .34646*.34646 (= 0.12004) due to the "card removal" effects discussed above.
The card removal effects can vary from situation to situation, so there is no easy way to come up with a simple rule of thumb to derive the exact probs of each player winning both runouts in all run-it-twice situations.