EV calculations for Texas Holdem

I'm looking at the tables you generated on the Wizard of Odds site for Texas Holdem and I don't understand the values shown in the expected value (EV) column. For example, for a 10 handed game the win probability for a pair of aces is shown as 0.3136. Here is how I would calculate the EV with that number:

P(win) = .3136
P(lose) = 1 - P(win) = .6840

Net win = 9
Net loss = 1

This assumes that each of the 10 players pays 1 to play, so the winner receives a total of 10, but since s/he contributed 1 of that, the winner's net win is 9. Now, here is how I would calculate the EV:

EV = P(win)*(Net win) - P(loss)*(Net loss) = (.3136)*9-(.6840)*1 = 2.1384

But the value shown in the EV column is 2.1071. Am I missing something here?

Another question I have about the same table regards the sorting method. At first it looks as though the hands are sorted on the win probability, but there are some exceptions. For example, the hand 74 suited is shown above J3 suited, despite having a smaller win probability (.1159 versus .1174).

Quote: chagota

But the value shown in the EV column is 2.1071. Am I missing something here?

Split pots. If you notice the Average Win, it's 9.91, not 10.00 (as it would be if it didn't split any pots with other hands). That's because the table rolls split pots into average win instead of having separate columns for each type of split (two-way, three-way, on up to ten-way). P_"win", then, is actually P_win+P_2tie+...+P_10tie. The way the Wizard computes EV in this case is (P_"win")(avg win)-1.

Quote: chagota

Another question I have about the same table regards the sorting method. At first it looks as though the hands are sorted on the win probability, but there are some exceptions. For example, the hand 74 suited is shown above J3 suited, despite having a smaller win probability (.1159 versus .1174).

It's sorted by EV. 74s will win more pots outright due to straight possibilities that J3s doesn't have, and therefore has better EV, but J3s catches more split pots (because of the jack) giving it the better "win" rate as defined above.

Oh, well, that was easy enough. Thanks!

On second thought, why doesn't the EV calculation use the total win probability?

Wait, scratch that. I'm confused now for real. Working backwards from the EV column we have:

EV = 2.1071 = p("win")*9- ( 1-p("win") ) = 10*p("win") - 1 ==> p("win") = .31071 which is less than the win probability given for a pair of aces (.3136). The idea here is that the larger probability includes the possibility of ties. But in that case the EV calculation excludes that possibility. But why would we exclude ties for the EV calculation?

Again, am I missing something here? Please advise.

Quote: chagota

Wait, scratch that. I'm confused now for real. Working backwards from the EV column we have:

EV = 2.1071 = p("win")*9- ( 1-p("win") ) = 10*p("win") - 1 ==> p("win") = .31071 which is less than the win probability given for a pair of aces (.3136). The idea here is that the larger probability includes the possibility of ties. But in that case the EV calculation excludes that possibility. But why would we exclude ties for the EV calculation?

Again, am I missing something here? Please advise.

Splits are there, they're just rolled up into the win probabilities and average pots. The effect is that the win probability goes up a bit and the average pot goes down a bit. In terms of the numbers given, (EV+1)/P("win") = Average pot. For AA, you have (2.1071+1)/(.3136)=9.91. The raw data isn't available for us to look at (unlike the two-handed page, where the outcomes are listed in the second half of the page), but if they were, your average pot size would be: