Thursday, April 23, 2009
Everyone is Wrong
At my weekly poker lunch meeting yesterday, an interesting topic came up and has caused a lot of heated discussion and confusion. It all started when SolarTed, one of the regulars at the group, brought in a new book by Alan Schoonmaker, titled "Poker Players Are Different". In the opening chapter, the author asks the following question: if you push all-in PF in a full-ring cash game with pocket aces, how many callers do you want to get?"
We went around the room and asked everyone what they thought, and the majority of players said that getting one caller was ideal. This certainly agrees with the conventional wisdom espoused in most poker books that states you want to get heads up with AA, because this maximizes your chance of winning.
The author, however, pointed out that maximizing your chance of winning isn't the goal. The goal is to maximize your expected value. These two things are not the same. In other words, the author said you want all nine opponents to call, as this maximizes EV.
This topic intrigued me, so I figured I should run the actual EV calculation to see who's right. (my original verbal answer at the time of the meeting, by the way, was I thought one wanted to see 2-3 players, as I figured this would maximize EV).
I ran pokerstove to calculate how AA does against 1, 2, 3, etc. opponents who hold random hands (I also ran it against opponents who held better than average hands, but the results didn't change significantly enough to matter).
As we all know, expected value equals the percentage of the time we win, multiplied by the amount we win, minus the percentage of the time we lose, multiplied by the amount we lose. Writ another way:
EV=(%winrate)($won) - ($loserate)($lost).
The classic coinflip problem best illustrates this: if we flip a coin, its chance of coming up heads is 50%, and tails 50%. If I bet a buck on the outcome, the EV for the coin toss is:
EV=(50%)($1) - (50%)(1$) = zero. If it's a trick coin, that comes up heads 75% of the time, and I bet on heads, my EV is:
EV=(75%)($1)-(25%)($1)=$.50 positive expected value.
Now, per PokerStove, AA holds up against a variable # of players as follows:
Against one opponent with $100 stack sizes, the expected value with AA is therefore:
Similarly, against two opponents:
EV=(73%)($200)-(27%)($100)=$119, and so on...
I put all this into a spreadsheet and created the chart shown above. The funny thing is that it turns out everyone was wrong, including Schoonmaker. Note how the curve maxes out around 7 players and then slowly starts falling back down. Said another way: to maximize EV when pushing with AA, you want 7 opponents, or in other words, the more the merrier... up to a point. If you're in a 9-handed game, if everyone calls, it's a good thing, but you're losing a tiny bit of EV vs. when all but one call. In even larger games, the effect would be larger.
In any case, one opponent is the worst possible answer, and Alan Schoonmaker's 9 opponents, while pretty damn good, is not ideal. The best result when shoving preflop with AA is to get 7 callers, thereby maximizing your EV. And maximizing EV, of course, is the ultimate goal of any poker decision you make at the table.
All-in for now....
Posted by bug at 11:15 AM