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View Full Version : Analyzing the cost of a mistake (TOP chapter)


frankg
02-15-2005, 12:35 AM
In Sklansky's TOP, chapter 24, section "Analyzing the cost of a mistake", he brings up a situation in no-limit hold-em that had me thinking. He gives an example in which you should not necessarily choose a play that is favored to be correct over 50% of the time if the cost of making a mistake is too great.

You are dealt QQ in early position and open the betting with a standard raise. Everyone folds except one player, who makes a huge reraise. You have a pretty good read on the player and know that he would only make a move like that with AA, KK, or AK. The odds work out that it's a 4-to-3 favorite your opponent has AK instead of AA or KK, so 4/7 of the time your QQ is the favorite, and 3/7 of the time it is the underdog. So, naturally, most people would at least call the re-raise (or go all-in). However, if your opponent DOES have AK, your QQ is only a 13-10 favorite pre-flop. So you will average winning 13 times, but the other 10 out of 23 times you will lose against AK. Furthermore, those 3 times out of 7 that your opponent has AA or KK you are a big 4.5-to-1 underdog.

Hence you cannot say "My queens are 4-to-3 favorites to be the best hand, so I must call". The math dictates that the times your opponent has AA or KK you hurt yourself so much that you don't gain enough of it back when he has AK, thus making it a -EV play.

The above is almost exactly the way Sklansky words this problem. I had to read it a few times to fully grasp this concept. It was very eye opening for me because I think in most cases I would either call or reraise with queens, even if i thought my opponent had AK. I realize that this is all a moot point if you don't have an accurate read on your opponent. This problem wouldn't apply to looser opponents also. But let me ask the experienced players here: would you fold your queens if you knew your opponent probably had AK? In a no-limit cash game? In a no-limit tournament? Short stacked?

Kaz The Original
02-15-2005, 01:24 AM
Note, if your opponent has AK, he is only going to hit the flop 1 in 3, so your EFFECTIVE odds vs AK are 2:1, right?

frankg
02-15-2005, 01:30 AM
Yes, but you can't be certain he has AK. You think he PROBABLY (60%) has AK, but if you're wrong than he has AA or KK. Do you still call?

einbert
02-15-2005, 01:14 PM
You can analyze this situation better by using Baye's theorem to figure out the equity of QQ versus your opponent's range of hands. A program like PokerStove can do this easily.

287,667,072 games 0.312 secs 922,009,846 games/sec

Board:
Dead:

equity (%) win (%) / tie (%)

Hand 1: 60.1426 % [00.60 00.00] { AA-KK, AKs, AKo }
Hand 2: 39.8574 % [00.40 00.00] { QQ }

So if you're getting 60:39 or about 3:2 from the pot, you should call. Otherwise, fold.

Yosemite Mark
02-15-2005, 05:59 PM
The basic principle here is: Your overall EV is: (payout from a good result * chance of the good result) minus (cost of a bad result * chance of a bad result). In your example, the payout of QQ vs AK is a small one (you are a small favorite), but the cost of the bad result (QQ vs AA or KK) is huge - you are a big underdog. The fact that QQ vs AK is more likely is not enough to make up for the diffence in payouts.

Here's another example: I offer a bet on a roll of a single 6-sided die. If it comes up 1 through 5, I will pay you $1. However, if it comes up 6, you have to pay me $100. Even though you are most likely to win this bet (odds are 5:1 in your favor), you would never take it, as the 1/6 chance of losing has such a huge cost.

Hope that makes the concept easier to understand....