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NSchandler
04-27-2005, 02:27 AM
I am a relative newcomer to poker, and just read through TOP. It's quite an amazing book, and while it is clearly an invaluable resource to those who want to improve their game, I feel that the section on game theory and bluffing requires a slight adjustment.

Specifically, it seems that the section on "optimum bluffing strategy" could have you bluffing slightly less than is optimal. Let me give an example.

Sklansky recommends bluffing with such a frequency that the odds you are bluffing are equivalent to the pot odds. That is, if there is a $100 pot, for example, and bets are $20, your opponent is getting 6:1 odds to call your bet. Therefore, you should present your opponent with 6:1 odds that you are bluffing, making him indifferent to calling or folding (the sign of an optimal mixed strategy).

Now, suppose you are playing Hold 'Em, and after the turn, you will have 1 of 2 hands. 7 out of 8 times, you will have a hand with 18 outs. The other 1 out of 8 times, you will have a cinch hand. Again assume that the pot is $100 and bets are $20. How often should you be bluffing with your hand that gives you 18 outs? It would seem that you should bluff with 3 additional cards, presenting your opponent with 18:3 or 6:1 odds that you are bluffing. The problem, however, is that you aren't truly bluffing 1 out of 7 times as you should be. You are bluffing less often. That is, 1 out of 8 times, you have a lock on the hand - you cannot bluff.

Think about your opponent now. He's thinking, "There's a 1/8 chance that I was drawing dead. If this is the case, he can't be bluffing. There is also a 7/8 chance that he had 18 outs. If this is the case, there's a 1/7 chance that he's bluffing. Putting these together, there's only a 1/8 chance that he's bluffing. Therefore, I'm not getting proper odds to call. I will fold."

Notice what happened. While you are indeed bluffing 1 out of 7 times with your 18-out hand as suggested, you are only bluffing 1 out of every 8 total hands because you're unable to bluff with your cinch hand.

So how often should you bluff with your 18-out hand? In this case, you should choose about 3.512 cards to bluff with (yes, again, I know you can't bluff with 3.512 cards - as a decent approximation, you could choose 3 cards to bluff with and 1 additional card that you will bluff with if you flip a coin and it lands heads-up).

If you bluff with 3.512 cards, now your opponent is being given about 6:1 odds. 1/8 times you again have the cinch hand and are never bluffing. The other 7/8 times, you are bluffing 3.512/21.512 = 16.33% of the time. Putting this together, you will be bluffing a total of 14.29% (16.33% * 87.5%) of the time, which is 1 out of every 7 times, giving your opponent 6:1 odds to call your bet.

If you do the math, you should find that bluffing with 3.512 cards here is superior to bluffing with only 3 (assuming your opponent can't tell whether you have the cinch hand or the 18-out hand).

Sklansky is not wrong - his solution is obviously the theoretically correct solution. However, since you can't bluff with cinch hands (or can't bluff often enough with near-cinch hands), you need to "transfer" bluffs you would like to make with these hands to your non-cinch hands in order to give your opponent the proper odds.

I don't want to be so presumptuous to think that I can find a serious flaw in a book as amazing as TOP. I have probably either overlooked something or Sklansky felt that discussing such a complicated scenario would not be worth it. But if I did make a mistake, I want to be corrected so I can re-align my thinking. And if not, I think this is a very interesting topic, as it opens up all sorts of issues (but that's for another post).

njs

jdl22
04-27-2005, 03:25 AM
Welcome to the forum.

I didn't quite follow everything you said, but here's an attempted response.

Suppose you have a cinch or 18 out hand with probability 1/8 and 7/8 respectively. Suppose your opponent has no knowledge of your outs, or it's some sort of stud type river. You know 6 cards so you have 28 cards that are not outs if you are drawing. Furthermore suppose you bluff with probability p given that you miss with your bluffing hands.

Let's look at all possible outcomes. 1/8 of the time the river wasn't relevant. 7/8*18/46=63/184 of the time you have caught up and with probability 7/8*28/46=49/92 you have whiffed on the final card. Just to check these, add them up and you should get 1 which is the case.

If you're bluffing with probability p when you miss then the total probability of a bluff is 49p/92. Since you are ahead with probability 1/8+63/184=86/184=43/92. Given a bet the odds that you are ahead are 43:49p so we need p such that 43/49p=6 since she will be getting 6:1. That p=43/(49*6)~.15. Multiplying this by 28 gives about 4.2.

You underestimated the number of cards. The reason you must bluff more often when you miss is that your opponent won't be able to tell whether you are drawing or not. Since this implies that you are ahead more often in her eyes you should bluff more.

jason1990
04-27-2005, 09:24 AM
Let me see if I understand your example correctly. On the flop, you have, for example, 6 outs to an unbeatable hand and (somehow) 41 outs to a strong 18-out drawing hand. So, for simplicity, we round these odds off and say that there is a 1/8 chance the turn will give you an unbeatable hand, and a 7/8 chance the turn will give you an 18-out draw.

The turn comes, and you make your drawing hand. You choose 3 cards other than your 18 outs and decide that if one of those cards comes on the river, you will bluff. This means that the conditional probability you will bluff on the river, given that you have the 18-out draw, is 1/7. However, the conditional probability you will bluff, given your holding on the flop, is (7/8)*(1/7)=1/8.

On the other hand, if you could choose 3.512 cards to bluff with on the river, then the conditional probability you will bluff on the river, given that you have the 18-out draw, is 3.512/21.512=1/6.1; whereas the conditional probability you will bluff, given your holding on the flop, is (7/8)*(1/6.1)=1/7. So which is right?

Is this your question? If so, the answer is easy to state, but difficult to implement. You must play your hand in such a way that the conditional probability you will be bluffing, given all the information at your opponent's disposal, is 1/7. When you use an "optimal" strategy in the game theory sense, then your opponent can know exactly what you're doing and it doesn't change a thing. So suppose your opponent knows exactly what you're doing. That is, he knows the frequency with which you do will do every action with every holding. Given this knowledge, he can now look at the board cards and at the betting sequence so far, and compute the conditional probability that you are bluffing, given all of this information. You want that computation to result in 1/7.

G. Ed Conly
04-27-2005, 08:57 PM
I think you'll find that the situation you're addressing is a semi-bluff. The "bluff" occurs after all the cards are out. See if this doens't clear up the "discrepancy".

SpeakEasy
04-28-2005, 03:12 PM
[ QUOTE ]
How often should you be bluffing with your hand that gives you 18 outs?

[/ QUOTE ]

If the seconds-hand on my watch is between the 10 and the 12, I will bluff. The remainder of times, I won't. Completely random determination.