Head Up Theory Question

[FishAndChips]

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If you anticipate having to fold at any point during the hand to one of your opponent's reraises, just don't raise yourself-- simply call. Then you at least get a showdown, and can not be bluffed out of a large pot.

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This argument isn't particularly persuasive because you could conceivably be costing yourself by not raising (even though you'd sometimes fold to a reraise).

Suppose that at a certain point in a certain game your opponent raises so that the pot is $95 and you could either fold, call $1, or raise $1 more. If you know that your opponent is playing an optimal strategy, you can put your opponent on a range of hands, and he should be bluffing with probability 1/96. It might be the case that you are holding the unique second-nuts, and if you raise, your opponent will call with the third-nuts, fourth-nuts, and fifth-nuts (each of which is an equally likely holding for him, at this point, as the nuts). But your opponent won't reraise for value again without the nuts (if you raise and he does reraise, there would then be a 1/100 probability he is bluffing).

If you just call, you'd be giving up EV. You need to raise for value because your opponent will call your raise with a worse hand three times as often as he will reraise with a better hand. So the question is, do you

(a) raise and always call a reraise;

(b) raise and always fold to a reraise; or

(c) raise and sometimes call, sometimes fold to a reraise.

The answer here is (c). Read the section "Using Game Theory to Call Possible Bluffs" in chapter 19 of TOP. In the scenario I've just described, your hand, the second-nuts, can only beat a bluff. If your strategy is to always call the reraise, your opponent could exploit it by never bluffing -- then you would be calling all the time with a certain loser. Of course since the pot is so big, you still have to call the vast majority of the time or else your opponent could exploit your strategy by bluffing more. When your opponent bluffs, he is risking $2 to win $97. Therefore your optimal mixed strategy at the end is to call 97/99 of the time and fold 2/99 of the time.

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Well I must return serve here, and say that your argument is not very persuasive either. There is no way you raise, and reraise, your opponent 94 times in $1 increments and then fold to another raise. You should stop raising BEFORE you get to the point where you raise and then fold. If you can't call a raise, why raise yourself. Yeah you may sacrifice slight EV on a value raise if he'd call three times out of four with a worse hand, but you lose a huge pot if you incorrectly fold the one time he raises.

Your proposed strategy could be greatly exploited. Look at it from your opponents perspective. If he knows your strategy, he is getting 94-1 on a bluff that will work 1 in 4 times!! I'd gladly sacrifice the EV of one additional raise 3 in 4 trys, if it meant never folding a winner in such a large pot.

As for your citing TOP ch. 19, it actually proves your strategy is flawed by showing that (as stated above) your opponent is getting 94-1 on a 3-1 bluff, and is therefore extremely -EV for you. You opponent shouldn't bluff 1 in 95 tries as you suggest in your reply, but should bluff 1 in 4. If you want your opponent to only bluff 1 in 95 times, you need to call 94 times out of 95! "If your opponent is getting 4-to-1 odds on a bluff, you must call randomly four out of five times to make bluffing unprofitable." (TOP pg.190)

Also, most examples about bluffing on the end deal with the "bluffer" knowing almost exactly what the other player's hand is, while the player deciding whether to call the possible bluff could not tell if the other player made his hand (and therefore was at a big informational disadvantage against the potential bluffer.) That is not the case here. Neither player has any exposed information, and no one's hand has changed value throughout the course of the betting. Without those things, bluffing here is completely different.

[M.B.E.]

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Your proposed strategy could be greatly exploited. Look at it from your opponents perspective. If he knows your strategy, he is getting 94-1 on a bluff that will work 1 in 4 times!!

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No, look at my post again. The opponent's bluff will work only 2 in 99 times, not 1 in 4.

[Stephen H]

Sure, let's take your strategy at the 1-5 card game. Just to be clear, here's how I interpret your strategy:

Going first:
1: Check/fold
2-4: Check/call
5: Bet/raise forever.

Going second:
1: Check behind or fold to a bet.
2-4: Check behind or call a bet.
5: Bet or raise forever.

Here is my counter strategy that has +EV over yours:
Going first:
1-3: Check, fold to a bet.
4: Bet, fold to a raise.
5: Bet (raise forever, but that will never come up given your strategy)

Going second:
1-3: Check behind, fold to a bet.
4: Bet, fold to a bet.
5: Bet.

It should be fairly obvious that the person going second has a positional advantage, so I'll show the hand chart for your strategy going second. Suffice it to say that your EV is even worse going first.
The BB code eats extra whitespace, so I'm not sure the best way to display the table. The columns are:
My card / Your card / Result

1 / 2 / You win $1 (check/check)
1 / 3 / You win $1 (check/check)
1 / 4 / You win $1 (check/check)
1 / 5 / You win $1 (check/bet/fold)
2 / 1 / I win $1 (check/check)
2 / 3 / You win $1 (check/check)
2 / 4 / You win $1 (check/check)
2 / 5 / You win $1 (check/bet/fold)
3 / 1 / I win $1 (check/check)
3 / 2 / I win $1 (check/check)
3 / 4 / You win $1 (check/check)
3 / 5 / You win $1 (check/bet/fold)
4 / 1 / I win $1 (bet/fold)
4 / 2 / I win $2 (bet/call)
4 / 3 / I win $2 (bet/call)
4 / 5 / You win $2 (bet/raise/fold)
5 / 1 / I win $1 (bet/fold)
5 / 2 / I win $2 (bet/call)
5 / 3 / I win $2 (bet/call)
5 / 4 / I win $2 (bet/call)

Note that while you win 10 out of the 20 combinations, you only win more than the ante when you have the 5 and I have the 4. In the case where you're going first, you will never win a bet from me; I'll only bet when I know I'm going to win. If you add up the results, you'll see that across the 20 possible hands, you'll win $11 and lose $15 for an EV of -$4/20, or -0.20. The rate for you going first is -$5/20, or -0.25. Combining the two gives your strategy a total EV of -0.45/hand versus my counterstrategy. As you can see, your theory that you should "break even" on your hands 2-4 is false; while you win 6 of the 12 hands, you only win $6 on your wins, and lose $11, far from breaking even. And you don't make more than the ante when you hold the nuts, so you aren't covering up the difference there. So, the "correct" strategy isn't as simplistic as that.

[xpsyuvz]

edit: I was trying to save StephenH some time, but it looks like he already posted while I was writing this.

Donkey69,

I came up with a counter-strategy to yours to show how it can be exploited ( -- and my opinion is that any strategy can be exploited if the opponent knows the strategy that is used).

First, just to be clear: let’s say the game requires a $1 ante. And we’re using a deck of 5 cards (1-5) where 5 wins.

Strategy for player-A will be:
When in the first position and player-A has:
1 -- then he folds
2, 3, or 4 -- then he checks and then calls (if he is raised).
5 -- then he raises indefinitely.
When in the second potion:
Basically he does the same as above (where for 2, 3, or 4 he checks or calls if raised.)

Strategy for player-B will be:
When in the first position, then with card
1-- Fold
2-- Check. If raised then fold.
3-- Check. If raised then fold.
4-- Raise. If re-raised then fold.
5-- Raise indefinably.
When in the 2nd position, then:
Basically do the same as above except for
4 -- If raised then fold. If player-A checks, then raise.


To look at the expected value we can play out the 40 different scenarios (20 where player-A is in the first position and 20 where player-A is in the second position). Here I’ll list the card combinations:

A in the 1st position vs. B in the 2nd
1 vs. 2 A loses -1 & B wins +1
1 vs. 3 A loses -1 & B wins +1
1 vs. 4 A loses -1 & B wins +1
1 vs. 5 A loses -1 & B wins +1
2 vs. 1 A wins +1 & B loses -1
2 vs. 3 A loses -1 & B wins +1
2 vs. 4 A loses -2 & B wins +2
2 vs. 5 A loses -2 & B wins +2
3 vs. 1 A wins +1 & B loses -1
3 vs. 2 A wins +1 & B lose -1
3 vs. 4 A loses -2 & B wins +2
3 vs. 5 A loses -2 & B wins +2
4 vs. 1 A wins +1 & B loses -1
4 vs. 2 A wins +1 & B loses -1
4 vs. 3 A wins +1 & B loses -1
4 vs. 5 A loses -2 & B wins +2
5 vs. 1 A wins +1 & B loses -1
5 vs. 2 A wins + 1 & B loses -1
5 vs. 3 A wins +1 & B loses -1
5 vs. 4 A wins +1 & B loses -1
(So far player-A wins +10 and loses -15 while player-B wins +15 and loses -10. And note how player-A’s losses against player-B’s 4 card outweigh his 4 card when up against B other cards.)

A in the 2nd position and B in the 1st position:
(I’m not going to bother typing this out but I think the results are that player-A wins +11 and loses - 15. And the only difference here is player-B’s 4 card raises into player-A’s 5 before folding.)

So the total expected value outcome is player-A loses $9.00 while player-B wins $9.00. (Here player-A doesn't breakeven like you suggested.)

And meanwhile I'm sure player-B's strategy is exploitable too... (If someone has an unexploitably strategy here, I'd be interested in hearing about it.)

[M.B.E.]

In the pot-limit game (assuming we're not allowed to raise less than the pot), we should continue raising as long as

p(3q-r+5) > 4,

where

p:= the probability our opponent is not holding card 1,000,000, given the action so far;

q:= the probability our opponent will reraise if he is not holding card 1,000,000 and we now reraise, given the action so far; and

r:= the probability our opponent will fold if he is not holding card 1,000,000 and we now reraise, given the action so far.

[FishAndChips]

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Your proposed strategy could be greatly exploited. Look at it from your opponents perspective. If he knows your strategy, he is getting 94-1 on a bluff that will work 1 in 4 times!!

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No, look at my post again. The opponent's bluff will work only 2 in 99 times, not 1 in 4.

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Well I wrote 1 in 4 based off of your statement: "If you just call, you'd be giving up EV. You need to raise for value because your opponent will call your raise with a worse hand three times as often as he will reraise with a better hand."

I assumed you meant you had to deal with a "raise and then fold scenario" (oh god no) because 3 out of 4 times the raise would be for value, but the other time it meant you were beat.

If you play correctly, you will not be able to put your opponent on hand 1,000,000 when you have 999,999 with great enough certainty (in relation to pot odds) to allow yourself to fold, unless you raise a number of times beyond the threshold point for hands greater than yours. Therefore you made a mistake raising that last time.

Anytime a raise is for value, a reraise (with no change in hand value in between) can not make you fold. If it does (therefore making your hand an underdog) your previous raise WAS NOT FOR VALUE!

[M.B.E.]

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Anytime a raise is for value, a reraise (with no change in hand value in between) can not make you fold. If it does (therefore making your hand an underdog) your previous raise WAS NOT FOR VALUE!

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The flaw in your thinking has to do with the first parenthesis there, "with no change in hand value in between". If your opponent reraises, then that gives you more information about your opponent's range of hands; in general it means that you can assign a higher probability to your opponent having the nuts than you assigned to that probability before his last reraise.

I'm pretty sure that what I said in this post (which obviously was not discussing the game David asked about at the outset of this thread) is correct. Essentially it just hypothesizes a scenario and then applies chapter 19 of TOP.

[FishAndChips]

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Anytime a raise is for value, a reraise (with no change in hand value in between) can not make you fold. If it does (therefore making your hand an underdog) your previous raise WAS NOT FOR VALUE!

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The flaw in your thinking has to do with the first parenthesis there, "with no change in hand value in between". If your opponent reraises, then that gives you more information about your opponent's range of hands; in general it means that you can assign a higher probability to your opponent having the nuts than you assigned to that probability before his last reraise.

I'm pretty sure that what I said in this post (which obviously was not discussing the game David asked about at the outset of this thread) is correct. Essentially it just hypothesizes a scenario and then applies chapter 19 of TOP.

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If you raise for value as the nth bet, and your opponent reraises for the n+1 bet, are you telling me you can deduce with enough certainty that you went from a favorite when called (that's why you were value betting) to having a less than pot odds chance of still having the best hand (all the while you and your opponent still hold the same hands you had before you bet)?!

I'm aware that each time your opponent raises it narrows down his range of hands, but if one additional raise makes you decide you are an underdog to a greater certainty than the odds offered from the pot, then your previous raise was a mistake. Don't raise for value if you can't even call a one unit reraise.

[M.B.E.]

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If you raise for value as the nth bet, and your opponent reraises for the n+1 bet, are you telling me you can deduce with enough certainty that you went from a favorite when called (that's why you were value betting) to having a less than pot odds chance of still having the best hand (all the while you and your opponent still hold the same hands you had before you bet)?!

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It's conceivable.

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I'm aware that each time your opponent raises it narrows down his range of hands, but if one additional raise makes you decide you are an underdog to a greater certainty than the odds offered from the pot, then your previous raise was a mistake.

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You keep saying that, and I understand why it seems obvious to you, but it is not a universal truth. In some scenarios in some games it holds true, in other scenarios in other games it does not.

[FishAndChips]

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If you raise for value as the nth bet, and your opponent reraises for the n+1 bet, are you telling me you can deduce with enough certainty that you went from a favorite when called (that's why you were value betting) to having a less than pot odds chance of still having the best hand (all the while you and your opponent still hold the same hands you had before you bet)?!

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It's conceivable.

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Many things are conceivable --like the fact that you could still have the best hand and should call. I don't understand how you could be so bad at reading your opponent's hand before you put in the 94th bet, but after the 95th bet, you know for certain you're beat. And if you thought before you bet, "Hmmm if I raise here and I'm reraised, I will fold. Maybe I should call since I'm no longer the favorite if I'm reraised. Ah bugger off, I know what I'll do here: I'll raise for value!" Sorry but this just seems incorrect.

Why, exactly, would you raise in this situation and not call?



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I'm aware that each time your opponent raises it narrows down his range of hands, but if one additional raise makes you decide you are an underdog to a greater certainty than the odds offered from the pot, then your previous raise was a mistake.

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You keep saying that, and I understand why it seems obvious to you, but it is not a universal truth. In some scenarios in some games it holds true, in other scenarios in other games it does not.

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Perhaps, but not in this game as it has been laid out thus far.

PS I like having this discussion with you. I do not take offense that you disagree with me. In fact, I enjoy it. I know that I have still have much to learn about poker. [img]/images/graemlins/smile.gif[/img]