Classic Type Game Theory Problem

But it seems to me that B's best odds is by ignoring what A does- just play the odds. If B has less than .5, he wants to redraw regaurdless- if A redraws, A is likely beating him, if A stands pat, A is surely beating him. So I see no reason for B to stand pat here, ever.

If B is greater than .5, and A redraws, he should stand pat- A is likely losing.

So the only bluff left is when B>.5 and A<B. A can try checking a loser, as you mentioned. But here I disagree- B knows that a redraw is unlikely to beat A. He knows a redraw would most likely make him a worse hand. For this to make sense for A to even try, B has to be very close to 100- closer to 100 than A is to 0. The more likely that is to be true, the less likely B is to win on a redraw anyway. So I say that B is better off just not playing that game and going by the numbers- if above .5, stand and ignore A.

[Xhad]

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So the only bluff left is when B>.5 and A<B. A can try checking a loser, as you mentioned. But here I disagree- B knows that a redraw is unlikely to beat A. He knows a redraw would most likely make him a worse hand. For this to make sense for A to even try, B has to be very close to 100- closer to 100 than A is to 0. The more likely that is to be true, the less likely B is to win on a redraw anyway. So I say that B is better off just not playing that game and going by the numbers- if above .5, stand and ignore A.

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If A never checks a loser, B is incorrect to stand pat after A stands pat whether his hand is .51 or .90. Yes it is extremely unlikely that I will outdraw you if I draw to a .90, but since I am NEVER ahead when you stand pat I am correct to take a nonzero chance of winning instead of a zero chance.

If you know that I know this, drawing to your .52 when I show a .58 is definitely wrong. I will always stay pat if you draw, that much you have correct, but if you stand pat I am forced to draw out of respect for the number of legitimate hands you could be standing pat with (even if I realize you could be doing this, by the way). You will outdraw my .58 only 42% of the time, while I will back into a loser 52% of the time if you trick me into drawing. By drawing you are turning a +EV situation into a -EV one!

In fact, while I haven't the patience to run the numbers right now, if you are in the A seat and take a strategy of perfectly telegraphing your hand every time (which is what you advocate), you completely negate the only advantage you have, meaning that B may be EV neutral or maybe even +EV since he has position on you. The fact that you can see my hand before the draw is irrelevant if I also always know the relative strength of your hand before the draw.

[Alex/Mugaaz]

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But it seems to me that B's best odds is by ignoring what A does- just play the odds. If B has less than .5, he wants to redraw regaurdless- if A redraws, A is likely beating him, if A stands pat, A is surely beating him. So I see no reason for B to stand pat here, ever.

If B is greater than .5, and A redraws, he should stand pat- A is likely losing.

So the only bluff left is when B>.5 and A<B. A can try checking a loser, as you mentioned. But here I disagree- B knows that a redraw is unlikely to beat A. He knows a redraw would most likely make him a worse hand. For this to make sense for A to even try, B has to be very close to 100- closer to 100 than A is to 0. The more likely that is to be true, the less likely B is to win on a redraw anyway. So I say that B is better off just not playing that game and going by the numbers- if above .5, stand and ignore A.

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A is better off standing pat on some hands that are lower than b's upcard.

[gumpzilla]

Jerrod Ankenmann's answer in the Poker Theory forum looks interesting, and I'm confident that he's substantially more knowledgable than I when it comes to game theory, so I'd go take a look.

[Jerrod Ankenman]

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I came to the conclusion that the ideal strategy for A is to stand pat on all hands where 1.0 - B < A.

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That is part of the optimal strategy in some cases, when b > [sqrt(2)-1], which is around .4142. There, B always draws when A does.

But the interesting cases are when b < [sqrt(2)-1], where it turns out A stands pat when his hand exceeds [1 - sqrt(1-2b)]. This value is above b, so it does contain an element of bluffing. Furthermore, B randomizes between standing and drawing. Use the usual indifference conditions for these cases.

alThor

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I posted the solution in the Poker Theory forum, but I've now seen TWO references to this [1 - sqrt(1-2b)] thing, and I'm trying to understand why on earth this would come up.

Suppose there existed a strategy where if b < .5, A could make money by standing pat with a hand less than .5. Then B could respond to this strategy by simply redrawing all the time with b < .5. Then B would make money because A would have stood pat with a hand weaker than .5

Hence, no such strategy can be optimal.

Jerrod Ankenman

[Xhad]

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If player A stands, however, this signifies to player B that player A's hand is greater, and that player B should now hit. To have an edge for this move as a bluff, player A must have a hand greater than .50, because it is very likely to induce Player B to hit.

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Not necessarily. If I have .49 and B has .53, and I know B will hit if I stay, then staying is clearly the correct move. I'm an underdog either way this hand but staying makes me a 49% underdog as opposed to a 47% underdog.

[BB King's]

<font color="red"> Are you sure? As b approaches 0.5 from the right, Y approaches infinity. Isn't Y supposed to be &lt;=1?
</font>

You are quite correct. I have a problem here. Let’s take a closer look.

Given: b=0.58. We/I get: x=0.4 and Y=1.05.
Which is obviously wrong since Y&lt;=1.

We have to solve the system for Y==1 … getting …
b’=2-sqrt(2)=0,586 and x’=0,414

Okay ?!?

BB King’s ... Formerly known as the pokerplayer formerly known as Jack

[Darryl_P]

So are you saying that player B should always draw when his hand is between 0.5 and 0.586 and player A stands? If so, how about an EV calculation? If A's EV is higher than 0.5076 then I would argue player B is not playing optimally since player B can assure an EV of 0.4924 under the strategy I proposed.

[BB King's]

<font color="red"> So are you saying that player B should always draw when his hand is between 0.5 and 0.586 and player A stands? </font>

Yes – I am !!! For the record he should also, always draw with a hand lesser than 0.5.

<font color="red">If so, how about an EV calculation?</font>

Here is the EV-Calculation’s.

EV(B/Draw)= (1-x)/2
EV(B/Stand)= (b-x)/(1-x)
EV(A/Draw)= 1-b
EV(A/Stand)= y*x

BB King’s ... Formerly known as the pokerplayer formerly known as Jack

[Darryl_P]

I meant an overall EV calculation, ie. what is player A's expectation for the game (I'm looking for a single number between 0 and 1) if both players adopt optimal strategies?

I've found a way for player B to achieve at least 0.4924 which means A cannot achieve more than 0.5076 (assuming my calculations are correct which I think they are)

I have a feeling that under your proposed solution A's EV would be higher which would indicate that B's strategy is not optimal.

Of course I could calculate it myself, but I figure since I went through the trouble of calculating it for my own proposed solution I figured I'd ask you to take the time to do it for yours.