Classic Type Game Theory Problem
 

I just thought of this problem recently when a player mistakingly exposed his pat lowball hand too soon in a triple draw game against an all in opponent.

I realized that the situation rephrased in rigorous terms is a simply stated classic type game theory problem, perhaps never addressed before. Just in case that's so, we'll call it the Sklansky Exposed Pat Hand Problem. It goes like this:

Player A and Player B are both dealt a real number from zero to one. Higher number wins. No betting except for antes. Player A looks at his downcard and decides whether to keep it or replace it. If he replaces it he gets that second card face down. After Player A acts, Player B has the same option. And of course his decison will be based partially on what A did. But the thing about this game is that Player B's first card is face up. So B knows that A's decision to replace was based on what A saw.

If there is $100 in the pot and both players play perfectly what is the EV for both players? What is the optimum strategy?

I'm going to put this question in the Poker Theory, Probability, and Science Math and Philosophy Forums at the same time.

Re: Classic Type Game Theory Problem
 

all i know so far is B is losing

interesting how bluffing factors into this game. you can force your opponent into making a -EV decision by making him throw away a winner and losing most of the time with his new draw

of course he will then catch on to that and start "calling" bluffs. are we talking about only ONE game here or do they keep playing and remembering past actions?

Re: Classic Type Game Theory Problem
 

I've been trying to solve this a while now, this is a lot harder than I thought.

Re: Classic Type Game Theory Problem
 

[ QUOTE ]
I've been trying to solve this a while now, this is a lot harder than I thought.

[/ QUOTE ]

I give up, my math knowledge is too limited.

I came to the conclusion that the ideal strategy for A is to stand pat on all hands where 1.0 - B < A. If this is true A is a favorite as long as B is less than somwhere between 0.63 to 0.66.

Re: Classic Type Game Theory Problem
 

[ QUOTE ]
I just thought of this problem recently when a player mistakingly exposed his pat lowball hand too soon in a triple draw game against an all in opponent.

I realized that the situation rephrased in rigorous terms is a simply stated classic type game theory problem, perhaps never addressed before. Just in case that's so, we'll call it the Sklansky Exposed Pat Hand Problem. It goes like this:

Player A and Player B are both dealt a real number from zero to one. Higher number wins. No betting except for antes. Player A looks at his downcard and decides whether to keep it or replace it. If he replaces it he gets that second card face down. After Player A acts, Player B has the same option. And of course his decison will be based partially on what A did. But the thing about this game is that Player B's first card is face up. So B knows that A's decision to replace was based on what A saw.

If there is $100 in the pot and both players play perfectly what is the EV for both players? What is the optimum strategy?

I'm going to put this question in the Poker Theory, Probability, and Science Math and Philosophy Forums at the same time.

[/ QUOTE ]Can i guess? A wins someplace between $66 and $75 a hand. I'll take the midpoint as a in the dark guess. 70.83cents a hand.

Re: Classic Type Game Theory Problem
 

The very basic strategy should be for Player A should hit if his hand is less than Player B's, and stand if his hand is greater. Every hit would then give his hand an EV of .5, and Player B should then hit if his hand is less than .5, and stand if greater.

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. Player B is also aware of the potential of this bluff, and should only hit on Player A's stands when Player B's are slightly above .50...let's say .50 to .75 or so. Anything greater than .75 yeilds a small expected return, especially when considering how likely it is that A is bluffing.

The EV of both strategies then becomes more difficult to calculate because it depends on psychology and the players.

Re: Classic Type Game Theory Problem
 

Not to be a nitpicker David, but is this a repetitive game or one time/occasional only? It matters.

No, it doesn\'t !!!
 

You have to find the optimal solution.

It doesn't matter wheter it is a repetitive game or a one time/occasional only.

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

Solution !!!
 

If (b<=0.5 and a<=0.5) then A=Draw and B=Draw
If (b<=0.5 and 0.5<=a) then A=Stand and B=Draw

If (0.5<=b and x<=a) then A=Stand and ‘B=Y*Draw+(1-Y)*Stand’
If (0.5<=b and a<=x) then A=Draw and B=Stand


x=sqrt(2*b-1) Y=(1-b)/x


ex.
Given: b=0.82 and a=0.81
Calculating: x=0.8 Y=0.225
A Stands Pat and B is Drawing 22.5% of the time.


EV-Calculation is left as an exercise to the reader.


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

Re: No, it doesn\'t !!!
 

[ QUOTE ]
You have to find the optimal solution.

It doesn't matter wheter it is a repetitive game or a one time/occasional only.

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

[/ QUOTE ]

It does matter as to whether you should always just use the mathematically optimal solution if there is one or a mixed strategy with opponents who have previously seen how you play. This is similar to knowing in triple draw lowball the frequency with which your opponent will attempt to freeze you on a worse hand or stand pat on a bluff on the 2nd draw and bet out on the end.

Re: Classic Type Game Theory Problem
 

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

[/ QUOTE ]

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

Re: No, it doesn\'t !!!
 

[ QUOTE ]
[ QUOTE ]
You have to find the optimal solution.

It doesn't matter wheter it is a repetitive game or a one time/occasional only.

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

[/ QUOTE ]

It does matter as to whether you should always just use the mathematically optimal solution if there is one or a mixed strategy with opponents who have previously seen how you play. This is similar to knowing in triple draw lowball the frequency with which your opponent will attempt to freeze you on a worse hand or stand pat on a bluff on the 2nd draw and bet out on the end.

[/ QUOTE ]

I'm pretty sure D.S is looking for the game-theory sound way of playing. You can always adjust from that basic strategy according to your opponents tendencies.

Re: Solution !!!
 

Are you sure? As b approaches 0.5 from the right, Y approaches infinity. Isn't Y supposed to be <=1?

What if you simply use Y=(1-b)/b ?

With this condition, player A will just barely have to decide to never bluff -- the typical condition in these types of problems.

When A has the ideal bluffing hand (ie. just a tiny fraction less than B's hand), his EV from bluffing (ie. standing) is Y*b

From drawing his EV is 1-b so the idea is to make these equivalent, ie Y*b = (1-b), or Y=(1-b)/b

Or am I missing something?

Re: Classic Type Game Theory Problem
 

Here's my take on it, show me where I'm wrong.

Terminology:

Avalue is Player A's card. Bvalue is player B's card.

Axioms:

First off- your odds of winnign a hand is 1-value. A 1 has no chance, a 0 has 100%, a .5 is even money.

Secondly- a randomly drawn card, assuming even distribution, has an average value over the long term of .5.
This means if a player draws, his hand basicly becomes a .5 (averaged over many many hands).



Strategy no bluffing:

Starting hand 1: A>B, B<.5. A should redraw. B should stand pat, as he is expected to win over A's .5. B wins 1-Bvalue percent of the time. Bvalue can be assumed to be the average value for the bottom half- .25. So B wins 75% of the time. This combo occurs .5*.75=.375 of the hands. So B gets .375*.75=.28125= $28.125 of expectation.

Starting hand 2: A<B, B<.5. A should stand pat. B sees the stand pat and redraws. A wins 1-Avalue percent of the time, B wins Avalue percent of the times. THis occurs .5*.25=.125 of the hands. B value's average hand is .25. A is less than him, so his average hand is .125. So B here gets .125*.125=.015625, or 1.5625 in expectation

Starting hand 3: B>.5. This is the tricky one. If A stands pat, B will redraw as he is beat. If A redraws, B has to redraw as A is now a .5. So B has to redraw either way. This puts him at .5 A will decide what to do based on his card. If he is <.5, he stands pat, if >.5, he redraws. When he redraws (50% of the time, or 25% of the total of the game), he has a .5 and thus a 0 expectation. When he is under .5, B will win 1-Avalue of the time. A here over the long run is .25 (middle of the range). This occurs .25 of the time. So B's expected value here is .25*.25=.0625=$6.25 in expected value.


In short, without bluffing B will win 35.9375% of the time. Huge advantage for A.


Adding in bluffing is more complicated, I'll try and post later. It may need to wait untila fter work.

Re: Classic Type Game Theory Problem
 

Bleh, work is boring today. Besides, the more I thought about it, bluffing is irrelevant.

Ok, adding in bluffing.

If B>.5, he will draw. So his expected card value is .5. If A is <.5, drawing on a bluff will hinder him. If A>.5, standing pat on a bluff will hinder him. So A should not bluff if B>.5. Thats ok, we're already huge favorites here. Lets not get too greedy [img]/images/graemlins/smile.gif[/img]


If A<B and B<.5, A shouldn't bluff. A redraw will take A up to .5, and make B a heavy favorite. A already has the high advantage here, can't waste it. B won't redraw on the bluff anyway, as chances are A redrew and lost.


If A>B and B<.5, A shouldn't bluff. He won't get B to redraw, as chances are he'd redraw to a worse hand.


In short, unless you add a round of betting at the end, bluffing is useless in this game. So I refer to my previous analysis.

Re: Classic Type Game Theory Problem
 

Umm, I just realised I played my hand for lowball, as the original post mentioned it. Reverse all the conditionals, but the math and logic work with the conditions reversed.

Re: Solution !!!
 

I just calculated B's EV under this strategy and it turns out to be 0.4924 -- determined as follows:

If B's hand is less than .5, his EV will be .375 (avg. of .25 and .5, straightforward to show why)

If B's hand is some number b >=.5 his EV will be given by:

b^2 + (1-b)*((1-b)/b)*((1-b)/2)

If A's first card is less than b, he draws and will miss with a probability of b -- that's the first way B can win and that's the first term in the above.

Or if A's first card is more than b (probability of 1-b), then B draws with probability (1-b)/b and wins with probability (1-b)/2 -- that's the second term in the above.

Now that we know B's winning probability as a function of his hand b, we just need to integrate the function from 0.5 to 1 (which gives the area under the graph) and multiply by 2 to scale it up since our interval has length 0.5 and not 1.

Evaluating that integral produces 0.609814 if I didn't make any arithmetical mistakes. Seems reasonable, though, since it basically says B's expectation will be 0.609814 if he's dealt a card higher than 0.5. If bluffing were not allowed it would be 0.625 and so being slightly less seems reasonable.

The final answer of 0.4924 is just the average of 0.6098 and 0.375

Thanks !!!
 

<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 got a point there. Back to HQ !

<font color="red"> When A has the ideal bluffing hand (ie. just a tiny fraction less than B's hand), his EV from bluffing (ie. standing) is Y*b </font>

No, it is Y*x ... We have to calculate the situation when A is in marginal bluffing-position.

<font color="red"> From drawing his EV is 1-b so the idea is to make these equivalent, ie Y*b = (1-b), or Y=(1-b)/b </font>

I somewhat aggree ... if we substitute Y*b with Y*x !?!

<font color="red"> Or am I missing something? </font>

Maybe not - maybe a little - but not much.

Thanks !!!

Re: Classic Type Game Theory Problem
 

[ QUOTE ]
In short, unless you add a round of betting at the end, bluffing is useless in this game. So I refer to my previous analysis.

[/ QUOTE ]

You are making a huge error in that you fail to realize that if A never "bluffs" by checking a loser, B should always draw no matter what his hand is whenever A stands pat. If B adopts this strategy, A can exploit it by intentionally checking if B has a pat hand if the probability of B making a worse hand is greater than the probability of A improving to beat B. I posted a more detailed analysis in the Poker Theory forum.

Re: Classic Type Game Theory Problem
 

Very interesting question. I haven't read the analysis of anybody else yet. I'll call the numbers of players A and B lowercase a and b, respectively. There are 2 antes in the pot, each player has paid 1.

B should never switch if A has switched and b &gt; .5, since then b will be likely to outperform A's second number. Likewise, if A has switched and b &lt; .5, B should always switch. So when A switches, B's strategy is easy.

It is when A stands pat that things get complicated for B, because this is the only area where A can "bluff." The simple strategy of switch whenever A stands pat fails to situations like a = .65, b = .7 where A benefits by tricking B into switching a winning number.

My guess is that most likely some kind of mixed strategy is going to be necessary here, and I don't have the experience to work out what's optimal in that regard. Let's assume that B adopts a slightly more complex strategy than above: B switches only when b &lt; .5 or when A stands pat and .5 &lt; b &lt; beta, where beta is some threshold value above which B always stands pat. If A knows B will employ this form of strategy, and happens to know beta, how does A proceed? We can break down into three regimes:

b &lt; .5: EV of 0 when A switches, since b will switch too, and EV of 2a - 1 when A stays, since b will switch. So, when b &lt; .5, A should stand pat if and only if a &gt; .5. Averaging over all a, A has an EV of .25 whenever b &lt; .5.

.5 &lt; b &lt; beta: B will stand pat if A switches, so switching nets A an EV of (1 - 2b)for switching and an EV of (2a - 1) for staying. 2a - 1 exceeds 1 - 2b for a &gt; 1 - b, so a should stand pat for all a &gt; 1 - b and switch for a less than that. Averaging this one is more complicated, so it's quite possible that I've made a mistake here, but I get that averaged over all a for a given b, a picks up an EV of (1 - b)^2 here. This analysis is only valid for b &gt; .5, so the best A can do over this range is .25, and the larger b gets the worse A does, which makes sense, so this looks plausible. We'll average over b at the end.

When b &gt; beta, A's choice is simple. B will never switch, so A's strategy becomes stand pat when winning, change when losing. The EV when switching is (1 - 2b), and the EV when standing pat is 1 (because A only stands pat when winning.) Averaged over all a, the EV in this range is (1 - 2b)b + (1-b) = 1 - 2b^2. As a sanity check, this gives EV of -1 for A when b = 1, as it should.

Okay, so now we know how A does on average for any given b. We need to average over all b for a given beta to figure out the value of the game for a particular strategy. I get (on the off chance that anybody has actually been following all of this thus far, this is another integral you probably want to check me on) that the average over all b for a given strategy is 1/6 + (beta^2)*(beta - 1).

At beta = 2/3, this EV gets pushed down to 1/54, which is the minimum. So for a fixed strategy, beta = 2/3 is the best B can do, and he's a slight loser there, just under 2 cents on the dollar by my calculation. I'm actually surprised he's not more of a loser. And mixed strategies are beyond me at the moment.

Re: 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&gt;.5 and A&lt;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.

Re: Classic Type Game Theory Problem
 

[ QUOTE ]
So the only bluff left is when B&gt;.5 and A&lt;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.

[/ QUOTE ]

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.

Re: Classic Type Game Theory Problem
 

[ QUOTE ]
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&gt;.5 and A&lt;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.

[/ QUOTE ]

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

Re: Classic Type Game Theory Problem
 

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.

Re: Classic Type Game Theory Problem
 

[ QUOTE ]
[ QUOTE ]
I came to the conclusion that the ideal strategy for A is to stand pat on all hands where 1.0 - B &lt; A.

[/ QUOTE ]

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

But the interesting cases are when b &lt; [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

[/ QUOTE ]

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 &lt; .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 &lt; .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

Re: Classic Type Game Theory Problem
 

[ QUOTE ]
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.

[/ QUOTE ]

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.

Is this better ?!?
 

<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

Re: Is this better ?!?
 

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.

Yes – I am !!!
 

<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

Re: Yes – I am !!!
 

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.

Re: Is this better ?!?
 

[ QUOTE ]
So are you saying that player B should always draw when his hand is between 0.5 and 0.586 and player A stands?

[/ QUOTE ]

It seems he should. In fact according to my math, B should never stay pat on a hand less than 2 - sqrt(2) unless he knows that A is playing incorrectly.

The reason is that the only potential "bluffing" situations are those in which b &gt; .5, but a + b &gt; 1 (any time one of the above is not true, A should not stand pat even if B will draw 100% of the time because A has a better chance of winning if he simply draws himself). But for all hands from .5 to 2 - sqrt (2), even if A is bluffing at EVERY SINGLE OPPORTUNITY B still has to draw out of respect for the number of legitimate hands A could be standing pat with.

Re: Is this better ?!?
 

[ QUOTE ]
even if A is bluffing at EVERY SINGLE OPPORTUNITY B still has to draw out of respect for the number of legitimate hands A could be standing pat with.

[/ QUOTE ]

I'm not sure he has to draw with 100% probability, though. It looks to be enough to draw with a high probability (equal to (1-b)/b, say).

It doesn't seem right for B to give up so much EV by drawing every single time if he doesn't have to.

Do you know the overall EV for each player under your assumed optimal strategies? If you say A's is more than .5076 then I'd like to pit my player B against your player A and see what happens in a simulation.

Re: Classic Type Game Theory Problem
 

[ QUOTE ]

Player A and Player B are both dealt a real number from zero to one.

[/ QUOTE ]

How?

Re: Is this better ?!?
 

I ended up calculating the overall EV under your strategy as well and it turns out to be 0.4938 for player B which beats my 0.4924

I couldn't do it analytically so I just made a big excel spreadsheet and did it numerically. At least I was able to double check my EV figure which turns out to be correct, but evidently not optimal as your strategy is better.

I should have known something was up if A couldn't gain anything by bluffing. The 0.4924 just seemed so good that I thought it couldn't be beat, but alas, it can. Live and learn.

Re: Classic Type Game Theory Problem
 

Player A's bluffing strategy has to include the actual value of B's card- ie the probability of drawing over B's card combined with the probablility of B drawing under A's card.

Re: Classic Type Game Theory Problem
 

[ QUOTE ]
How?

[/ QUOTE ]

I think it's pretty obvious that he means from a uniform distribution, and if generating real numbers (because of a countable computable/uncountable uncomputable issue) bothers you, then use a uniform distribution over some large (a trillion will probably do nicely) number of evenly spaced rationals over that range.

I think it's kind of interesting that nobody has commented on Jerrod's solution in this thread.

Re: Is this better ?!?
 

I should add that I did my EV calculation above based on BBB's summary in the other forum. Did you catch the modification to A's strategy for values of B's upcard between 0.5 and 0.586?

Re: Classic Type Game Theory Problem
 

[ QUOTE ]
[ QUOTE ]

Player A and Player B are both dealt a real number from zero to one.

[/ QUOTE ]

How?

[/ QUOTE ]

Through the magic of toy game technology, son.

:P

Jerrod

Re: Is this better ?!?
 

[ QUOTE ]
[ QUOTE ]
even if A is bluffing at EVERY SINGLE OPPORTUNITY B still has to draw out of respect for the number of legitimate hands A could be standing pat with.

[/ QUOTE ]

It doesn't seem right for B to give up so much EV by drawing every single time if he doesn't have to.

[/ QUOTE ]

He's not giving up that much EV because the hands in that range aren't that likely to back into losers that often.

Basically what it comes down to is that for a patbluff to work, B not only has to draw when he has the best hand, but he also has to back into a worse hand than he already and have that hand not still be winning. The hands in the .5-&gt;.59-ish range I suggest just aren't that far above random to begin with.

Of these hands, the hand for B with the widest patbluff range is 2-sqrt(2), which is about .59. That range is everything down to about .41. Patbluffing anything lower than that is -EV for A even if you fall for it 100% of the time (which is why I said draw "unless you know A to be playing incorrectly"), so that means that if you draw to hands in this range you are falling for a profitable bluff at most about 30% of the time, and a significant percentage of those times you will still be ahead of those hands even if you fall for it.

[ QUOTE ]
Do you know the overall EV for each player under your assumed optimal strategies?

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No I don't because I haven't completely finished the problem for the higher ranges of hands. =X However, I am convinced that I have the optimal strategy up to b = 2 - sqrt(2). The math involved can be found here. Note that the inequality is basically EV(catch bluff) &gt; EV (draw) and that 2 - sqrt(2) was the minimum value of b necessary to make it true if A is bluffing as much as he can and still have it be theoretically profitable.

Re: Classic Type Game Theory Problem
 

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How?

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I think it's pretty obvious that he means from a uniform distribution, and if generating real numbers (because of a countable computable/uncountable uncomputable issue) bothers you, then use a uniform distribution over some large (a trillion will probably do nicely) number of evenly spaced rationals over that range.

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It has nothing to do with being bothered by uncountability or computability. There is no uniform distribution on [0,1].