Hypothetical Heads Up Gambling Situation

[jay_shark]

Yup , you have to bluff a certain percentage of the time .

That's not too difficult to do .

Another question you could ask is what should player B's calling range be .

I sort of answered that already but you have to work it out for all the ranges .

[roommate]

if you are content with "a certain percentage of the time" than you are missing the true point of nicho's post...

this is multi level... and there is not perfect way for either player to play without a read on the opponent.

[jay_shark]

Yup I agree . The first time I read the question I didn't read it carefully .

I could figure this out but the solution would be ugly . The first problem you posed had a pretty nice result .

[thechainsaw]

[ QUOTE ]
I verified that my conjecture is true for all n .

Pretty cool .

[/ QUOTE ]

ten bucks says your conjecture is false - now THAT's poker

shove or fold?

[jay_shark]

Hey chainsaw , try it yourself .

The first problem has been solved .
Work it out for n=3 , then n=4 and you'll see a pattern emerge .

The final list Nicho provided is accurate .

[Nichomacheo]

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I could figure this out but the solution would be ugly . The first problem you posed had a pretty nice result .

[/ QUOTE ]

Anyone else man enough? [img]/images/graemlins/wink.gif[/img]

[thechainsaw]

[ QUOTE ]
Hey chainsaw , try it yourself .

The first problem has been solved .
Work it out for n=3 , then n=4 and you'll see a pattern emerge .

The final list Nicho provided is accurate .

[/ QUOTE ]

what is your answer for n=3?

[jay_shark]

Chainsaw , you need to ask yourself the following .

Since there is no forced action from the blinds , all you need to know is what's the probability you'll be dealt a better hand than the top one third with two more deals .

Here are the possibilities :

You can hit a better hand on the next deal which will happen a third of the time , or you can hit a better hand on the last deal .

Let x1 be the probability you get dealt a better hand than the minimum top third on your first deal .

Let X2 be the probability you get dealt a better hand on your second deal .

P(x1)=1/3
P(x2)=1/2*1/3

Add both results and you get 50 % .

Do you see why I multiplied 1/2 by 1/3 ?
This is because if you're dealt a hand in the bottom 50% of hands on the first deal , then you are obviously going to re-draw on the next hand .

[joeg]

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Here, knowing the % winning chance of a given hand against any other RANDOM hand is enough.

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This isnt true, take aces, it is your top 100% hand and wins 80% of the time, making a decision based on what % available hands your hand falls into is false because the realationship isnt linear.

As for knowing that you have a hand that will win 75% of the time, that alone isnt enough to tell you if it is playable, you need to relate it to how that percentage changes, and how likely you are to have a bigger edge, what that edge will be etc. This is no simple matter and certainly isnt given by a simple forumula based on either x% won or top y%. You can probably hack together a play hands that win x% of the time type system, but proving it correct, or generating a proper strategy would be very complicated indeed, much more complicated than any of the methods given here.

[jay_shark]

Joe , the first problem isn't as complicated as you think because there are no blinds in play . All you want to know with each hand dealt is the probability you will be dealt a better hand on your remaining attemps .

The answer we gave is precise and it depends on the top x % of hands for each hand dealt .