River bets

[cmyr]

I think it's worth noting a bit of the human element in this spot, which is that villian is more likely to put you on a busted flush draw betting a straight card then a busted straight draw betting a flush card, i.e. I think you get called more when you actually hit then you do when you hit your bluff.

[grizy]

you missed the point of my question... which is what's the likelihood hero has the hand he's representing from villain's perspective.

[Aisthesis]

Ok, let's say set-man calls river 2 with a probability of p1, which I'll first assume is less than 50%. And let's call the probability that LP bets river 2 p2. First question is the value of p2, for which my intuition is that it should be 100% given that p1 < 50%. Pot is $30.

Here's the EV:

1) LP checks behind with a probability of 1-p2, and his EV is 0 in that case.

2) LP bets pot with a probability of p2, and his EV is ((1 - p1)*30) - (p1*30) = 30 - (60*p1).

Well, obviously if p1 < 50%, then the maximum EV is setting p2 to 100%. And if p1 > 50%, then the maximum EV is setting p2 to 0%. If p1 = 50%, then it doesn't matter what you choose for p2, since the EV is always 0 for LP.

What I don't understand here, if we're restricting ourselves to river 2, is where you get the $36 on a bluffing rate of 20%.

That would mean that 80% of the time, LP checks behind and loses for an EV of 0.

20% of the time, he bets pot of $30. Half the time there he loses $30, and half the time, he wins $30. So, EV there is also 0 as I see it--as it would be if he bluffed 100% of the time or 0% of the time.

[Troll_Inc]

[ QUOTE ]

First question is the value of p2, for which my intuition is that it should be 100% given that p1 < 50%. Pot is $30.


[/ QUOTE ]

You have to consider both rivers together as if they were one.

Setman won't know which draw LP hit. So he doesn't have the luxury of changing his frequency between rivers 1 and 2. If he is going to call LP's bet 50% of the time, then he has to call it 50% on both river 1 and 2*.

LP has the luxury of deciding how often he is going to bet each river.





*cmyr makes a good point for practical application once an answer is determined. Setman might very well give credit more for someone drawing to a flush than a straight.

[Aisthesis]

I am. I set p1 the same regardless, and there are some other types of blank rivers (namely 5 and 6).

Ok, given p1 (and still considering only PSBs), we know that p2 is 100% if p1 < 50%, and p2 is 0% if p1 > 50%.

Now, let's consider an optimal value for p1 under the assumption that p1 is the probability of a call on BOTH rivers, and given p2, which is only the frequency of the bluff (whichever river LP makes, he definitely bets 100% of the time).

Ok, now on river 1, the EV for setman looks like this: On a fold, of course, it's 0, but he calls with a prob of p1, and loses 30*p1.

On river 2, LP only bets with a prob of p2. If he checks, setman wins 30. Subtotal here is thus (1-p2)*30 for setman.

If he bets, then setman wins 60 (the only relevant number, since a fold is 0). EV there is 60*p2*p1.

So, that means the total EV for setman (both rivers) is: -(30*p1) + 30 - (30*p2) + 60*p2*p1 = 30*(1 - p1 - p2 + 2*p1*p2).

To maximize this value, we only need to look (given p2) at 2*p1*p2 - p1 = p1*((2*p2) - 1). But that's still easy, p1 = 1 if p2 > 50%, and p1 = 0 if p2 < 50%, and it doesn't matter if p2 = 50%.

So, basically, assuming that those are the only two interesting rivers and that setman has to call the same number of times on both, the optimal values for both players are 50%.

Hence, LP bets river 1 always and bets river 2 50% of the time. Setman calls both rivers 50% of the time.

EV for LP is thus $45 on river 1 and 0 on river 2 for a total (for just these two rivers) of $45/2 = $22.50

EV for setman: He loses 15 on river 1 for an EV there of -$7.50

River 2: Half the time, LP checks and setman wins $30. (subtotal = $15)

The other half he wins $60 when he calls, which is again half the time. So, that's another $15. So, total EV on river 2 is $30.

Adding the two together, EV for setman is $7.50.

Ok, I think I'm starting to get your point. While setman is way ahead, by bluffing half the time on the draw you don't make, a fairly bad draw gets a bigger share of the pot than the hand that's ahead.

I think I'll try the same calculation on a half-pot bet...

[Aisthesis]

Ok, now on HPBs (half-pot).

Setman calls with some probability p1. LP becomes indifferent to bluffing river 2 when p1 = 2/3.

Now let's look at p2. We need to make it invulnerable to strategy changes for setman.

On river 1, setman now loses only p1*15.

On river 2, setman wins 30 when LP checks. Subtotal there is (1-p2)*30.

When LP bets, and he calls, he wins 45, for a subtotal of 45*p1*p2.

So, now the total EV for setman is -(p1*15) + ((1-p2)*30) + 45*p1*p2 = 15*(2 - p1 - (2*p2) + (3*p1*p2)).

To maximize this value, given p2, we need only look at 3*p1*p2 - p1 = p1*(3*p2 - 1).

So, the point of indifference for setman is when p2 = 1/3.

Hence, setman calls both bets 2/3 of the time, and LP bluffs 1/3 of the time.

Now let's look at the EV for both players.

On river 1, LP wins $30 1/3 of the time and $45 2/3 of the time, hence 10 + 30 = 40. So, his EV here is $20 (50% of the time we get river 1).

On river 2, LP breaks even, so total EV is $20, and EV for setman should thus be $10.

Ok, LP can do better on these rivers with a PSB than with HPB.

Of course, it's still not clear that his call on the turn was justified, because there are all the blanks as well as the paired boards. Moreover, setman should probably call MORE frequently if one of the less plausible straights hits. Figuring those in should give setman more than half of the $30 in the pot.

This has gotten kind of interesting, though. I think that is an argument for PSBs.

One other aspect: We're assuming setman really does have a set here. What if he instead is betting a big draw and, knowing that LP likes to bluff, will CR the nuts when he makes his hand?

[grizy]

time to find the perfect bayesian equilibrium.

sorry, drunk, and had to throw that out there.

[grizy]

[ QUOTE ]
You have to consider both rivers together as if they were one.

[/ QUOTE ]

[ QUOTE ]
*cmyr makes a good point for practical application once an answer is determined. Setman might very well give credit more for someone drawing to a flush than a straight.

[/ QUOTE ]

These two statements are inherently contradictory, do you see why?

[Troll_Inc]

[ QUOTE ]


[ QUOTE ]
You have to consider both rivers together as if they were one.

[/ QUOTE ]

[ QUOTE ]
*cmyr makes a good point for practical application once an answer is determined. Setman might very well give credit more for someone drawing to a flush than a straight.

[/ QUOTE ]

These two statements are inherently contradictory, do you see why?

[/ QUOTE ]

Not really.

The first deals with the math. redux of a simple poker problem, and the second deals with actual issues faced in the play of a specific hand.

This post is probably better for the theory or probability forum, but there are limitations that they would have with it. Just as posting it here has downsides, e.g. the point of the exercise clearly goes over the head of someone like you. (E.g. see your point about the turn.)

[RoundTower]

I've given up reading this thread but

http://forumserver.twoplustwo.com/showfl...;gonew=1#UNREAD