acehole60's avatar


1 points

In A8 hand in BB, first hand on top right table. Board: AJ52J. Any merit to donk (bluffing) river?

Oct. 7, 2014 | 4:02 a.m.

I should say, that all equities above are "ex-showdown equities", i.e. the expectation each hand has in the postflop betting. For instance, when X holds a K and folds to a bluff he will lose the pot (-P), since he was 'entitled' to it.

July 19, 2013 | 7:59 a.m.

Hi guys,

I'm trying to solve a variant of the AKQ game and it seems like I'm doing something wrong. I would love any comments you might have.

The game is as follows: We have two players, X and Y, and consider the half-street game, i.e. X checks to Y, Y can bet or check, X then decides to call or fold. Suppose X's distribution is purely Kings (bluffcatchers), while Y's distribution is AAQ (i.e. 2/3 winners and 1/3 losers). The pot is of size P and betsize is 1. What is the solution to this game? I do as in Mathematics of Poker page 145:

To find X's calling frequency (c), we make Y indifferent to bluffing. The expected value of Y bluffing with a Q is

<Y, bluff Q> = (X calls, lose 1) + (X folds, win P) = c*(-1) + (1-c)*P = P - c(1+P)

The expected value from Y checking back a Q is

<Y, check Q> = 0.

Setting these equal we find the usual ratio of calls

c = P/(1+P)

which implies the usual ratio of folds 

(1-c) = 1/(1+P) = \alpha.

In the same way, we solve the game for Y's bluffing frequency with a queen (b). X's expected value for calling with a K is

<X, call K> = (Y has aces, lose 1) + (Y has Q and bluff, win 1) = (2/3)*(-1) + (1/3)*b*1

and X's expected value from folding is

<X, fold K> = (Y has Q and bluff, lose P) = (1/3)*b*(-P)

Setting these equal we find

 (2/3)*(-1) + (1/3)*b*1 =  (1/3)*b*(-P) => b(1+P) = 2 => b = 2/(1+P)

Meaning that Y bluffs b of his queens. In the AKQ game we find b=1/(P+1) = \alpha and we see that what is in the nominator of b is the amount of aces to queens in Y's range (so we could generalize the game and say that Y's distribution is x1 aces and x2 queens and arrive at b = (x1/x2)/(P+1)).

So, at first sight the game is solved and we get basically the same solution as in the ordinary AKQ game (atleast for X's calling frequency, whereas Y is of course able to bluff more queens as his range gets stronger). BUT, I feel like I have made a mistake somewhere, since it feels as though Y's range should play into the calling frequency of X (so, I would expect c in this game to be different from c in the AKQ game); to see this, we can entertain the situation where the pot is not too big and Y's range is 99% aces and 1% queens. Now, even if Y bluffs all queens, bluffcatching from X's point of view will clearly be dominated by folding (since Y's range is so strong, we will be giving her a bet too often because her range is so strong)...this hypothetical situation seems to me to suggest that my analysis above is wrong. But I can't seem to find the error.

This obviously became tl;dr, but I hope somebody will chime in and help me with my mistake. Hopefully, we can generate some discussion, as I think solving these asymmetric games are key to applying game theory to actual poker hands. I'd be willing to discuss whatever game theory/toy game related as well.


July 18, 2013 | 11:14 p.m.

Comment | acehole60 commented on Phil Galfond Q&A

It's also in the link I posted. Starts at ~40mins.

June 27, 2013 | 7:20 a.m.

Comment | acehole60 commented on Phil Galfond Q&A

June 26, 2013 | 11:17 p.m.

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