# Last Round Poker Play as Linear Programming

(I wish I remembered more from my graduate algorithms class. Unfortunately we didn't have a textbook or even a course reader, just whatever Serge Plotkin felt like talking about that day.)

I've been trying to think about how to solve last-round play in 2-7 lowball when both players draw one. (I peeked ahead in Chen and Ankenman and they do solve the [0, 1] game for a single street and multiple raises but I didn't find anything about applying the results to a discontinuous game yet. Still, it will be interesting working through to that point.)

It occurred to me that finding the optimal play against a particular opponent strategy is just a linear programming problem. Not that I expect this to be an original insight or anything.

Player A's strategy is defined as, for each card he draws, a probability distribution of each of his possible lines: check/fold, check/call, check/raise/fold, etc. (If you are interested in this sort of thing, check out my strategy calculator allowing only a single line per card.) This can be expressed as a set of variables xc,l where c is the rank of the card drawn, l is the particular line.

The constraint is, of course, that for each c the sum over l of xc,l = 1. This can be normalized by giving a "default" line which is used whenever one of the other lines is not, so that the constraints are all of the form

x2,KF + x2,KC + ... + x2,BRC <= 1

And note that all the x's are >= 0.

Player B's strategy is defined the same way but his preferences must be treated as constants to yield a linear system.

The function to be maximized is, of course, A's expected value. This is just the sum over all terms of the form P(c && d) * xc,l * yd,m * V(c,d,l,m) --- the probablity that A will get card rank c and B will get card rank d, times the probability that A will have picked strategy l and B will have picked strategy m, times the outcome from A's perspective of those choices. One of the xl's is actually 1 minus the sum of all the other lines due to our normalization above but that's still a linear term. If we treat the y's as constants then this is linear. Easy!

Unfortunately it's not clear to me how to solve for both A and B simultaneously; that appears to be inherently nonlinear. Maybe after I finish The Mathematics of Poker.
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