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

*x*where

_{c,l}*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

*x*= 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

_{c,l}*x*<= 1

_{2,KF}+ x_{2,KC}+ ... + x_{2,BRC}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) * x*--- the probablity that A will get card rank

_{c,l}* y_{d,m}* V(c,d,l,m)*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

*x*'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

_{l}*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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