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The Simplest Game of Poker...

Take a deck of three cards--High, Middle, and Low. Deal one card each to two players, Opener and Responder. Both players ante. Opener checks or bets. If Opener checks, Responder then checks or bets. If Opener bets, Responder then folds or calls. If Opener checks and Responder bets, Opener than folds or calls.

This gives us three choices for Opener: cf, cc, b-. Since there are three cards the Opener then has twenty-seven pure strategies:

  • HcfMcfLcf
  • HcfMcfLcc
  • HcfMcfLb-
  • HcfMccLcf
  • HcfMccLcc
  • HcfMccLb-
  • HcfMb-Lcf
  • HcfMb-Lcc
  • HcfMb-Lb-
  • HccMcfLcf
  • HccMcfLcc
  • HccMcfLb-
  • HccMccLcf
  • HccMccLcc
  • HccMccLb-
  • HccMb-Lcf
  • HccMb-Lcc
  • HccMb-Lb-
  • Hb-McfLcf
  • Hb-McfLcc
  • Hb-McfLb-
  • Hb-MccLcf
  • Hb-MccLcc
  • Hb-MccLb-
  • Hb-Mb-Lcf
  • Hb-Mb-Lcc
  • Hb-Mb-Lb-

This gives us four choices for Responder: cf, bf, cc, bc. With three cards this gives us 64 pure strategies for Responder, and thus 64 x 27 = 1728 cells in the outcomes matrix for each possible value of ante and bet.

Anybody calculated the von Neumann-Morgenstern mixed-strategy equilibrium for this game?