If neither player folds – either both players check, or there is a bet and a call – then both cards are revealed and the player with the higher card takes all 4 chips.
Presumably you mean to say that if both players check, the player with the higher card takes the 2 chip pot, not 4 chips?
Anyway here is the unique (up to choice of x) Nash equilibrium, if my calculations are correct:
Heh, I tried to game-theory it all out and got nonsensical answers (my set of equilibria wasn’t convex!), and I wasn’t about to redo it, so I’m glad someone managed to determine [what I’m presuming is] the correct Nash equlibirum...
It is a little surprising to me that nyjnlf purpxvat jura fgnegvat jvgu n 2 vf gur bayl sbeprq pubvpr jvgubhg na boivbhf ernfba sbe vg. Gur bgure 6 sbeprq pubvprf ner rnfl gb qrgrezvar jvgubhg univat gb qb zhpu zngu, ohg abg gung bar. Naq gur bgure 5 pubvprf gung qb erdhver qbvat zngu gb qrgrezvar qba’g raq hc nf 1f be 0f.
But yeah, as Zvi says, I guess part of the whole question is how far into donkeyspace you want to go. i would’ve stuck with Nash, if I could figure it out, but...
Presumably you mean to say that if both players check, the player with the higher card takes the 2 chip pot, not 4 chips?
Anyway here is the unique (up to choice of x) Nash equilibrium, if my calculations are correct:
1 tbrf svefg: org jvgu cebonovyvgl k, arire pnyy
2 tbrf svefg: arire org, pnyy jvgu cebonovyvgl k+1/3
3 tbrf svefg: org jvgu cebonovyvgl 3k, nyjnlf pnyy
1 tbrf frpbaq: org jvgu cebonovyvgl 1⁄3, arire pnyy
2 tbrf frpbaq: arire org, pnyy jvgu cebonovyvgl 1⁄3
3 tbrf frpbaq: nyjnlf org, nyjnlf pnyy
Heh, I tried to game-theory it all out and got nonsensical answers (my set of equilibria wasn’t convex!), and I wasn’t about to redo it, so I’m glad someone managed to determine [what I’m presuming is] the correct Nash equlibirum...
It is a little surprising to me that nyjnlf purpxvat jura fgnegvat jvgu n 2 vf gur bayl sbeprq pubvpr jvgubhg na boivbhf ernfba sbe vg. Gur bgure 6 sbeprq pubvprf ner rnfl gb qrgrezvar jvgubhg univat gb qb zhpu zngu, ohg abg gung bar. Naq gur bgure 5 pubvprf gung qb erdhver qbvat zngu gb qrgrezvar qba’g raq hc nf 1f be 0f.
But yeah, as Zvi says, I guess part of the whole question is how far into donkeyspace you want to go. i would’ve stuck with Nash, if I could figure it out, but...
Jryy, bapr lbh xabj gung frpbaq-1 arire pnyyf naq frpbaq-3 nyjnlf orgf naq pnyyf, lbh xabj gung svefg-2: purpx naq pnyy fgevpgyl qbzvangrf svefg-2: org (fvapr vg qbrf gur fnzr ntnvafg 3 naq fyvtugyl orggre ntnvafg 1).
Ohg vs nf svefg-2 ntnvafg frpbaq-1 lbh purpx, frpbaq-1 zvtug org naq (fvapr lbh qba’g xabj vg’f 1 engure guna 3) lbh zvtug sbyq va erfcbafr (-1 sbe lbh), zrnavat lbh raq hc qbvat jbefr guna lbh jbhyq unir unq lbh org (+1 sbe lbh). Fb V qba’g frr ubj vg qbzvangrf?
Guvax bs purpx-naq-pnyy nf n cevzvgvir npgvba, gura lbh xabj lbh jvyy abg sbyq. Vg’f gehr gung vg’f abg boivbhf gung purpx-naq-sbyq qbzvangrf org.
Ah, makes sense now, thanks!
It’s worth noting that against the actual field I faced, you can do much better than Nash.