The following simple game has one solution that seems correct, but isn’t. Can you figure out why?

**The Game**

Player One moves first. He must pick A, B, or C. If Player One picks A the game ends and Player Two does nothing. If Player One picks B or C, Player Two will be told that Player One picked B or C, but will not be told which of these two strategies Player One picked, Player Two must then pick X or Y, and then the game ends. The following shows the Players’ payoffs for each possible outcome. Player One’s payoff is listed first.

A 3,0 [And Player Two never got to move.]

B,X 2,0

B,Y 2,2

C,X 0,1

C,Y 6,0

The players are rational, each player cares only about maximizing his own payoff, the players can’t communicate, they play the game only once, this game is all that will ever matter to them, and all of this plus the payoffs and the game structure is common knowledge.

Guess what will happen. Imagine you are really playing the game and decide what you would do as either Player One, or as Player Two if you have been told that you will get to move. To figure out what you would do you must formulate a belief about what the other player has/will do, and this will in part be based on your belief about his belief of what you have/will do.

**An Incorrect Argument for A**

If Player One picks A he gets 3, whereas if he picks B he gets 2 regardless of what Player Two does. Consequently, Player One should never pick B. If Player One picks C he might get 0 or 6 so we can’t rule out Player One picking C, at least without first figuring out what Player Two will do.

Player Two should assume that Player One will never pick B. Consequently, if Player Two gets to move he should assume that C was played and therefore Player Two should respond with X. If Player One believes that Player Two will, if given the chance to move, pick X, then Player One is best off picking A. In conclusion, Player One will pick A and Player Two will never get to move.

**Why the Game Has No Solution**

I believe that the above logic is wrong, and indeed the game has no solution. My reasoning is given in rot13. (Copy what is below and paste at this link to convert to English.)

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Va fhz, gb svaq n fbyhgvba sbe gur tnzr jr arrq gb xabj jung Cynlre Gjb jbhyq qb vs ur trgf gb zbir, ohg gur bayl ernfbanoyr pnaqvqngr fbyhgvba unf Cynlre Gjb arire zbivat fb jr unir n pbagenqvpgvba naq V unir ab vqrn jung gur evtug nafjre vf. Guvf vf n trareny ceboyrz va tnzr gurbel jurer n fbyhgvba erdhverf svthevat bhg jung n cynlre jbhyq qb vs ur trgf gb zbir, ohg nyy gur ernfbanoyr fbyhgvbaf unir guvf cynlre arire zbivat.

Update: Emile has a great answer if you assume a "trembling hand."

If we multiply player 2's utility function by 100, that shouldn't change anything because it is an affine transformation to a utility function. But then "P1: B, P2: Y" would maximize the sum. Adding values from different utility functions is a meaningless operation.

You're right. I'm not actually advocating this option. Rather, I was comparing EY's seemingly arbitrary strategy with other seemingly arbitrary strategies. The only one I actually endorse is "P1: A". It's true that this specific criterion is not invariant under affine transformations of utility functions, but how do I know EY's proposed strategy wouldn't change if we multiply player 2's utility function by 100 as you propose?

(Along a similar vein, I don't see how I can justify my proposal of "P1: 3/10 C 7/10 B". Where did the 10 come from? "P1: 2/7 C 5/7 B" works equally well. I only chose it because it is convenient to write down in decimal.)