Wine In Front Of Me

"All right: where is the poison? The battle of wits has begun. It ends when you decide and we both drink, and find out who is right and who is dead."

"But it's so simple. All I have to do is divine from what I know of you. Are you the sort of man who would put the poison into his own goblet, or his enemy's?"

-The Princess Bride

So begins the game of "wine in front of me", or WIFOM, for short. In gaming, it's any kind of game or subgame, especially a psychological one, in which a player is given a set of apparently equal choices where one or more is completely wrong. In such games one often may try to use what he knows of his opponent to make a better choice. However, in some situations this leads to recursive reasoning: "But that's just what he wants me to think, so I'll do the opposite. But maybe that's what he wants me to think, so I'll not do the opposite. But maybe that's what he wants me to think..."

In Mafia, WIFOM arguments are often a Scum tactic used to distract the Town. The scum will make an unusual play at night, which would lead to a situation that would 'clear' them (because players will think, "Why would a scum do that?"). These arguments are sometimes used by Newbies and should be avoided in favor of clearer arguments.

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Calculating WIFOM

(mixed strategy)

Let's take the black-and-white stone example from the beginning.

Analyzing from Player 1's perspective, there are infinite possibilities for him to implement. Aside from the two obvious possibilities, player 1 can randomize his choice - he can flip a coin, roll a dice, or use any method available to select a random number from 0% to 100%. Player 1 can therefore assign the variable X to represent the probability he implements of choosing the white stone.

For any given percentage that player 1 implements, exactly one of three conditions is true: Player 2 is better off choosing the black stone, Player 2 is better off choosing the white stone, or it doesn't matter what player 2 chooses. When it doesn't matter what player 2 chooses, the expected return of player 2 choosing the white stone must be equal to the expected return of player 2 choosing the black stone (otherwise, it would matter which stone player 2 chooses). This leads to a simple equation that can be used to solve for p (the odds of player 1 choosing the white stone at which it doesn't matter what player 2 chooses).

Expected return given that Player 2 chooses white = Expected return given that Player 2 chooses black

-1 * p + 2 * (1-p) = 2 * p + -3 *(1-p)

-p + 2 - 2p = 2p - 3 + 3p

2-3p = 5p-3 A quick bit of algebra reveals the solution to be x = 5/8. This means that, when Player 1 selects his stone, if he chooses randomly but weights his decision such that he chooses the white stone 5/8 of the time, then no matter what player 2 does, player 1's expected return will be 1/8 of a dollar.

This means that a game that may seem equal is actually unbalanced in favor of player 1.

Similar calculations can be performed from player 2's perspective. Let Y equal the odds of player 2 selecting the white stone such that it doesn't matter what player 1 chooses.

(y)(1) + (1-y)(-2) = (y)(-2) + (1-y)(3) y - 2 + 2y = -2y + 3 - 3y 3y-2 = 3-5y

The intersection is now at (5/8, -1/8)

An interesting point is that, even though the WIFOM game being played is asymmetrical, the players' equilibrium probability is the same - 5/8 White. This holds true for any WIFOM game, symmetrical or asymmetrical. The two players should decide using the same logic, despite the fact that they get different returns from the game.