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Wine In Front Of Me: Difference between revisions

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What we end up with is a bizzare guessing game where either player can choose to "force a draw" - that is, ensure that the expected return is the average of all possible expected returns.
What we end up with is a bizzare guessing game where either player can choose to "force a draw" - that is, ensure that the expected return is the average of all possible expected returns.
[[Category:Logical Fallacies]]
[[Category:Theory]]

Revision as of 20:42, 8 June 2007

"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, but this often leads to recursive reasoning, as anything that one player could think of is something another player could think of, in the form of "But that's just what he wants me to think, so I'll do the opposite...but maybe that's what he want's me to think...so I'll not do the opposite...but maybe that's what he want's me to think...etc, etc.". Such games are basically random, and therefore should be carefully considered before they are engaged in.

In Mafia, WIFOM arguments are usually 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. Also, these arguments are often used by Newbies and should be avoided if possible, in favor of clearer arguments.

WIFOM as Game Theory

In Game Theory, a WIFOM game is a weighted, head-to-head guessing game with no dominant strategy for either side. It occurs when, in an otherwise random guessing game, one of the choices carries an inheret advantage or drawback. For example:

-Two players have a black stone and a white stone each. They each secretly select a stone, then they simultaneously reveal their stone. If the stones match as White, player 1 owes player 2 $1. If the stones match as Black, player 1 owes player 2 $3. If the stones don't match, player 2 owes player 1 $2. Which stone is correct for each player to choose?

The initial reaction of player 1 would be to select the White stone, as he then only risks losing $1 instead of $3. This, however, would be obvious to player 2, who would then select the White stone in order to match player 1. Player 1 should therefore select the black stone to not match player 2, which means player 2 should choose the black stone, which means player 1 should choose the white stone... reducto ad absurdum.

Ironically, the Princess Bride scene from which WIFOM derives its name is not, in fact, WIFOM - it's a simple 50/50 guessing game. It would only be WIFOM if there were an inheret disadvantage to poisoning the wine that happened to lay closer to oneself.

Abstract examples of WIFOM: -Two people play rock-paper-scissors. If the game ends in scissors-beats-paper, the Paper player must pay the scissors player $1. If the game ends in paper-beats-rock, the Rock player must pay the paper player $2. If the game ends in rock-beats-scissors, the Scissors player must pay the Rock player $3. When you play this game, should you choose rock, paper, or scissors? -A weighted coin is flipped that lands on Heads 75% of the time. Player 1 writes down his prediction of the result on a piece of paper. Player 2 writes down his prediction of what player 1 will select on a piece of paper. If player 2 guesses correctly, he wins. Otherwise, if player 1 correctly guesses the outcome of the coin, he wins. If neither of these occur, the game is a draw. What should each player guess?

Practical examples of WIFOM: -When chasing down a dangerous criminal, a policeman comes to a fork in the road. To the left is a dark alleyway where the criminal, if followed, would have a decent chance of escaping anyways. To the right is a well-lighted boulevard where the criminal, if followed, would surely be caught. However, the policeman doesn't know which way the criminal went, and if he guesses wrong, the criminal will easily have enough time to make a clean getaway. If you were the criminal, which way would you go? If you were the policeman, which way would you guess the criminal went? -A large business is attempting to run a smaller competitor out of the market. The smaller business must begin production on one of two products: The Smidge, which will bring in $1 million worth of income, or the Widget, which will bring in $2 million worth of income. If they select a different product than the large business, they will gain enough money to compete directly with the larger business. If they select the same product, they will be run out of business. What product should each company manufacture?

Mafia examples of WIFOM: -There is an outed cop and a hidden doctor and mafia. The mafia is aware of the existance of the hidden doctor. As the doctor, who should you save? As the mafia, who should you kill? -It is the final four, with two confirmed innocents and two non-confirmed innocents left alive. You are one of the confirmed innocents, and you strongly believe player A to be scum. The other confirmed innocent strongly believes player B to be scum. At night, player B is nightkilled. Who should you vote for? As scum, should you kill the person accusing you or the person not accusing you?

Calculating WIFOM

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

Analyzing from Player 1's perspective, there are infinite possiblities for him to impliment. Aside from the two obvious possiblities, player 1 can randomize his choice - he can flip a coin, role 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 X (the odds of player 1 choosing the white stone at which it doesn't matter what player 2 chooses).

Expected return given that p2 chooses white = Expected return given that p2 chooses black

(x)(-1) + (1-x)(2) = (x)(2) + (1-x)(-3)

-x + 2 - 2x = 2x - 3 + 3x

2-3x = 5x-3

A quick bit of algebra reveals the point of intersection to be at (5/8, 1/8). 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 originally seems equal on the surface is, in fact, 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, symetrical or asymetrical - The two players should make their decisions in the exact same manner, despite the fact that they get different returns from the game.

What we end up with is a bizzare guessing game where either player can choose to "force a draw" - that is, ensure that the expected return is the average of all possible expected returns.