Bob "Hey Alice let's play paper-scissor-rocks :) I will always play paper."

Alice "What if you don't play paper?"

Bob "If I play don't play paper you win if it's a draw [under original rules] and you draw if you lose [under original rules]. If I win you have to obey my order."

Alice "What if I get a draw?"

Bob "Then you will have to do my little favour :P"

Alice "What favour?"

Bob:

The payout table is as follows

It turned out that Alice had scissor and Bob had rock --- a draw --- with consequences equivalent to a lose as Bob defined. The analysis from Alice: based on random choices there are a 2/3 chance to win for rocks and scissors, but Bob claimed to play scissor so she'd better play scissor. Bob made a conclusion that the risk for Alice is invariant despite the rules:

--------------------------------------------------

Consider the following payout matrix.

$P_1 = \begin{bmatrix}1&0&1\\1&1&0\\-1&1&0\end{bmatrix}$, $P_2 = \begin{bmatrix}1&-1&1\\1&1&-1\\-1&1&-1\end{bmatrix}$

Where $P_1$ is the matrix that Alice thought to be but $P_2$ is the real one. The matrix $P$ is defined as follows: $[P]_{ij}$ is the amount you win if you play the j-th option (vertical) and the opp plays the i-th option (horizontal). (Some textbooks may switch between you and your opp but it's just a matter of perspective on whether you, the analyzer is taking part in the game.)

By strong duality we know that the optimal strategy for both Bob and Alice should have the same expectation. Now suppose Alice has a probability vector $\vec{x}$ to play the three choices with payout matrix $P$. Then the expected outcome for Bob is given by $P\vec{x}$ for different choices from Bob. He as the payer would of course want to minimize the paid. Then we can set up the following linear program for Alice, to maximize the gain when Bob minimizes her gain among the three choices.

$\max (\min ([P\vec{x}]_i)) ~~~\text{subject to}~~~\sum [\vec{x}]_i = 1, \vec{x}\geq \vec{0}$.

And by adding dummy variable we have:

$\max z ~~~\text{subject to}~~~P\vec{x} - z\vec{b}\geq \vec{0},~\sum [\vec{x}]_i = 1,~\vec{x}\geq \vec{0}$

Where $\vec{b} = (1,1,1)^T$. This will be the optimal strategy for Alice.

Using $P_1$ we have $\vec{x}^* = (0,0.5,0.5)^T$ with expected gain $0.5$. Using $P_2$ we have the same optimal vector, but the expected gain is zero. It can be concluded that Alice had the perception that she has an advantage in the game, while actually she had not.

**The matrix with four -1 and five 1, turns out to be a fair game (mathematically).**

-----------------------------------------------

Gambling at high stakes are of course a totally different game. There are not enough games for one to apply central limit theorem so that the result converges so that they can play in the most probable way. In these games all factors including the psychological status shall be considered and here Alice obviously did not have a good enough mind to do such analysis, so she eventually lose the game.

Can you think of other non-trivial matrix which is fair as well?

Can you think of other non-trivial matrix which is fair as well?

## No comments:

## Post a comment