Game Theory: 2 person — Zero Sum Games
Concept of the Game:

There are two players. For our purposes we will call them Rowland and Collin.
The game is played in stages where each stage has three phases.
1. Each player determines his/her next move.
2. Both players reveal their move at the same time.
3. The "winner of the stage" is determined and the payoff takes place
Each player has a finite set of possible moves. (The choices open to each player may be the same, as is the case in a Rock/Paper/Scissors game, or each player may have a different set of choices.)
The choices open to each player remain the same for each stage of the game.
Based on the choices of each player the winner is determined and so is the payoff, the exchange where the loser pays the winner wins some amount.
Each player has a strategy for making moves. That strategy is based entirely on an overall frequency distribution of the possible moves open to the player.

An Example

We have two players, Rowland and Collin.
Rowland has 3 possible moves, called a, b, and c.
Collins has four possible moves, called r, s, t, and u.
The payout matrix is
Payout
Table Collin's Choices

r s t u

Rowland's
Choices a 6 4 -1 -8

b -3 1 -4 6

c -2 -5 5 2

In the table positive values represent wins for Rowland and losses for Collins while negative values represent wins for Collins and losses for Rowland. Thus, if Rowland plays a and Collins plays u the result in the table, -8, means that Rowland pays Collins 8 units (dollars, pennies, Euros, pieces of rice, or whatever).
Each player has a strategy. In this example we will show a strategy where Rowland favors playing a overall, and favors c over b. We do this by having the frequency distribution for Rowland's moves be 50% for a, 22% for b, and 28% for c. Collin, on the other hand, has a strategy represented by the distribution 40% for r, 30% for s, 20% for t, and 10% for u. We could rewrite the table above to show these strategies:
Table
with
strategies Collin's Choices

r s t u

0.4 0.3 0.2 0.1

Rowland's
Choices a 0.5 6 4 -1 -8

b 0.22 -3 1 -4 6

c 0.28 -2 -5 5 2

It is important to note that the strategies are merely the frequency distribution of the possible choices for each player. Therefore, the sum of all the strategy options for a player must be 1.

The strategies mean that overall, if the game is played through many stages, we can estimate the profit/loss for each possible outcome of the game. Thus, Roland plays a about 50% of the time while Collin plays u end, only 10% of the time. Looking at all twelve of the possible outcomes, we therefore expect that the a - u combination will happen about (50%)(10%) or (0.5)*(0.1) or 0.05 or 5% of the time. Since the a - u combination results in a loss for Rowland of 8 the expected outcome or value for that combination is (0.05)*(-8) or -0.4 units. To find the expected value for any cell by multiplying Rowland's strategy frequency distribution times the frequency distribution of Collin's choice times the win/loss value from the table.

Therefore, we can construct a new table giving the expected value in each cell. That new table, having the expressions in each cell, is:
Table of
expected
values:
expressions Collin's Choices

r s t u

0.4 0.3 0.2 0.1

Rowland's
Choices a 0.5 0.5*0.4*6 0.5*0.3*4 0.5*0.2*(-1) 0.5*0.1*(-8)

b 0.22 0.22*0.4*(-3) 0.22*0.3*1 0.22*0.2*(-4) 0.22*0.1*6

c 0.28 0.28*0.4*(-2) 0.28*0.3*(-5) 0.28*0.2*5 0.28*0.1*2

And the same table with the expressions evaluated is:
Table of
expected
values:
computed Collin's Choices

r s t u

0.4 0.3 0.2 0.1

Rowland's
Choices a 0.5 1.2 0.6 -0.1 -0.4

b 0.22 -0.264 0.066 -0.176 0.132

c 0.28 -0.224 -0.42 0.28 0.056

If we add up all twelve of the expected cell values (the twelve cells with background color YELLOW we will get the expected value of the game for the given strategies. Of course, if Rowland or Collin decides to have a new strategy then the expected values in each of the cells may change as may the expected value for the game with the new strategies.

The computations to arrive at the expected value for the game are significant. However, we might consider the following matrices:

which represent the Rowland's strategy (R), the payoff table (P, and Collin's strategy (C). In looking at these three matrices, there is no surprise about P. It looks just as we would expect. Rowland choices correspond to rows of the matrix and Collin's choices correspond to columns of the matrix. However, Rowland's strategy, R, appears as a 1 x 3 matrix. Whereas the rows of P correspond to Rowland's options, it is the columns of his matrix strategy, R, that correspond to his options. The same question arises with respect to Collin's strategy matrix, C. There, whereas the columns of P correspond to Collin's options, it is the rows of C that correspond to those same options. Why do we do this?

The answer comes from the actions of matrix multiply R * P, which we can do since R is 1 x 3 and P is 3 x 4, then we are multiplying the row that is R times each of the columns in P. In that way we are applying the frequency distribution held in R to each of the corresponding column entries in P. For example, the first frequency distribution value is 0.50. If we were to multiply R * P we would be multiplying that 0.50 times each of the four top row entries in P. That is, we would be applying that frequency distribution value to each of the four different entries in the first row of P.

In a similar fashion, looking at P * C, which is legal since that would be a 3 x 4 matrix times a 4 x 1 matrix, we would be multiplying each row of P times the column of values that is C. Again, taking the first (i.e., the top) entry in C we would be applying each of the first entries in each of the rows of P to the frequency distribution in the top entry of C. That is, we would be applying Collin's first strategy value to each of the entries in P that represent the payoffs associated with Collin's first option.

In short, if we multiply R * P * C we will not only have performed all 12 of the computations seen in the matrix above, we will also have added all of them together. In fact, if we do perform R * P * C the result is a 1 x 1 matrix with the single value Thus, the expected value of the game given the two strategies, is 0.75. In fact, for any given two-person, zero sum game with a payoff matrix P with two players, Rowland with strategy R and Collin with strategy C, as long as the payoff matrix has rows corresponding to Rowland's options and columns corresponding to Collin's options and Rowland's strategy is a single row and Collin's strategy is a single columnm, then the expected value of the game for those two strategies is the single value in R * P * C.

Next we look at a miscarriage of fairness. If one of the players is aware of the other player's strategy, then the player in the "know" can come up with a "pure" strategy (always choosing a particular action) to minimize loss / maximize gain depending on the game. We will consider two situations. First, what if Rowland knows Collin's strategy. In that case, the most we can say about the matrices is that P and C have not changed, but R = [a b c ;]. But that means that we can perform P * C. That will leave us with

and if we multiply this out we get (13/5)a + (-11/10)b + (-11/10)c Rowland wants to have the largest possible value of this expression. Furthermore, a, b, and c all have to be between 0 and 1 inclusive. We find the largest coefficient for the terms in the expression, (13/5), and we set its corresponding frequency, in this case a, to be 1 and all the other values to be 0. Thus, the "pure" strategy is to always choose a and, in so doing, the new expected value of the game will be (13/5) or 2.6, far better than the original 0.75.

The second case has Collin knowing Rowland's strategy. In that case the most we can say about the matrices is that R and P have not changed, but C = [r s t u ]^T. But that means that we can perform R * P. That will leave us with

and if we multiply this out we get (89/50)r + (41/50)s + (1/50)t + (-53/25)u Collin wants to have the smallest (hopefully most negative since Collin wins with a negative payoff) possible value of this expression. Furthermore, r, s, t, and u all have to be between 0 and 1 inclusive. We find the smallest (most negative) coefficient for the terms in the expression, (-53/25), and we set its corresponding frequency, in this case u, to be 1 and all the other values to be 0. Thus, the "pure" strategy is to always choose u and, in so doing, the new expected value of the game will be (-53/25) or -2.12, far better, from Collin's perspective, than the original 0.75.

Payout Table		Collin's Choices
Payout Table		r	s	t	u
Rowland's Choices	a	6	4	-1	-8
	b	-3	1	-4	6
	c	-2	-5	5	2

Table with strategies			Collin's Choices
			r	s	t	u
			0.4	0.3	0.2	0.1
Rowland's Choices	a	0.5	6	4	-1	-8
	b	0.22	-3	1	-4	6
	c	0.28	-2	-5	5	2

Table of expected values: expressions			Collin's Choices
			r	s	t	u
			0.4	0.3	0.2	0.1
Rowland's Choices	a	0.5	*0.50.46*	*0.50.34*	*0.50.2(-1)*	*0.50.1(-8)*
	b	0.22	*0.220.4(-3)*	*0.220.31*	*0.220.2(-4)*	*0.220.16*
	c	0.28	*0.280.4(-2)*	*0.280.3(-5)*	*0.280.25*	*0.280.12*

Table of expected values: computed			Collin's Choices
			r	s	t	u
			0.4	0.3	0.2	0.1
Rowland's Choices	a	0.5	1.2	0.6	-0.1	-0.4
	b	0.22	-0.264	0.066	-0.176	0.132
	c	0.28	-0.224	-0.42	0.28	0.056