When dealing with a model, we use the letter  for the mean. We write or, more often, replacing p by, Instead of , we can also write E(X ). ( Think of this as being what we expect to get on average ). This notation comes from the idea of the mean being the E xpected value of the r.v. X. Mean of a Discrete Random Variable S1 Reminder about Expectation

e.g. 1. A random variable X has the probability distribution P x (X = ) p Find (a) the value of p and (b) the mean of X. Solution: (a) Since X is a discrete r.v., (b) mean,

54Col max – Row min 21 B A Mixed Strategies Using probabilities to determine the best mixed strategy 1) 2  2 games maximin minimax Max(row/min) 2 ≠ Min(col max) 4 so not stable Check to see if there is a stable solution

5– B A Suppose player A chooses option 1 (row1) with probability p and option 2 (row 2) with probability 1– p If B chooses option 2 then the expected win for A is: E(win) = 2  p + 5(1 – p) = –3p + 5 The expected pay–off is deduced using E(X) =  x  P(X=x) from S1 If B chooses option 1 then the expected win for A is: E(win) = 4p – 3(1 – p) = 7p – 3 p 1–p

A`s expected pay-off p B chooses 1 B chooses 2 7p – 3 = –3p p = 8 p = 0.8 These 2 lines are plotted as shown and intersect at the highest minimum winnings when: 7p – 3 = –3p p = 8 p = 0.8 E(win) = –3p + 5 E(win) = 7p – 3

So if A chooses option 1 with probability 0.8 then his expected win is: E(win) = 7  0.8 – 3 = 2.6 E(win) = 7p – 3 If B chooses option 1 then If A chooses option 1 with a value of less than 0.8 then: If B chooses option 2 thenE(win) = –3p + 5 E(win) = 7p – 3 E(win ) ≥ 2.6 E(win) ≤ 2.6 The expected win of 2.6 is the greatest that A can guarantee. The value of the game is 2.6.V(A) = 2.6

The same analysis can be carried out for player B but considering the losses. Remember the matrix shows A`s gain so to find B`s gain change the signs. 5– B A A`s gain –532 –2–41 21 B A B`s gain

Suppose player B chooses option 1(col 1) with probability q and option 2 (col 2) with probability 1 – q –532 –2–41 21 B A If A chooses option 2 then the expected win for B is: The expected pay–off is deduced using E(X) =  x  P(X=x) from S1 If A chooses option 1 then the expected win for B is: E(win) = –4q – 2(1 – q) = –2q – 2 E(win) = 3  q – 5(1 – q) = 8q – 5 q 1–q

The intersection of these 2 lines occurs when: –2q – 2 = 8q – 5 10q = 3 q = 0.3 If q = 0.3 then B expects to lose 2.6 as we expected from considering A`s winnings. So to minimise his loses B should choose option 1 with probability 0.3 and option 2 with probability 0.7 E(win) = –2q – 2 = -2.6