# Essay On Say P Is The Probability Of Column Using Left And 1-P Choosing Right,

The payoff of Row choosing up will be, p10+1-p4=6p+4

The payoff of Row choosing down will be, p1+1-p8=8-7p

## Given that one is a better response,

6p+4>8-7p

## 13p<4

p<413

## Therefore,

The best response(BR) for row is, BRr=1 if p<4130,1if p=4130 if p>413

This should be the same for column, say again using p as row’s probability of choosing Up, and 1-p for down,

Payoff of column by choosing q for left, 2q+71-q=7-5q

## Payoff of column by choosing q for right,

9q+5-5q=4q+5

## Again only one could be a best response therefore, because right gives a higher average payout,

7-5q<4q+5

## 9q>2

q>29

Therefore, BRc=0 if q<290,1if q=291 if q>29

## The Mixed strategy equilibrium will be the intersection of the two functions BRcand BRr,

Therefore the equilibrium is at 29,413

Using the student number,

Again, let p be the probability for column to choose left and 1-p for column to choose right,

Then in row choosing up, p2+1-p8=8-6p

And in row choosing down, p5+1-p2=3p+2

2p+8-8p>5p+2-2p

## 9p<6

p<23

BRr=1 if p<230,1if p=230 if p>23

## Using the same method for the column player,

4q+6-7q<6q+3-3q

## 6q>3

q>12

BRc0 if q>120,1if q=121 if q<12

## The mixed equilibrium will be at,

23,12

The graph will be,

Where the vertical axis is p and the horizontal axis is q.

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