Hey guys =]
I really need help with this question, I have some answers to make it easier for you to help if you want.
Here is the question...
The table gives the pay-off matrix for a zero-sum game between two players, Amy and Bea. The values in the table show the pay-offs for Amy.
Amy makes a random choice between strategies P and Q, choosing strategy P with probability 'p' and strategy Q with probability '1-p'
_________| Strategy X | Strategy Y | Strategy Z |
Amy Strategy P | 4 | -2 | 0 |
Strategy Q | -1 | 5 | 4 |
i) Write down and simplify an expression for the expected pay-off for Amy when Bea chooses strategy X. Write down similar expressions for the cases when Cea chooses strategy Y and when she chooses strategy Z.
ii) Using graph paper, draw a graph to show Amy's expected pay-off against p for each of Bea's choices of strategy. Using your graph, find the optimal value of p for Amy
Amy and Bea play the game many times. Amy chooses randomly between her strategies using the optimal value for p.
iii) Showing your working, calculate Amy's minimum expected pay-off per game. Why might Amy gain more points than this on average?
iv) What is Bea's minimum expected loss per game? How should Bea play to minimise her expected loss?
My Correct/Incorrect Answers
ii) I drew my graph and found that her maximum value for p occurs when
iii)Using the maximum value for p, found above. I used it in the following to find the Expected pay-offs:
and found that the minimum expected pay-off is 1.11...
Amy may gain more points than this as Bea may not choose randomly and play strategies X and Z more.
iv) Bea's minimum expected loss is when...
I plug this value for p into the following
Therefore, Bea's minmum expected loss = 1.5
and Bea should play strategies X and Y with probability, p=0.5 .
I hope this is easy to understand, my apologies for the messy table as I could not get it right here. Thankyou anyone who reads this and more thanks for anyone who helps =]
Thankyou guys =]