Econ assignment question

[maniacut]

I'm using backward induction or the rollback method to get to B's decision nodes but from there I can't figure out what A is going to choose, or how many (if any?) equilibriums there are.

[tabako]

If you use backward induction to know what B is going to play, assume A has this information as well. Just do the same thing that you did to figure out what B will play. A will pick the choice that maximizes his payoff, assuming B plays optimally.

There will be multiple equilibrium if, for example, B is indifferent between two choices. In this case, just assume he picks one or the other, and then use backward induction from there to find out what A would do under either circumstance. These separate cases will make up the different equilibriums.

[maniacut]

Ok I think I've got it. I have 3 equilibrium paths, A(x) B(z) w payoff (6,3,5)... A(y) B(z) w payoff (7,3,4)... A(y) B(t) C(h) w payoff (4,3,7).

And I've denoted the complete strategies as follows
Equilibrium strategy of A = { x, y}
B = { zz, zt }
C = { hh }

Is this right?

[tabako]

A(y) B(t) C(h) w payoff (4,3,7)

This can't be an equilibrium because if A does strictly better if he goes x instead of y. I understand this might seem sort of counterintuitive because A can't know whether B will z or t because he is indifferent between the choices, but the equilibrium concept doesn't "care" about this. I don't know the best way to explain this, but I know its not an equilibrium solution.

Your other two paths are right.
I haven't seen the complete strategies notated as you presented them, but it makes enough sense.

[J-Mac]

I might be way off, but isn't A's utility if he goes 'y' only 5.5, since B is indifferent between his two choices, one of which gives A utility of 4 and one which gives A a utility of 7?

Therefore, since going 'x' has a utility of 6 for A, wouldn't the only equilibrium be AxBz for (6,3,5)?

[tabako]

[ QUOTE ]
I might be way off, but isn't A's utility if he goes 'y' only 5.5, since B is indifferent between his two choices, one of which gives A utility of 4 and one which gives A a utility of 7?

Therefore, since going 'x' has a utility of 6 for A, wouldn't the only equilibrium be AxBz for (6,3,5)?

[/ QUOTE ]

For pure strategies, you assume that the indifferent player does one or the other, and then find the equilibrium in accordance with this (so there will be two if B is indifferent between two choices).

If you are allowing mixed (i.e. randomization between choices), then you will have the same two equilibrium as in pure strategies, but they will each apply for a given value of p, where p is the probability B assigns to playing a certain choice. So p is on the interval [0,1] which is subdivided for the two equilibrium.