FAQ on Homework 2
1. Question: On the example of page 48, why do they say TR is a Nash Equilibria? Is that a typo?
Answer: I think they are right about TR being a Nash Equilibrium.
If the other person is playing R, I get a zero either way (so I have no motivation to change). If I am
playing T, the other person gets a zero either way, so they have no motivation to change. This is a
degenerate case (and not pareto optimal), but still in Nash Equilibrium.
OK, I can buy that, that would make sense. So we really are talking about motivation to change by
considering ONLY our own utility, and not the fact that we could lower the other person's utility (we
are not motivated to move if we can lower the other player's utility, only if we can raise our own).
Right. We only switch if WE do better.
2. (Questions in blue, answers in red) Several of the problems talk about the "pure strategy" Nash
Equilibrium. I don't know what that means. They use the term “pure strategy to distinguish from
“mixed strategy”. In other words, if both people only have options of saying “I’ll to Option 1” instead
of “I’ll do option 1 half the time and option 2 half the time. As far as I understand, "pure strategy"
means as a player I am always going to choose this strategy, I won't deviate from it no matter what you
do. Not really. A pure strategy is just the type of options you are allowed. For example, when you are
asked if you want pie or cake, you have to pick which one in a pure strategy. In a mixed strategy, you
can half a third of a piece of pie and two thirds of cake. If that is correct, I don't understand what is
meant by a "pure strategy Nash Equilibrium". I read pages 68-71 several times, and got online and
tried to find examples to help me understand, but I'm not sure I've got it (especially frustrating/funny to
me is page 71 has an example that says, "It is easy to see that the matrix game [...] has no Nash
equilibrium in pure strategies"-- *grin* I read these pages several times and each time I got there I'd
say to myself, "No, I'm sorry, but for me it isn't." Anyway, as best as I can figure (guess) to get a pure
strategy Nash equilibrium I do the following: For player 1 I look at the best choice (highest value) in
each column, For player 2 I look at the best choice in the row... anywhere these values overlap (best
row choice was best column choice... I have a "pure strategy" Nash equilibrium. Perhaps my
problem is I don't see why we use the term "pure strategy"... I mean, what I just described looks to me
to be the Nash Equilibrium as we learned how to determine in class (using your arrow method). The
arrow method is for pure strategies. For mixed strategies, we do the computations like in class
yesterday. ... Why throw on the term "pure strategy"? What benefit does that distinction give us?
3. (Questions in blue, answers in red.) I'm trying to understand question 2. Particularly, the phrase,
"Each agent simultaneously name share they would like to hae (s1 and s2)". We both yell out, “I want
50 cents!” or “I want 75 cents.” Does this mean s1 is the share of the dollar player 1 wants, and s2 is
the share player 2 wants? yes. Or do both players state two different shares they think would be
acceptable ("Either 40% or 55% of this dollar would be acceptable to me."). Also, in this case should
"0<=s1, s2<=1" be "0 <= s1<= s2 <= 1"? I would assume that s1 and s2 both have to be greater than
or equal to 0 and less than or equal to 1 (since we're talking about 1 dollar or a percentage, however we
want to look at it). Yes, this is just shorthand for 0 <= s1 <= 1 AND 0 <= s2 <= 1.
4. (Questions in red, answers in blue) Lastly, the book talks about both elimination of strictlydominated strategies, and just elimination of dominated strategies. But I can't find anywhere where it
distinguishes between the two (what one gives you that the other doesn't). Are they identical? They
are close. You get a little more power with iterated elimination. For example, perhaps I have no one
strategy that is dominated, but after the other person eliminates one of their strategies, now I do have a
dominated strategy. Does that make sense? When I find a dominated strategy, I look for a row (if I
am player 1) in which every single utility is less than (or equal to, for weak domination) the utility of
another row. Such a row can be eliminated. I may have NO such strategy initially, but a dominated
row is created when the other player eliminates some of his columns (due to domination).
Or perhaps does it just go back to the original question I asked you... if I were to eliminate just
dominated strategies (not strictly-dominated) I would potentially be eliminating some Nash
equilibrium, because given two choices with the same utility for a certain player (like the example on
page 48) they may both be Nash equilibria because I'm not motivated to move from one or the other,
but one may have been in a dominated row/column. Is there anything more to the difference between
these elimination methods I should consider. Yes, for the examples in the assignment, sometimes you
have to eliminate a row which is weakly dominated (every entry is the same or lower than another
row’s entries). You might eliminate equilibrium in this case – so if you needed to find ALL Nash
equilibrium it might be dangerous.
5. Can you shed a little light on the set up for number 4?
Answer: It is like there are 10 types of prizes of varying values. Assume, a prize of type10 is the best
and a prize of type 1 is the worst. We both get a prize that we don’t show to others. All prizes occur
with the same frequency, so we don’t assume there are more of the bad prizes. We are both asked if
we want to exchange our prize. If we both want to exchange, we do. Otherwise, we keep what we
were given. Create a strategic form game to see what decisions each player will make.
6. For problem 2, should "s1+s1<=1" be "s1+s2<=1"?
Answer: Yes, that is a typo.
7.
And continuing the question about the difference between "elimination of strictly dominated" and "elimination of
dominated", are you saying that "elimination of strictly dominated" has nothing to do with iterated elimination?
No, they are the same thing. Iterated elimination and just plain elimination can be slightly different if
you consider that at first you may not be able to eliminate anything, but can after the other person does,
you are able to..
You might choose the term “strictly dominated” if you only removed a strategy if EVERY payoff was
less than that of another row/column. You might use the term “weakly dominated” if every payoff was
less than or EQUAL to that of another row/column.
8. I was understanding that you could use either method (elimination of strictly dominated, or just elimination of
dominated) to do iterated elimination, but didn't understand why you'd choose one or the other (or more
correctly, why you'd choose strictly dominated elimination, since what you could eliminate with "strictly" would
be a subset of what you could eliminate using the less strict "non-strictly" dominated elimination). For example,
on the following matrix:
0 1 2
0 2 3
0 3 4
Using dominated elimination I could eliminate rows 1 and 2 because they are both dominated by 3. But using
strictly-dominated elimination I could not, because 1 and 2 are NOT strictly dominated by row 3. Is this not the
difference between strictly dominated elimination and dominated elimination?
Yes, but I wouldn’t worry about the distinction. I think we usually don’t care which you use
9. And after looking at your FAQ I have a new question. I interpreted this statement in problem 4, "If both
players agree then the prizes are exchanged" to mean that if they both wanted to exchange AND if they both
wanted to keep (they agreed) then they would exchange. It sounds like I assumed incorrectly, and that it is only
if they agree to exchange that they exchange. Right?
They must agree to exchange. If one or more does not want to exchange, they don’t.