Conference rusure::math

In the first game, is the n that the opponent gets the n he played, or the n you
played?

Call the opponent B, and the protagonist A.  As I understand the payoff
determination in .0, the payoff if A chooses i and B chooses j (call these
strategies Ai and Bj respectively) is i-j unless |i-j| = 1, in which case it is
1-2i (for i>j) or 2i+1 (for i<j).  That is:

   B0 B1 B2 B3 B4 B5
A0  0  1 -2 -3 -4 -5
A1 -1  0  3 -2 -3 -4
A2  2 -3  0  5 -2 -3
A3  3  2 -5  0  7 -2
A4  4  3  2 -7  0  9
A5  5  4  3  2 -9  0
A6  6  5  4  3  2 -11

Now A0 is dominated by 1/2(A1+A3), B0 by 1/2(B1+B3), and A1 by 1/2(A4+A6).
So we can strike those three strategies, leaving the 5x5 submatrix whose upper
left corner contains -3 and whose lower right corner contains -11.

Considering the diagonal elements and B's strategies, we see that no B strategy
can be dominated by a linear combination of others unless it is B1; to be as
good as B1 against A2 such a combination would have to be pure B5, which
wouldn't dominate vs. A4.  So in the reduced matrix no pure B strategy can be
eliminated.

Similarly, considering the largest values in each row, no A strategy can be so
dominated unless it is A5.  Against B1, a mixed strategy would have to contain
at least (1/2)A6 to dominate, but then it would be worse than A5 against B5.  So
no A strategy can be eliminated either.

I leave the bit of actually calculating A's and B's mixtures to others.

Assuming by "analogy" with (2) that the payoff for the opponent is what the
opponent plays, the payoff matrix (A for protagonist, B for opponent) is

   B1 B2 B3 B4
A1  2 -2 -3 -4
A2 -1  4 -3 -4
A3 -1 -2  6 -4
A4 -1 -2 -3  8

It is trivial that none of A's strategies are dominated by a linear combination
of the others, and also that no pure B strategy dominates any other.  Suppose
that Bj (j/=1) is dominated or equaled by some combination of strategies (WLOG
these don't include Bj).  Then I claim that *same* combination dominates or
equals B1.  Against Aj, this is certainly true, because the combination doesn't
include Bj.  Against any other A strategy, the payoff of Bj is -j < -1, so if
the combination dominates Bj it has a payoff less than both -1 and 2, hence the
payoff is better than against B1.

Since the combination which beats or equals Bj can't be pure B1, we can use it
to get a combination of non-B1 strategies that beats B1.

Now consider a combination that beats or equals B1; say it is pB2+qB3+rB4 where
p+q+r = -1.  From B's point of view, the payoff vs. A2, A3, and A4 respectively
cannot be better than 4p-4(1-p), 6q-4(1-q), and 8r-3(1-r).  So we have

    8p-4<=-1, 10q-4<=-1, 11r-3<=-1; p<=3/8, q<=3/10, r<=2/11.

Thus 1 = p+q+r < 0.475+0.300+0.182 = 0.957, a contradiction.

Now: the payoff of B's best strategy must have the same payoff vs. A1 as vs. A2;
we get convenient cancellations leading to p1=2*p2.  Similarly considering A2
vs. A3 and A3 vs. A4 we get 2*p2=3*p3, 3*p3=4*p4, so B's options must be played
in the ratios 12:6:4:3.  B's solution is 1/25(12B1+6B2+4B3+3B4), with payoff
-12/25.  A's solution I leave for others (I only promised half a solution...)

The game suggests a pretty conjecture which is unfortunately false.  If you
extend the game so each player has 6 options, then the strategy B1 is dominated
by a combination of B2..B6 played in the ratios 30:20:15:12:10.

The proper mix for A (I omit a lot of detail) is 1/75(13A1+19A2+21A3+22A4), with
payoff -12/25.

	I agree with the solution to game (1). Since the payoff is negative,
then from the point of view of this model, Malkavian Prank is not a good card
to play. However, there are other, pragmatic considerations which can make it
handy or fun.

	The solution to game (2) I agree with as far as it goes. However, if
you just invert the 5x5 matrix, then you encounter a problem. (No, the matrix
is *not* singular.) What is the problem? This is the most interesting part of
the analysis.

	I would strongly recommend MAPLE to help here.

Tara
Andrew.

    What's the problem?
    
    It's a straight forward two player zero sum game. If player 1 plays a
    mixed strategy {pi}, i=1..5 and player 2 plays {qi}, i=1..5 then player
    1 is looking for min/p max/q sum/ij pi.Mij.qj and player 2 for 
    max/q min/p sumij pi.Mij.qj, both with a constraint of sum/i pi = sum/j
    qj = 1 in the region pi,qj >= 0.
    
    Elimination of dominated strategies should ensure that the inequalities
    are binding in only one of the optimisations (ie one player's strategy
    has positive wieghts for all pure strategies) so this one can be solved
    from (eg) pi.Mij = {1,..,1}. Substituting back these specific values,
    the other player's strategy can be foud as a simple LP problem, and may
    have zero weights for some pure strategies.
    
    Excel gives me {0.1515, 0.1380, 0.3392, 0.1571, 0.2142} for the player
    2 strategy, I'll post the LP solution after the week-end.
    
    Andy.

Andy,

	I don't think it's as simple as you suggest.

>    Excel gives me {0.1515, 0.1380, 0.3392, 0.1571, 0.2142} for the player
>    2 strategy, I'll post the LP solution after the week-end.

	P2 can do this, and guarantee himself a certain payoff. However, if
you look at the P1 strategy which yields the same payoff, you find (I think)
that one of the components is negative. What does this mean? It means that
P2 can do even better.
    
	My solution is:

	P1: {0,0,359,114,255,0,70}/798
	P2: {0,0,201,224,219,154}/798

with a payoff of 110/399.

	The point (which I had forgotten/overlooeked) is: it's not just
dominated pure strategies which can be simplified out. When we spot a dominated 
mixed strategy, this also enables us to reduce the dimensionality of the
strategy space.

Regards from sunny Johannesburg,
Andrew.