Optimal Strategies for Random Tournament Games

NORTH- HOLLAND
Optimal
David
Strategies for Random Tournament Games
C. Fisher
and Richard
B. Reeves
Department of Mathematics
University of Colorado at Denver
Denver, Colorado 80217-336~
Dedicated
to Professor John Maybee on the occasion of his 65th birthday.
Submitted
by J. Richard Lundgren
ABSTRACT
Given a tournament T, the tournament game is: Two players independently
pick a node of T. If both pick the same node, the game is tied. Otherwise,
the player whose node is at the tail of the arc connecting the two nodes wins.
Fisher and Ryan showed that this game has a unique optimal strategy. For a
tournament game on a random n node tournament, we find the probability that
the optimal strategy uses k nodes with nonzero probability.
Fisher and Ryan [l, 21 studied a generalization
of “scissors, paper, and
stone” called tournament games. Given a tournament,
two players independently
pick a node.
If both pick the same node, the game is tied.
Otherwise,
the player whose node is at the tail of the arc connecting
the
two nodes wins. Fisher and Ryan showed there is a unique optimal strategy
for this game which uses an odd number of nodes with nonzero probability.
Further,
they showed that the nonzero probabilities
in this strategy are
fractions with odd denominators
(see Figure 1). Laslier, Laffond, and Le
Breton [3] independently
proved these results in a different setting.
For any tournament
T, let S(T) be the nodes used with nonzero probability in the optimal strategy for the tournament
game on T. Let an n
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DAVID C. FISHER AND RICHARD
B. REEVES
FIG. 1. The optimal strategy of a tournament. The strategy is to play one of the
nodes according the probabilities shown above. This strategy wins as least as often as
it loses against any node an opponent might play. Since the game is symmetric, no
strategy could win more often it loses. So the strategy is optimal.
node
random
tournament
be a tournament
among the 2(y) tournaments
tournament
T?
THEOREM.
on n nodes.
picked with equal
How large is S(T)
Let T, be an n node random
prob(lS(T,)I
Then
( >
if Ic isodd,
0
if k is even.
;
= k) =
tournament.
2+-l)
probability
for a random
Proof.
Since IS(
is odd, prob((S(T,)I
= k) = 0 for even k.
For odd k, we must find the number of tournaments
T with ITI = n
and IS(T)1 = k. There are (i) c h oices for S C T so that S(T) = S. Let
N1 and Nz be the number of ways of interconnecting
S and S’ (the nodes
of T that are not in S), respectively.
Fisher and Ryan showed that there
are 2(kT1) tournaments
U with
IUI = k and IS(U)1 = k. So
Nl
Since
IS’1 =
=
2(‘;7.
n - k and there are no restrictions
N2 =
2(“;‘).
within
S’, we have
RANDOM TOURNAMENT
GAMES
85
Let Nrz be the number of ways of connecting S’ to S. Let a E S’,
and let p be the probability that the optimal strategy beats a. With no
restrictions, there are 2k ways to connect a to S. However, p must be at
least i. Further, since the probabilities in the optimal strategy are fractions
with odd denominators, no subset of S has probability 3. So for each way
of connecting S to a with p > 4, there is a way (by reversing the arcs
between a and S) that p < i. Thus there are 2”-’ ways to connect S to a
so that p > i. Therefore IV12= 2(k-1)(“-k),
and
prob(lS(W = k)
= number of n node tournaments with IS(T)/ = k
number of n node tournaments
=
(;)2(*1’)2(k-l)(n-k)2(“;‘)
(2NlNd%
2(t)
=
2(Y)
XI
What is the average size of S(T) for an n node tournament? Using the
theorem,
E(IS(T,)I)
=
2
k(;)2-(“‘1
k=O
= n2-(n-1)
5
(;I
:>
k=O
k is odd
k is odd
(” I, 1) = n2-(n-l)2n-2
= n2-(7+l)
= ;.
k is even
So on the average, the optimal strategy uses half of the nodes.
REFERENCES
1
2
3
D. C. Fisher and J. Ryan, Optimal strategies for a generalized “scissors,
paper and stone” game, Amer. Math. Monthly 99:935-942 (1992).
D. C. Fisher and J. Ryan, Tournament games and positive tournaments, J.
Graph Theory, to appear.
G. Laffond, J. F. Laslier, and M. Le Breton, The bipartisan set of a tournament game, Games Econom. Behav. 5:182-201 (1993).