SLIDE 1 Tournaments
(with Mate) Here G is a connected graph with n vertices, and each vertex has a strategy taken from {1, . . . , m} at random. At each time step an edge is chosen; both vertices take on the lower of the two strategies with probability p and the higher with probability 1 − p. Since G is connected, eventually we will reach a state where only
- ne strategy remains. Write S for the strategy that is left; then
P(S l) can be computed by coarse-graining the strategies into those at most l and those exceeding l.
SLIDE 2
Tournaments
Write a0 for the number of vertices initially in {1, . . . , l}, and let ar be the number after the rth time a significant edge is chosen. This is a random walk where ar = ar−1 + 1 with probability p and ar = ar−1 − 1 with probability 1 − p, independent of which edges are chosen, and independent of G.
SLIDE 3
Tournaments
Write a0 for the number of vertices initially in {1, . . . , l}, and let ar be the number after the rth time a significant edge is chosen. This is a random walk where ar = ar−1 + 1 with probability p and ar = ar−1 − 1 with probability 1 − p, independent of which edges are chosen, and independent of G. We can find the probability that this random walk hits 0 before n, and take a weighted average over a0 (which is Binomial) to get P(S = l) = (l − 2lp + mp)n − (l − 1 − 2lp + 2p + mp)n (m − mp)n − (mp)n .
SLIDE 4
Tournaments
The distribution of final results does not depend on the graph, but the time taken to reach consensus does.
SLIDE 5
Tournaments
The distribution of final results does not depend on the graph, but the time taken to reach consensus does. Conjecture For any n, m and p, the complete graph has the smallest expected time to consensus.
SLIDE 6
Tournaments
The distribution of final results does not depend on the graph, but the time taken to reach consensus does. Conjecture For any n, m and p, the complete graph has the smallest expected time to consensus. We can prove For any n and p, if m = 2 the complete graph has the smallest expected time to consensus of any regular graph.
SLIDE 7
Tournaments
What about the worst case? Not the path, at least for p close to 0 (or 1). De- fine a sundew to be a graph formed from Km by adding n−m pendant edges, dis- tributed as evenly as possible to the m vertices. When m = n/2 and p = 0 this has expected time between e(G)(log n − 1) and e(G)(log n + 3), so Θ(n2 log n). The worst-case expected time for the path when p = 0 is (n − 1)2, and in fact we can show that the overall expected time for the path is O(n(log n)2) (prob not tight).
SLIDE 8
Tournaments
PRISM simulations for a range of plausible graphs suggest that a sundew is the worst case for p close to 0, and a tadpole (formed by adding a pendant path of length n − m to Km) is the worst case for p close to 1/2
SLIDE 9
Tournaments
Can we prove that the expected time is always monotonic on [0, 1/2]? Not obvious even in simple cases.
SLIDE 10
Tournaments
Can we prove that the expected time is always monotonic on [0, 1/2]? Not obvious even in simple cases. For a simple random walk in {0, . . . , n}, moving left or right with probability p and 1 − p, we can show that the time to reach an endpoint is monotonic on [0, 1/2] provided the starting distribution is symmetric.
SLIDE 11
Tournaments
Can we prove that the expected time is always monotonic on [0, 1/2]? Not obvious even in simple cases. For a simple random walk in {0, . . . , n}, moving left or right with probability p and 1 − p, we can show that the time to reach an endpoint is monotonic on [0, 1/2] provided the starting distribution is symmetric. The same applies if there are equal delays at all points – if we move left, move right, or remain with probabilities rp, r(1 − p), 1 − r. What if the value of r depends on where we are, but is symmetric on {0, . . . , n}? This would imply monotonicity for a complete graph.
SLIDE 12
Rock-paper-scissors
(with Mate, Chris C) Look at similar process with three strategies in cyclic order: rock always beats scissors, scissors always beats paper, paper always beats rock. For general graphs, the starting position will affect the outcome. We’ll concentrate on the well-mixed situation (complete graph).
SLIDE 13 Rock-paper-scissors
From simulations, it appears that all strategies are roughly equally likely to eventually dominate unless the starting state is very
SLIDE 14 Rock-paper-scissors
From simulations, it appears that all strategies are roughly equally likely to eventually dominate unless the starting state is very
The eventual winner is determined by the first strategy eliminated; if Rock is initially dominant it is likely Scissors will be eliminated, allowing Paper to win.
SLIDE 15 Rock-paper-scissors
If almost all the population are of one type, the numbers of the
- ther types change like an urn process which starts with w white
balls and b black. At each step a ball is drawn; if white, it is removed, if black, two black balls are returned. Eventually (at time Tw,b) all white balls are eliminated.
SLIDE 16 Rock-paper-scissors
If almost all the population are of one type, the numbers of the
- ther types change like an urn process which starts with w white
balls and b black. At each step a ball is drawn; if white, it is removed, if black, two black balls are returned. Eventually (at time Tw,b) all white balls are eliminated. E(Tw,b) = ∞, but we can bound the probability that the process has ended by a given time. Lemma For any w, b and t with t ≥ w, wb t + wb ≤ P(Tw,b > t) ≤ wb t + b + 1 − w .
SLIDE 17
Rock-paper-scissors
It follows that if all but o(n1/3) are Rock, Scissors will be eliminated whp before any new Scissors arise, allowing Paper to win.
SLIDE 18
Rock-paper-scissors
It follows that if all but o(n1/3) are Rock, Scissors will be eliminated whp before any new Scissors arise, allowing Paper to win. In fact, writing r for the number of Rock players, etc, rps is a sub-martingale, implying that if ps = o(n) Rock will not be eliminated first (whp).
SLIDE 19
Rock-paper-scissors
It follows that if all but o(n1/3) are Rock, Scissors will be eliminated whp before any new Scissors arise, allowing Paper to win. In fact, writing r for the number of Rock players, etc, rps is a sub-martingale, implying that if ps = o(n) Rock will not be eliminated first (whp). Would conjecture that if ps = o(n) Paper wins whp but if min(rp, ps, sr) = ω(n) then the probabilities of winning approach (1/3, 1/3, 1/3).
SLIDE 20 Rock-paper-scissors
Also considered an alternative model with non-overlapping
- generations. We have a fixed population of size n in generation 0.
To create an individual in generation t + 1, choose two (independent) individuals from generation t, and take the winner. Do this n times (independently) to get the next generation.
SLIDE 21 Rock-paper-scissors
Also considered an alternative model with non-overlapping
- generations. We have a fixed population of size n in generation 0.
To create an individual in generation t + 1, choose two (independent) individuals from generation t, and take the winner. Do this n times (independently) to get the next generation. Now simulations suggested that starting from fixed proportions of each strategy (eg 2:1:1) the eventual winning strategy is determined whp.
SLIDE 22
Rock-paper-scissors
By n = 8000 the chance of Paper winning was over 99%.
SLIDE 23
Rock-paper-scissors
By n = 8000 the chance of Paper winning was over 99%. However, can prove that eventually (n ≈ 108), Scissors overtakes. It looks like the most likely winner rotates infinitely many times.
SLIDE 24
Rock-paper-scissors
What if we have more than three strategies?
SLIDE 25
Rock-paper-scissors
What if we have more than three strategies?
SLIDE 26
Rock-paper-scissors
What if we have more than three strategies? The first elimination no longer determines the result, but once a strategy is eliminated two others can be lumped together. If Rock is eliminated, then we reduce to the three-strategy model on Scissors, Spock, {Paper, Lizard}.
