Tournaments (with Mate) Here G is a connected graph with n vertices, - PowerPoint PPT Presentation

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 one 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 .

Tournaments Write a 0 for the number of vertices initially in { 1 , . . . , l } , and let a r be the number after the r th time a significant edge is chosen. This is a random walk where a r = a r − 1 + 1 with probability p and a r = a r − 1 − 1 with probability 1 − p , independent of which edges are chosen, and independent of G .

Tournaments Write a 0 for the number of vertices initially in { 1 , . . . , l } , and let a r be the number after the r th time a significant edge is chosen. This is a random walk where a r = a r − 1 + 1 with probability p and a r = a r − 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 a 0 (which is Binomial) to get P ( S = l ) = ( l − 2 lp + mp ) n − ( l − 1 − 2 lp + 2 p + mp ) n . ( m − mp ) n − ( mp ) n

Tournaments The distribution of final results does not depend on the graph, but the time taken to reach consensus does.

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.

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.

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 K m 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( n 2 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).

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 K m ) is the worst case for p close to 1 / 2

Tournaments Can we prove that the expected time is always monotonic on [0 , 1 / 2] ? Not obvious even in simple cases.

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.

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.

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).

Rock-paper-scissors From simulations, it appears that all strategies are roughly equally likely to eventually dominate unless the starting state is very one-sided.

Rock-paper-scissors From simulations, it appears that all strategies are roughly equally likely to eventually dominate unless the starting state is very one-sided. 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.

Rock-paper-scissors If almost all the population are of one type, the numbers of the other 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 T w,b ) all white balls are eliminated.

Rock-paper-scissors If almost all the population are of one type, the numbers of the other 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 T w,b ) all white balls are eliminated. E ( T w,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 wb t + wb ≤ P ( T w,b > t ) ≤ t + b + 1 − w .

Rock-paper-scissors It follows that if all but o ( n 1 / 3 ) are Rock, Scissors will be eliminated whp before any new Scissors arise, allowing Paper to win.

Rock-paper-scissors It follows that if all but o ( n 1 / 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).

Rock-paper-scissors It follows that if all but o ( n 1 / 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) .

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.

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.

Rock-paper-scissors By n = 8000 the chance of Paper winning was over 99%.

Rock-paper-scissors By n = 8000 the chance of Paper winning was over 99%. However, can prove that eventually ( n ≈ 10 8 ), Scissors overtakes. It looks like the most likely winner rotates infinitely many times.

Rock-paper-scissors What if we have more than three strategies?

Rock-paper-scissors What if we have more than three strategies?

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 } .