OR/MS Today - February 2003 March Madness - Appendix A: Inside the Black Box For any team i, using the model of Kaplan and Garstka [2] it is possible to estimate Fi1, Fi2, and Fi3 which are the probabilities of winning the tournament, losing the tournament final, or reaching the Final Four but losing, respectively. The conditional expected payoff given ownership of team i, denoted by pi, is then equal to pi = .6Fi1 + .2Fi2 + .1Fi3 Note that while (since there is exactly one champion and exactly one runner-up), (since there are two losing semifinalists — the sum of the Fi3's for the first 32 teams in the tournament equals one, as does the sum for the remaining 32 teams in the tournament). The conditional expected payoff pi is thus the fraction of the total pool prize you can expect if you are the sole owner of team i (for the sum of the pi's must equal one). We assume that Mike has estimates of the pi's thanks to his covert partnership with Ed. To model bidding behavior, the specific assumption we employ is that each sucker allocates 1,000 points across teams independently in accord with the same multinomial distribution. That is, each team i has a characteristic probability ri of receiving any point from any sucker. The beauty of this assumption is that the marginal distribution of the number of points bid on team i by any given sucker is binomial with 1,000 trials and success probability ri. For Mike to win ownership of team i, he needs to bid more than the highest amount offered over n suckers. Let Bi represent the (random) maximum bid placed on team i over all n suckers, and let bi equal Mike's bid on team i, i = 1, 2,...64. The probability distribution of Bi is simply the probability distribution of the maximum of n independent binomial variables. Thus, if Xi is the binomial random variable formed from 1,000 trials with success probability ri, then the probability that Mike wins sole ownership of team i having bid bi is given by Pr{Bi < bi} = [Pr{Xi < bi}]n Mike's betting problem is to select the bids for all teams. Assuming that Mike has estimates for the pi's and the ri's, Mike can maximize his expected winnings (and by side payments, Ed's as well!) by selecting the bids (the bi's) to subject to bi > 0 and integer for i = 1, 2,...64 The problem above is a knapsack problem, and can be easily solved via dynamic programming (as we do in Excel). Return to More March Madness Table of Contents OR/MS Today Home Page OR/MS Today copyright © 2003 by the Institute for Operations Research and the Management Sciences. All rights reserved. Lionheart Publishing, Inc. 506 Roswell Rd., Suite 220, Marietta, GA 30060 USA Phone: 770-431-0867 | Fax: 770-432-6969 E-mail: lpi@lionhrtpub.com URL: http://www.lionhrtpub.com Web Site © Copyright 2003 by Lionheart Publishing, Inc. All rights reserved.