You are in a contest where you are scheduled for three matches competing alterna

Question

You are in a contest where you are scheduled for three matches competing alternately against two opponents (call them Strong (S) and Weak (W)). So you could compete in order against S-W-S, or W-S-W, for opponents "S" and "W". You "win" the contest if you beat both opponents - so, this demands that you are successful in at least two consecutive matches. Suppose the probabilities of defeating each opponent are known and the probabilities do not change during the contest. Also presume that P (You defeat "W") > P (You defeat "S"). a) In general, which order would you choose (S-W-S or W-S-W) to give yourself the best chance of winning the contest? b) Now suppose P(You defeat "S") = .4 and P(You defeat "W") = .6, what is your probability of winning the contest? c) Which paired probabilities create the greatest difference to the probability of winning the game given your choice of strategies? So, we're looking for P(You defeat "W") and P(You defeat "S") that gives the highest payoff difference. d) Suppose the game changes to include five matches and you must win three matches consecutively to win the game. Does your basic strategy (selection of opponent order) change? If so, why? If not, why not?

Explanation / Answer

a) Let P(W) be w and P(s) be s where w>s

We would choose the strategy SWS because it will give a better chance of winning

P(winning) = sw +(1-s)ws for SWS and ws+(1-w)ws for WSW

P(winning = ws(2-s) and ws(2-w) Since 2-s is greater than 2-w

b) Probability = 0.4*0.6 (2-0.4) = 0.384

d) Selection of order will not change since this is an iteration of the first case only

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