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 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 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)

Since P(W) > P(S), I choose to have more W's in my contest. So I choose W-S-W

b)

P(SWS) = 0.4*0.6*0.4 = 0.096

P(WSW) = 0.6*0.4*0.6 = 0.144

P(WIN) = 1/2*P(SWS) + 1/2 P(WSW) = 0.12

c)

From above calculations, P(WSW) has high probability of winning when compared to P(SWS)

d)

No. It does not chage since the wiining against W or S does not change and are independent of order.

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