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) Tocalculated the probability to win with the first combination SWS, we know that to win it is necessaty to win at least two consecutive matches. That give us two posibilites, to win the first and the second or to win the second and the third. More over to calculated the total proability to win, ther is a third posibility that is to win the three matches. Knowing that proability doesn´t change and that every macth cna be considered as independent, we have:

P(win)= P(winS)*P(winW)+P(winW)*P(winS)+P(winS)*P(winW)*P(winS)

For the second combination we use the same logic and get:

P(win)= P(winW)*P(winS)+P(winS)*P(winW)+P(winW)*P(winS)*P(winW)

from both equation we can noticed that we have two equal terms P(winW)*P(winS)+P(winS)*P(winW)

only varies P(winS)*P(winW)*P(winS) and P(winW)*P(winS)*P(winW), knowing that P(winW)>P(winS)

then the second combination give us the greater probability.

b) now knowing the probabilities we substitute and get

P(SWS)=0.24+0.24+0.096=0.576

P(WSW)=0.24+0.24+0.144=0.624.

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