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Fifty Challenging Problems in ProbabilityOpen Ended Class Creator: quantumelixir Status: Under Construction |
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Lesson 2: Problem 02: Successive WinsSuccessive Wins To encourage Elmer's promising tennis career, his father offers him a prize if he wins (at least) two tennis sets in a row in a three-set series to be played with his father and the club champion alternately: father-champion- father or champion-father-champion, according to Elmer's choice. The champion is a better player than Elmer's father. Which series should Elmer choose? The series that Elmer should chose is Father-Champion-Father. The champion has the higher chance to win, thus we stick him in the middle since he has the best chance to win and we play with father twice to improve the chances we will win with father. Since we need to win the game 2 times in a row we assume a win in the middle game and that gives us two times to win with father either of which will give us the necessary two games together. Again the problem for me is how to prove this mathematically. Hints welcome. 
Since these are supposed to be challenging problems, I initially was skeptical of what seemed like the obvious choice (FCF). I believe this skepticism is right. I think one is better off going with CFC. In either case, in order to win the prize, the second game MUST be won. Therefore, it is preferable to put a less challenging opponent there. This can be worked out algebraically as well. Notate and conquer: let
p = probability of win vs. father
q = probability of win vs. champ
so that p>q. Now compare P(win|FCF) and P(win|CFC). I'm not following your logic. It looks like you're trying to figure out the probability for success for all 3 matches. The criteria here is two successive matches. Thus you should evaluate P(win|FC) P(win|CF) and P(win|CF) P(win|FC), instead of P(win|FCF) and P(win|CFC). It's not a very precise definition, I'll have to work on that. By P(win|FCF) i mean the probability of winning the prize given that Elmer decides on FCF. What is P(win|FCF)? There are two ways to win the prize, either Elmer wins the first two matches which has probability pq, or Elmer loses the first and wins the second two matches which has probability (1-p)pq. So: P(win|FCF) = pq + (1-p)pq = 2pq - p^2q Then compute P(win|CFC) and compare. Heh, lost me there. Going to have to take a closer look at it and get back to you. It seems I misunderstood the premise of this question. I figured they were playing in pairs against other opponents rather than playing against the father and tennis pro. It's apparent to me that you'd want to play CFC in that case to increase the chances of actually winning against the tennis pro. Still waiting on word on whether or not we can proceed. I have the book now so it's just a matter of contacting the right people.
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