Trying to avoid actually grading semester exams, I was playing around with a problem from F. Mosteller's classic, "Fifty Challenging Problems in Probability." Problem two concerns a three-set tennis match in which the player, a youth named Elmer, will alternately play against his father and the club pro, given the club pro is a better player than the father. He will win a prize if he can win two consecutive matches of the three. The question is whether he is better of playing the sequence father-pro-father or pro-father-pro.

The counter-intuitive part is that his odds of winning are better if he plays the better play, the pro, more often. (This reminds me of Parrondo's paradox, which I will try to write about soon) Here is a simple explanation (I hope). If we let f represent the probability he wins against his father, and p the probability he beats the pro, then to get two consecutive wins, he must win the first two or the last two, so we need the probability win, win added to the probability of Lose, win, win.

For the sequence father-pro-father the probability of success is fp + (1-f)pf. We can distribute the 1-f to get fp + fp - fpf, and factoring fp out of each term we get fp(2-f).

If we do the same with the sequence pro-father-pro we just interchange the p and f to get pf(2-p). Now since we are given that p is smaller than f (he is LESS likely to beat the better player) we see that 2-p must be larger than 2-f, and so the pf(2-p) is the higher probability.

The advantage when actually calculated is very small. For example, Ifthe probability of beating his father is .4, and the probability of beating the pro is .2, his probability of winning in the father-pro-father sequence is about .128 . By switching the order to play pro-father-pro his probability of success increases to .144(which is twice what it would be if he only played a three set match against the pro).

What happens if we change these probabilities of success, but keeping the order so that the pro is better than the father? Letting p remain at .2 and raising his level against his father to .5 improves his chance of success to 18%. In fact, if we substitute the value .2 into the expression pf(2-p) we get .2f(1.8).... the probability of success is a linear equation, .36f. If we let his probability against his father go to one, you can see tha the has a 36% chance of winning if he goes Pro-father-pro. But if you look at the other sequence, fp(2-f), and again substitute in the .2 value against the pro, the equation is quadratic, .4 f - .2f^{2}. Visualizing the graph of this negative quadratic, we can see that he vertex will occur when f= 1, so that is the maximum, and when f=1 we get a probability of success of .2, exactly what his probability against the pro in a single game would be. This makes sense if you consider that if he ALWAYS beats his dad, the match really depends on the one game in the middle against the pro.

When the order is p-f-p-f two wins can happen with probabilities pf + (1-p)fp + (1-p)(1-f)pf. Now compare the probability when the order is reversed, fp + (1-f)pf + (1-f)(1-p)fp. Note that all the terms except the second are the same. Once more the fact that 1-p must be larger than 1-f (because f is greater than p) leads us to conclude the best order is to play the pro first.

## 1 comment:

It seems more obvious if you look at it this way:

No matter what, he has to beat his dad once and the pro once to get the prize.

He is less likely to beat the pro than his dad.

Therefore, he has better chances the more "shots" he has to beat the pro. So playing the pro twice (where neither outcome is 100% critical to winning the prize) is better overall odds.

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