The opening chapter of Julian Havil's wonderful book

*Nonplussed*presents three tennis paradoxes. The best (most surprising) of these is that if you are playing a strong server (the probability that the server wins a point on their serve is just over 90%) then you are more likely to break serve from 40-30 or 30-15 down than you are at love all. (There is an assumption that the probability of winning the point is independent of the score.)

Of course in any knock-out tournament, even if the better player wins every match it is by no means certain that the top two players will meet in the final. If they are in the same half of the draw, they meet earlier and, in a random draw with 2^n players, the probability of that is (2^(n-1)-1)/((2^n)-1) which is only just under 1/2. Seeding addresses this problem. But seeding can create its own issues. Suppose we have three top players, equally good, all some way better than any of their other competitors. Then two will be in the same half of the draw, and the third will have double the chances of winning of each of the other two. If this victory is then used to determine the seeding for the next tournament, that player will carry this advantage forward, and will win more tournaments than their equally-good adversaries.

The system for challenging line-calls brings in a new dimension. In each set each player can make at least three challenges when they believe a ball was wrongly called out (or in). Successful challenges don't count, but unsuccessful ones are deducted from the allowance of three. If the call is shown to be wrong, then depending on the circumstance the point is replayed or the point is won by the player challenging. The challenge must be played immediately - if you think the ball was out and you return it, you've lost your right to challenge.

How sure do you have to be that the call was wrong before you should challenge? There are obviously a lot of factors that influence the decision. How many challenges one has left - presumably one wants one in reserve for the dubious call in the decisive game still to come. The stage of the set - if the set is almost finished and you haven't used any challenges, there is little to be gained by saving them. The importance of the point - if the disputed call is resulting in a crucial service break you might challenge even if you are sure the call is correct! The gain from a successful challenge - if you are going to win, rather than lose, a point, the challenge is more worthwhile than if you are going to replay it. Yesterday Andy Murray challenged a call that his first serve was out - would that be a better challenge than a second serve called out? I would guess that the threshold estimate of probability of success before you challenge is strongly affected by all these circumstances.

There are other factors too. Sometimes you might just want a moment to regroup, and it might be worth using a challenge just to get a short break. And how do you decide whether to challenge or to play the ball? If you think your opponent's shot is just out, but you can return it, do you stop and challenge or do you play on? The decision must be made instantly!

It would be fascinating to now how much professional tennis players think about these things when using their challenges. I doubt if they do probability calculations in their heads, so do they have heuristics and if so, how good are they? Are some players better tacticians in using challenges than others?

You might enjoy a couple of the posts at http://education.theage.com.au/content/maths/gjj0s8

ReplyDeleteThanks. That's very nice!

ReplyDelete"If they are in the same half of the draw, they meet earlier and, in a random draw with 2^n players, the probability of that is (2^(n-1)-1)/((2^n)-1) which is only just under 1/2."

ReplyDeleteHow did you come up with that equation?

Probability that B is in the same half of the draw as A if there are 2^n players:

ReplyDeleteof the 2^n -1 players who are not A, 2^(n-1)-1 are in the same half of the draw as A, and 2^(n-1) are in the other half. So the probability that B is in the first group is [2^(n-1) - 1]/[2^n -1].

Hope that explains it!

Tony

I understand that there are (2^n)-1 players who are not A, but why is it that 2^(n-1)-1 are in the same half of the draw as A? In particular, why is the exponent (n-1)?

ReplyDeleteThe draw divides the players into two equal halves, with 2^(n-1) in each. (Sorry, typing maths as text without using exponents does make it a lot more difficult to follow!) So A's half of the draw contains 2^(n-1)-1 players other than A, and the other half contains 2^(n-1): the probability that B is in the former is [2^(n-1)-1)]/[2^(n-1)-1 + 2^(n-1)].

ReplyDeleteFor example, if there are 8 players, three are in the same half of the draw as A and four aren't. So the probability that B is in the same half of the draw as A is 3/7.

That makes perfect sense. Thank you so much. I am in the process of setting up a tennis blog answering interesting tennis questions using regression techniques, and I would love to hear your thoughts/suggestions. I will leave a link here once it's up and running (:

ReplyDelete