How to beat Roger Federer at Wimbledon

This Riddler problem is about the game of tennis!

Your wish has been granted, and you get to play tennis against Roger Federer in his prime in the Wimbledon final. You have only a 1 percent chance to win each point, but Roger, sporting gentleman that he is, offers to let you name any score and begin the match at that point. (So, if you’ve entertained a fantasy of storming back after being down three match points in the fifth set, now’s the time to live it.) What score can you name that gives you the best chance to win, and what is your chance of winning the title?

A rough (and approximate) solution:
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A more detailed (and exact) solution:
[Show Solution]

6 thoughts on “How to beat Roger Federer at Wimbledon”

  1. Great explanation Laurent! I had come to the conclusion that being up 2 sets and 6 points on the tie break is the most advantageous position to be in as well (but didn’t do any of the messy math). One of the things I was wondering about was whether or not being up 2-0 or 3-1 would be a preferable situation. I know the odds of winning unless if you don’t win the pre-arranged 6 match-point tie-break are infinitesimally small, but is there a difference between to two cases? I would imagine that a smaller number of remaining points favors the weaker player but I also might be completely wrong in that regard. Would love to hear if you come up with anything!

  2. Laurent

    Sorry it’s been a while since I did my stats course so can you explain why my answer of using the binomial distribution (6C1 0.01^1 (1-0.01)^5 ) is wrong?


    1. The quantity you wrote down is the probability of winning one point and losing the other five. The reason this isn’t quite right is that you don’t always play six points. It’s match point, so if you win a point at all the game ends and you don’t play any more points.

      The probability you’ll win the first point is just $0.01$. The probability you’ll win the second point is $(1-0.01)(0.01)$. The probability you’ll win the third point is $(1-0.01)^2(0.01)$. And so on. If you sum these up (geometric series), you get $1-(1-0.01)^6$, which makes sense because $(1-0.01)^6$ is the probability that you lose every point. One minus this is the probability that you don’t.

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