You hear on the radio that your favorite tennis player won her latest match 6-4 6-4 without ever having her serve broken.

In each set, she won the 5 games that she served and one of the ones
that she didn't.  So, our best assumption is that she breaks her
opponents serves 20% of the time and her opponent breaks her serve 0%
of the time.

If she served second, then there is a probability of P that she won
all five of her games and broke her opponent's serve once to win 6-4.
But if she served first and won 6-4, then she must have won her last
game, which broke her opponent's serve, and the probability of winning
all of her games and breaking exactly one specific game of her
opponent's is P/5.

So, since the odds are five times as likely that the circumstances
came from serving second as first, the probability that she served
first is 1/6.

-- Matthew Daly <mwdaly@pobox.com>

[ Editor: this counts one set only.  The serve alternates through the
match, and a 6-4 set has an even number of games (10), so whoever
serves first in the first set also serves first in the second.  The
factor of 5 then applies to each set: if p is her chance of breaking
serve and she holds with probability h, serving first gives likelihood
(h^5 p (1-p)^4)^2 and serving second (5 h^5 p (1-p)^4)^2, a ratio of
1:25 whatever p and h are.  With even prior odds, the probability that
she served first is 1/26. ]

[ Editor: further material on this puzzle, edited separately from other collected postings (batch 14); it may repeat some of the above. ]

[ Editor: problem as stated there: ]
Your favorite tennis player won her latest match 6-4 6-4 without ever
having her serve broken; what is the probability that she served the
first game of the match?

You hear on the radio that your favorite tennis player won her latest
match 6-4 6-4 without ever having her serve broken.  Based only on this
information, what is the probability that she served the first game of
the match?  You may assume that her probability of winning a given game
depends only on who is serving.

Followup question: what if the score was 6-4, 6-3 or 6-3, 6-4 instead?

-- Jonathan Dushoff <dushoff@gate.sinica.edu.tw>


Jonathan Dushoff <dushoff@gate.sinica.edu.tw> wrote:

    Posters have offered solutions of 1/6, based apparently on the
    assumption that the answer is the same as if there were only one set
    [Editor: "only set" in the original],
    and 1/2, based explicitly on the assumption that serve alternates by
    set rather than by game.  Both assumptions are incorrect.

    [On the followup question:]  A poster has offered the solution of 1,
    based on the assumption that you can only have one service break per
    set.  This assumption is, if not incorrect, at least inappropriate.

There is only one way for the first server to win a set 6-4 without being
broken, which is to break serve in game 10.  The second server can do this
with a break in game 1, 3, 5, 7, or 9.  For a 6-4, 6-4 win, the first
server is limited to breaks in games 10 and 20.  The second server can
break in 1,11; 1,13; 1,15; 1,17; 1,19; 3,11; 3,13; 3,15; ... 19,19
[Editor: presumably 9,19] which gives 25 ways, so the probability that
the winner served the first game is 1:25 (or if you prefer 1/26).

The first server can win a set 6-3 with a break in game 2, 4, 6, or 8.
There is no way the first server can win 6-3 by breaking twice.  The
second server can win 6-3 only with 2 service breaks and those must be in
games 1,9; 3,9; 5,9; 7,9.  Thus there are 4 ways for each server to win a
set 6-3.

For a 6-4, 6-3 win, the first server serves the first game of the second
set.  There are 4 ways for the first server to win 6-4, 6-3 and there are
16 ways for the second server to win 6-4, 6-3, so the probability is 4/20
or 1/5.

[ Editor: by the counts above, the second server has 5 ways to win the
first set 6-4 and 4 ways to win the second set 6-3, i.e. 5 x 4 = 20 ways,
not 16, giving 4/24 = 1/6. ]

For a 6-3, 6-4 win, the player who serves the second game serves first in
the second set (game 10).  There are 16 ways the person who serves the
first game can win 6-3, 6-4 and there are only 4 ways the second server
can win 6-3, 6-4.  Therefore the probability that the 6-3, 6-4 winner
served first is 4/5.

[ Editor: likewise the first-game server has 4 x 5 = 20 ways (4 for the
6-3 set, then 5 as second server of the 6-4 set), not 16, against 4 x 1 =
4 for the other player, giving 20/24 = 5/6. ]

-- Steve Sosensky <steves@loop.com>