# p chance of winning tennis point -> what f(p) chance of winning game?

In Wii Tennis, I have fixed $\,\,p\,\,$ chance of winning a given point.

What is my chance $f(p)$ of winning the entire game?

If $p=0.5, f(p)=0.5$ by symmetry, but I believe $f(0.51) > 0.51$

EDIT: to clarify, the rules of Wii Tennis are the same as regular tennis.

EDIT: would a Markov Chain be useful here?

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How many points are you playing up to in each game? –  Nicolas Villanueva Jun 10 '11 at 15:50
For any reasonable set of rules, you will be right in thinking that $f(0.51)>0.51$. But given the rules, we should be able to come up with a more precise result. I guess you may find it hard to believe that there are people who do not play Wii Tennis. But it is true. –  André Nicolas Jun 10 '11 at 15:57
You may find this post and its followup worthwhile: gowers.wordpress.com/2009/07/04/a-mathematician-watches-tennis –  Byron Schmuland Jun 10 '11 at 16:11
Can you please fix the text - the word "ball" should be used instead of game in a number of places - which initially made me wonder if you had done the calculation for an entire set. –  user12904 Jul 3 '11 at 17:47
@Frederick I'm not understanding your comment? –  barrycarter Jul 9 '11 at 22:02

Let me assume that Wii tennis follows the rules of tennis - where game goes through points $0, 15, 30, 40$ and/or deuce, advantage, and won game. Let me define $q=1-p$.

After 1st ball, chance is $p$ that you lead $15-\text{to}-0$ and $q$ for $0-15$. After 2nd, there is $p^2$ that it's $30-0,\,\, 2pq$ that it's $\,\,15-15, q^2$ for $0-30$. After 3rd, $p^3$ for $40-0, 3p^2q$ for $30-15, 3pq^2$ for $15-30$ and $q^3$ for $0-40$. After 4th game, chances are $p^4$ that you already won, $4p^3q$ for 40-15, $6p^2q^2$ for 30-30, $4pq^3$ for 15-40, $q^4$ you already lost.

After 5th game (you keep on playing even if it's already decided), the probability is $p^4+4p^4q$ that you already won, $10p^3q^2$ for 40-30, $10p^2q^3$ for 30-40, and $q^4+4pq^4$ that you already lost. After 6th game, it's $p^4+4p^4q+10p^4q^2$ that you already won, $20p^3q^3$ that it's deuce, and $q^4+4pq^4+10p^4q^2$ that you already lost.

From the deuce, the probability of your winning is $$p^2 + 2p^3 q + 4p^4 q^2 + 8p^5 q^3+\dots = \frac{p^2 }{1-2pq}$$ The coefficients are power of two because you may divide the rest of the game to fluctuations away from deuce - to someone's advantage and back (two types for each excursion) - and then finally two victories by the same player.

So your total probability of winning is $$f(p)=p^4+4p^4q+10p^4q^2 + \frac{p^2}{1-2pq}\times 20p^3q^3 = \frac{p^4(15-34 p +28 p^2-8p^3)}{1-2p+2p^2}$$ As a check, $f(1/2)=1/2$. Also, $f(0.51) = 0.52498501$. It's greater than $0.51$ but not too much greater. It can probably be proved that regardless of the detailed rules of the game, if there is a game composed of several points, $f(p)$ will be greater than $p$ for $p>1/2$.

The slope of the tanh-like graph is just around 2.5 in the middle but the graph becomes nearly horizontal for $p$ close to 0 or 1. For example, $f(0.9)=0.9985522$. If you win 90% of the balls, you win 99.9% of the games. When both athletes have very different skills, it makes almost no sense to remember the score because the better player will win.

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Thanks for the great edits, @Chris! You're a co-author. ;-) That's exactly what I saw in Mathematica. –  Luboš Motl Jun 10 '11 at 18:27
Ian Stewart's Game, Set, And Math has a great explanation with lots of diagrams of this formula. I haven't checked to see whether it's exactly the same, but the graph looks similar enough. –  yatima2975 Jun 10 '11 at 20:11
I was initially unsure of your answer, since it seemed really complex. However, if you're at deuce, the win chance is "Simplify[Solve[x == p^2 + 2*p*q*x, x] /. q -> 1-p]", and Mathematica confirms that doesn't simplify much. As a note, your numerator simplifies to (-3 + 2 p) (5 - 8 p + 4 p )^2, but this doesn't really any value to the answer. –  barrycarter Jun 11 '11 at 0:11
Your formula is also derived in section 4.7 entitled Bernoulli and the game of tennis of Problems and Snapshots from the world of probability by Blom, Holst, and Sandell. –  Byron Schmuland Jun 22 '12 at 17:47