Probability of player winning a tennis match

[Bobmek]

Hey guys, this is just a problem I have been playing around with in my spare time, but it occurs to me that someone might have been working on this before. So I figured I'd ask the wonderful people here for help. The problem is:

Assume that the probability of player A winning a point is p and the probability of player B winning a point is q=1-p. What is the the probability P(p) of Player A winning a 3 set match?

 

[awkward]

$$p^3 + 3 p^2 q$$

 

[mathman]

$$p^3 + 3 p^2 q$$

You've given the probability of winning a 3 point game, not a 3 set match.

 

[awkward]

You've given the probability of winning a 3 point game, not a 3 set match.

You're right, I wasn't paying attention to the whole match/game/point thing. The answer I gave would be correct if p were the probability of winning a game.

Finding the probability of winning a game from the probability of winning a point is the more complex question.

 

[QuendeltonPG]

Hmm couple of thoughts.

For a each game you could describe it as a binomial tree going from (0,0) to (15,0) or (0,15) and so on. Of course you will have to remove the paths that are no longer feasible once the game has been won. Then you will have to take into account that at 40,40 you either move to a node either (adv,40) or (40,adv) from which you either move back to (40,40) or the game ends which i'd solve by conditioning.

Similar idea would then allow you to work from 0,0 games to (1,0) or (0,1). Including the possibility of a tie break and the prob of winning a tie break.

What have you come up with so far?

 

[chiro]

Hey guys, this is just a problem I have been playing around with in my spare time, but it occurs to me that someone might have been working on this before. So I figured I'd ask the wonderful people here for help. The problem is:

Assume that the probability of player A winning a point is p and the probability of player B winning a point is q=1-p. What is the the probability P(p) of Player A winning a 3 set match?

Sounds like you could use a variation of the Gambler's ruin problem and apply it in this context. Basically you would find the probability of one player ruining against another and use that as a basis for seeing the probability of person a or b winning the set.

 

[willem2]

You need to deal with the fact that games or sets can last arbitrarily long.

If the game-score is 30-30 or deuce or 5-5 in a tiebreak, there's a probabilty of p^2 that you win the next 2 games, and a probability of 2pq that the games get shared.
If the probablity to win when at deuce is D, then you have

D = p^2 + 2pqD

Once you have D it is easy to calculate the probabilities at 40-30 or advantage,
you win with probability p, or get to a deuce with probabiltiy q.

Finally you can draw a finite tree that must end in a -0 win, -15 win, advantage,
deuce, disadvantage or a loss, and sum the winning probabilities of all those.
You'll get a larger tree for a tie break

And once you have that you can do the same for the set scores.
I doubt that the result would fit on a single page, altough finding
P_game(p), P_tiebreak(p) and then P_set(P_game, P_tiebreak) and
P_match(P_set) is doable.

 

[diggy]

This is an example of a problem that is a hassle to analytically calculate, but trivial to simulate given an explicit value for p.

 

[diggy]

Ok -- here is a plot with

Y
% games won in green
% sets won in orange
% matches won in blue

X
% chance to win the point

Sorry for the bumpiness from low sampling, but I don't have time to remake it with higher stats.

 

[awkward]

If your probability of winning a point is $$p$$ and your opponent's probability of winning is $$q=1-p$$, then the probability of your winning a tennis game is

$$p^4 + 4 p^4q + 10 p^4q^2 + 20 p^5q^3 + \frac{40p^6q^4}{1-2pq}$$

In order to win, you must win at least 4 points and be 2 points ahead. Let $$W(x,y)$$ be the event that you win with a score of x points (you) to y (your opponent). We break your possible wins into 5 cases.

Case 1: A win at (4,0). Clearly
$$P(W(4,0)) = p^4$$.

Case 2: A win at (4,1). In order to get to (4,1), the score must reach (3,1) and then you must win a point. To reach (3,1) is a Binomial event with probability $$C(4,1)p^3q$$; so
$$P(W(4,1)) = 4 p^4q$$.

Case 3: A win at (4,2). In order to get to (4,2), the score must reach (3,2) and then you must win a point. The probability of reaching (3,2) is $$C(5,2)p^3q^2$$; so

$$P(W(4,2)) = 10p^4q^2$$.

Case 4: A win at (5,3). In order to get to (5,3), the score must reach (3,3) and then you must win 2 points. The probability of reaching (3,3) is $$C(6,3)p^3q^3$$; so

$$P(W(5,3)) = 20 p^5q^3$$

Case 5: A win after the score reaches (4,4). In order to reach (4,4), the score must reach (3,3) and you and your opponent must each win one point. The probability of reaching (3,3) is $$C(6,3)p^3q^3$$; the probability that you will each win one of the next two points is $$C(2,1)pq$$; and the probability that you will win once the score hits (4,4) is $$\frac{p^2}{1-2pq}$$ (an exercise left for the reader). So the probability of a win in Case 5 is
$$\frac{40 p^6q^4}{1-2pq}$$.

Adding the probabilities for cases 1-5 results in
$$P(\text{you win}) =p^4 + 4 p^4q + 10 p^4q^2 + 20 p^5q^3 + \frac{40p^6q^4}{1-2pq}$$