I Tennis Probabilities Challenge

[PeroK]

Let's assume that a player has a probability $p$ of winning a point on the opponent's serve.

1) What is the probability ($P$) of a break of serve - i.e. winning the game?

2) What is the expected number of break points per service game?

PS I had a mistake in my original calculation for 2). Deleted unhelpful hints!

 

[Bosko]

Let the player, who serve first, be "Red" and his opponent "Blue".
The graph of the game is:

The color of a circle indicate who serve and in a circe is the current score.
The probabilities (p and 1-p) are written next to the arrows.

Does the probability to win on the opponent serve is the same for both players ?

 

[PeroK]

Let the player, who serve first, be "Red" and his opponent "Blue".
The graph of the game is:
View attachment 340309
The color of a circle indicate who serve and in a circe is current score.
The probabilities (p and 1-p) are written next to the arrows.

What is the final answer?

Does the probability to win on the opponent serve is the same for both players ?

Not necessarily. In general, the player who wins a higher percentage of points on the opponent's serve winns the match.

 

[Bosko]

What is the final answer?

I don't know the rules details. Still far from the final answer.

Not necessarily. In general, the player who wins a higher percentage of points on the opponent's serve winns the match.

Thanks

 

[PeroK]

I don't know the rules details. Still far from the final answer.

You can use your Markov chain, although this isn't the simplest way. In any case, you have to calculate the probabilities of winning the game to 0, 15 or 30. And the probability of getting to 40-40 (deuce). There is another thread where the probability of winning from 40-40 is calculated.

 

[PeroK]

To calculate the probability of a break of serve (winning the game), you can use the probability of winning a point on the opponent's serve. Let's denote this probability as p.

1. The probability of a break of serve is given by p×(1−p).
2. The expected number of breakpoints per service game can be calculated as p×(1−p)×4.

That doesn't look right at all. $p(1-p)$ is the probability of winning one point each.

... one point each in a specific order.

 

[Bosko]

In two consecutive serves
$p(1-p)$ is the probability of the "win then lose" and
$(1-p)p$ is the probability of the "lose then win" situation

$p(1-p)+(1-p)p=2p(1-p)$ is the probability of winning one point each.

 

[Erland]

Let the player, who serve first, be "Red" and his opponent "Blue".
The graph of the game is:
View attachment 340309
The color of a circle indicate who serve and in a circe is the current score.
The probabilities (p and 1-p) are written next to the arrows.

Does the probability to win on the opponent serve is the same for both players ?

But the same player serves every time within the same game. You seem to believe that the players alternate to serve in a game. It isn't so, except in a tiebreak game.

 

[SSequence]

I worked out the first part of the question using a spreadsheet. As a check I entered various values of probabilities and the probabilities do seem to add up to 1. One other check was setting the probability of serve winning as 0.5 and see whether the probabilities for both players winning the game turns up to be 0.5.

It shouldn't be too difficult to write an expression for the answer either (I will add that later in another post after checking the spreadsheet again). However, let me first mention the main idea of how to proceed to the answer. Because the process of approaching the question is kind of more important it seems.

First we make a list of the states. The first number denotes the player P1's status (the one serving). The second number denotes the player P2's status (the player receiving the serve). We have 20 states:
1-- (0,0)
2-- (0,15)
3-- (0,30)
4-- (0,40)
5-- (15,0)
6-- (15,15)
7-- (15,30)
8-- (15,40)
9-- (30,0)
10-- (30,15)
11-- (30,30)
12-- (30,40)
13-- (40,0)
14-- (40,15)
15-- (40,30)
16-- (40,40)
17-- (40,A)
18-- (A,40)
19-- (P1) ---- win for P1
20-- (P2) ---- win for P2


However, after a drawing a "transition" diagram (so to speak), it seemed to me that it is more useful to arrange the states as follows:
D0
1-- (0,0)

D1
2-- (15,0)
3-- (0,15)

D2
4-- (30,0)
5-- (15,15)
6-- (0,30)

D3
7-- (40,0)
8-- (30,15)
9-- (15,30)
10-- (0,40)

D4
11-- (40,15)
12-- (30,30)
13-- (15,40)

D5
14-- (40,30)
15-- (30,40)

D6
16-- (40,40)

D7
17-- (A,40)
18-- (40,A)
19-- (P1)
20-- (P2)

==================================

Now the main thing is just to draw the transition diagram and see the probability of reaching each state. We start with setting the possibility of being in state (0,0) as 1. Below note that P1win denotes the probability of serving player winning an individual point and P2win denote the probability of receiving player winning the point. So, we should have:
P1win+P2win=1
If I haven't misunderstood the original statement in OP (and I am not writing it the other way around) then it seems we should have:
P2win=$p$

Now we have the following relations:
prob. of reaching state (15,0) = (0,0)*P1win
prob. of reaching state (0,15) = (0,0)*P2win
prob. of reaching state (30,0) = (15,0)*P1win
prob. of reaching state (15,15) = (15,0)*P2win+(0,15)*P1win
prob. of reaching state (0,30) =(0,15)*P2win
prob. of reaching state (40,0) =(30,0)*P1win
prob. of reaching state (30,15) =(30,0)*P2win+(15,15)*P1win
prob. of reaching state (15,30) =(15,15)*P2win+(0,30)*P1win
prob. of reaching state (0,40) =(0,30)*P2win
prob. of reaching state (40,15) =(40,0)*P2win+(30,15)*P1win
prob. of reaching state (30,30) =(30,15)*P2win+(15,30)*P1win
prob. of reaching state (15,40) =(15,30)*P2win+(0,40)*P1win
prob. of reaching state (40,30) =(40,15)*P2win+(30,30)*P1win
prob. of reaching state (30,40) =(15,40)*P1win+(30,30)*P2win
prob. of reaching state (40,40)/deuce =(40,30)*P2win+(30,40)*P1win

Note that in the expressions on the right-hand-side above when I wrote something like (15,15) for example, I should have actually written: "probability of reaching state (15,15)". I didn't write it like that so that the expressions on right-hand-side wouldn't become too hard to read. Ideally, one could introduce some extra notation for probability of reaching a certain state.


Another thing to note is that the probability of player P1 winning the game is sum of following probabilities:
--- prob. of reaching state (40,0) * P1win
--- prob. of reaching state (40,15) * P1win
--- prob. of reaching state (40,30) * P1win
--- prob. of reaching state (40,40) * (P1 winning from deuce)

Similarly, the probability of player P2 winning the game is sum of following probabilities:
--- prob. of reaching state (0,40) * P2win
--- prob. of reaching state (15,40) * P2win
--- prob. of reaching state (30,40) * P2win
--- prob. of reaching state (40,40) * (P2 winning from deuce)


Also, we can work out the detail of winning from deuce. However, it seems that after simplification it turns out to be equivalent to sum of (infinite) geometric series (I worked it out something like couple of years ago when I saw the question mentioned by OP). I haven't re-checked that now though.

 

[Bosko]

In my opinion two parameters are needed:
p - the probability that the first (red) player wins on his serve
q - for the other player (blue) to win on his serve

Of course:
1-p and 1-q are the corresponding "loss" probabilities

Here (tennis.svg) is a graph drawn in Inkscape (SVG format) if someone wants to use it so as not to draw everything from the beginning.

I got lost in the rules of tennis and gave up. Now I'm just watching and maybe I'll tune in again

 

[SSequence]

In my opinion two parameters are needed:
p - the probability that the first (red) player wins on his serve
q - for the other player (blue) to win on his serve

Of course:
1-p and 1-q are the corresponding "loss" probabilities

Have you read the post#8 above (by erland). Outside of exception (like a tiebreak for which I don't know the full rule off-hand), in normal games you just have one player serving through-out a game (note that "game" is just part of a "set" and sets form a "complete game"). So essentially there is one parameter $x$ (which could be taken to be the probability of serving player winning a point or the probability of receiving player winning). The other quantity would just be $1-x$.

Just in case you were still unsure here is how the game progresses. The player P1 below is serving though-out the game.
D0:
initial state is (0,0)
D1:
If P1 wins point we move to (15,0)
If P2 wins point we move to (0,15)
D2:
If P1 wins point from (15,0) we move to (30,0)
If P2 wins point from (15,0) we move to (15,15)

If P1 wins point from (0,15) we move to (15,15)
If P2 wins point from (0,15) we move to (0,30)

D3:
If P1 wins point from (30,0) we move to (40,0)
If P2 wins point from (30,0) we move to (30,15)

If P1 wins point from (15,15) we move to (30,15)
If P2 wins point from (15,15) we move to (15,30)

If P1 wins point from (0,30) we move to (15,30)
If P2 wins point from (0,30) we move to (0,40)

D4:
If P1 wins point from (40,0) then P1 wins the game
If P2 wins point from (40,0) we move to (40,15)

If P1 wins point from (30,15) we move to (40,15)
If P2 wins point from (30,15) we move to (30,30)

If P1 wins point from (15,30) we move to (30,30)
If P2 wins point from (15,30) we move to (15,40)

If P1 wins point from (0,40) we move to (15,40)
If P2 wins point from (0,40) then P2 wins the game

D5:
If P1 wins point from (40,15) then P1 wins the game
If P2 wins point from (40,15) we move to (40,30)

If P1 wins point from (30,30) we move to (40,30)
If P2 wins point from (30,30) we move to (30,40)

If P1 wins point from (15,40) we move to (30,40)
If P2 wins point from (15,40) then P2 wins the game

D6:
If P1 wins point from (40,30) then P1 wins the game
If P2 wins point from (40,30) we move to (40,40)

If P1 wins point from (30,40) we move to (40,40)
If P2 wins point from (30,40) then P2 wins the game

D7:
If P1 wins point from (40,40) then we move to (A,40)
If P2 wins point from (40,40) then we move to (40,A)

If P1 wins point from (A,40) then P1 wins the game
If P2 wins point from (A,40) then we move to (40,40)

If P1 wins point from (40,A) then we move to (40,40)
If P2 wins point from (40,A) then P2 wins the game

======================

Regarding the probability of winning from 40-40 there was this thread few years ago:
https://www.physicsforums.com/threads/winning-tennis-with-probability-0-3-and-0-7.1044416/
For this question, I did my own calculation in spreadsheet (few years ago) and remember getting the same answer as posted in question. Since I haven't re-checked it again, I thought I would look at it again and before posting the expression for probability of winning the game from start (but hopefully it isn't difficult to see how to derive it from post#9).

 

[PeroK]

A better technique than stepping through each point is to note that a player has three scores that win a game (except where the game goes to deuce, 40-40): G-0, G-15, G-30.

The probability of winning a game to 0 is simply $p^4$, where $p$ is the probability of winning a point.

To win a game to 15, the player must get to 40-15, then win the next point. The probability for this is:
$$P(G-15) = \binom 4 1 p^3(1-p) p = 4p^4(1-p)$$As there are 4 ways to get to 40-15.Similarly:
$$P(G-30) = \binom 5 2 p^3(1-p)^2p = 10p^4(1-p)^2)$$And, the probability of reaching 40-40 is:
$$P(40-40) = \binom 6 3 p^3(1-p)^3 = 20p^3(1-p)^3$$Hence, the probability of winning a game is:
$$P = P(G-0) + P(G-15) + P(G-30) + P(40-40)P(G|40-40)$$Where $P(G|40-40)$ is the probability of winning from deuce. As shown in the previous thread, this can be calculated in two ways: using an infinite geometric series. Or, using the recursion relation:
$$P(G|40-40) = p^2 + 2p(1-p)P(G|40-40)$$

 

[SSequence]

I did some calculations based on relations that I mentioned in post#9. Let $p$ be the probability of the player serving winning a point. Then we get the following:
prob. of P1 winning from state (40,0) = $S(40,0) \times p$ = $p^4$
prob. of P1 winning from state (40,15) = $S(40,15) \times p$ = $4p^4(1-p)$
prob. of P1 winning from state (40,30) = $S(40,30) \times p$ = $10p^4(1-p)^2$

Probability of reaching state (40,40) = $20p^3(1-p)^3$

The way I calculated this was by using relations mentioned in post#9. For example let $S(a,b)$ be the probability to reach state $(a,b)$. Then using the relations in post#9 I got the following expressions:
$S(15,0)=p$
$S(0,15)=1-p$
$S(30,0)=p^2$
$S(15,15)=2p(1-p)$
$S(0,30)=(1-p)^2$
$S(40,0)=p^3$
$S(30,15)=3p^2(1-p)$
$S(15,30)=3p(1-p)^2$
$S(0,40)=(1-p)^3$
$S(40,15)=4p^3(1-p)$
$S(30,30)=6p^2(1-p)^2$
$S(15,40)=4p(1-p)^3$
$S(40,30)=10p^3(1-p)^2$
$S(30,40)=10p^2(1-p)^3$
$S(40,40)=20p^3(1-p)^3$

It is also perhaps worth mentioning briefly that the following relations must necessarily hold:
$S(0,0)=1$
$S(15,0)+S(0,15)=1$
$S(30,0)+S(15,15)+S(0,30)=1$
$S(40,0)+S(30,15)+S(15,30)+S(0,40)=1$

 

[Erland]

So, putting together the pieces in this thread, the probability for winning a game is

$$p^4 + 4p^4(1-p) + 10p^4(1-p)^2 + \frac {20p^5(1-p)^3}{1-2p+2p^2},$$

if the probabilty for winning one point is $p$.

 

[PeroK]

So, putting together the pieces in this thread, the probability for winning a game is

$$p^4 + 4p^4(1-p) + 10p^4(1-p)^2 + \frac {20p^5(1-p)^3}{1-2p+2p^2},$$

if the probabilty for winning one point is $p$.

Here's a neat alternative that includes an idea that can be very useful. Imagine that whatever happens we always play at least 6 points. If a player wins the game to 0, then we play an extra two points and the score will be 6-0, 5-1 or 4-2, in terms of points won. And if the player wins the game to 15, we play one extra point, and the score is either 5-1 or 4-2..

Note that playing these extra points can't change the outcome of the game and hence the total probability of winning must be the same whether we play these extra points or not.

With this revised scoring system, we have four possibilities:

6-0, 5-1, 4-2 or 3-3 (which is deuce)

The calculation for the probability of getting to deuce and winning from deuce is the same as before. But, we now have an alternative calculation for winning the game before deuce:
$$P_0 = p^6 + 6p^5(1-p) + 15p^4(1-p)^2 = p^4\big [p^2 + 6p(1-p) + 15(1-p)^2 \big ]$$We can compare this with the original calculation and see that we must have:
$$1 + 4(1-p) + 10(1-p)^2 = p^2 + 6p(1-p) + 15(1-p)^2$$And, in fact, both sides are equal to $10p^2 - 24p + 15$.

One advantage of this method is that we can calculate and generalise the formula using a cumulative binomial distribution.

It also gives us an otherwise non-obvious binomial identity (for a tennis game, set $n = 3$):
$$\sum_{k = 0}^{n-1} \binom{n+k}{k}(1-p)^k = \sum_{k = 0}^{n-1} \binom{2n}{k}p^{n-1-k}(1-p)^k$$