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.