[bushindo]

I'm sure we're all familiar with the rules of tennis scoring. In case you're not, see this link

http://en.wikipedia.org/wiki/Tennis_scoring

Now, a game in tennis consists of points. According to wikipedia, "a game consists of a sequence of points played with the same player serving, and is won by the first player (or side) to have won at least four points and at least two points more than his or her opponent".

Assume that each point being contested is a binomial event, with player A winning with chance .55. Let's assume that there is no such thing as serve advantage. What is the chance for player A to win each game? Don't forgot to account for the possibility of deuces.

[Guest]

By each game you mean the whole game or individually?

[bushindo]

By each game you mean the whole game or individually?

By a game, i mean the discrete components of sets. For each game, both players start with 0 points. They then contest for points, which we assume is a binomial process. The first player to get to 4 points wins the game. However, if both players gets to 3 points, it is called a deuce. During a deuce, the first player to score a point gets something called advantage. If the player with advantage wins the next point, he wins the game. If he loses the next point, the game goes back to deuce. And so on.

[Guest]

But u said other is no thing as serve advantage... SO no one gets a preference or it stays a deuce? (I ask a lot befor i get into it, sorry :S)

[bushindo]

But u said other is no thing as serve advantage... SO no one gets a preference or it stays a deuce? (I ask a lot befor i get into it, sorry :S)

Sorry, serve advantage is the statistical edge that the player who is serving has over the non-serving player. I'm just saying the chance of each player winning a point is the same regardless of whether is he has the serve or not.

That is different from advantage, which happens when a player gets an point after deuce.

Edited May 24, 2009 by bushindo

[Guest]

I am not certain about this, as I used a bit of approximation in my calculations, but here is my best guess.

Win: 62.5%

Lose: 37.5%

[Guest]

im getting a never ending i guess kind of recursion multiplication but when i approximated it i got

60%

not too sure tho trying more stuff

[Guest]

i wrote a program to find the probability up to the 20th deuce i figure thats good enough and got

.5935776732

i only got .357056 for the other player so im missing 4% somewhere

[Guest]

Some brute-force old school math:

(0.55)^4 //Winning 4 in a row

+ (0.55)^4*(0.45)*5C4 //Winning 4 of 5

+ (0.55)^4*(0.45)*6C4 //Winning 4 or 6

+ (0.55)^5(0.45)^3 //Getting to and Winning the 1st Duce

+ (0.55)^6(0.45)^4 //Getting to and Winning the 2nd Duce

.

.

+ (0.55)^5(0.45)^3*[(0.55)*(0.45)]^n //Getting to and Winning the nth Duce

.

.

+ tend to 0 //Getting to and Winning the infinite Duce

Solving....

= 0.5753 + Summation n = 0 to n = infinity {(0.55)^5*(0.45)^3*[(0.55)*(0.45)]^n}

This is a converging summation there fore S = a/1-r

= 0.5753 + (0.55)^5*(0.45)^3

-----------------------

1- [(0.55)*(0.45)]

= 0.58144

58.144% chance of winning

Edited May 24, 2009 by adiace

[bushindo]

Some brute-force old school math:

(0.55)^4 //Winning 4 in a row

+ (0.55)^4*(0.45)*5C4 //Winning 4 of 5

+ (0.55)^4*(0.45)*6C4 //Winning 4 or 6

+ (0.55)^5(0.45)^3 //Getting to and Winning the 1st Duce

+ (0.55)^6(0.45)^4 //Getting to and Winning the 2nd Duce

.

.

+ (0.55)^5(0.45)^3*[(0.55)*(0.45)]^n //Getting to and Winning the nth Duce

.

.

+ tend to 0 //Getting to and Winning the infinite Duce

Solving....

= 0.5753 + Summation n = 0 to n = infinity {(0.55)^5*(0.45)^3*[(0.55)*(0.45)]^n}

This is a converging summation there fore S = a/1-r

= 0.5753 + (0.55)^5*(0.45)^3

-----------------------

1- [(0.55)*(0.45)]

= 0.58144

58.144% chance of winning

Your math for the first three situations are right, but the deuce part isn't.

Thelogical1, your approximation is quite close. Care to elaborate on your method?

The deuce part is very similar to a previously solved problem here on the forum

http://brainden.com/forum/index.php?showtopic=8255

[Guest]

i think your math is off adiace as explained below

i think the math should be to win the 4

.55^4

to win in 5

.55^4*.45*4c3

because he has to win the last one otherwise he won in 4 and your double counting

to win in 6

.55^4*.45^2*5c3

same reason he has to win last one

then there is no winning in 7 so then you take the first deuce and in a recursion like problem (i used recursion to solve i dont remember any formula)

.55^3*.45^3*6c3(.55^2+2*.45*.55(.55^2+2*.45*.55(.55^2+...........

first part is self explanitory the rest is win in 2 or tie again then win in two or tie so on so on

only way to win left but as stated above i missed something because i am missing almost 5% sowhere and i dont know where

[Guest]

Your math for the first three situations are right, but the deuce part isn't.

Thelogical1, your approximation is quite close. Care to elaborate on your method?

The deuce part is very similar to a previously solved problem here on the forum

http://brainden.com/forum/index.php?showtopic=8255

To win a duce one must consider the probability in entering a duce.

So (0.55)^3*(0.45)^3 is essential.

Now to this we multiply the condition for A to win, that is to win twice in a row. Other wise he loses or enters the next duce.

so (0.55)^3*(0.45)^3 * (0.55)^2

So regardless of which duce the second condition remains same, i.e (0.55)^2

so we have

(0.55)^2

x

{

(0.55)^3*(0.45)^3

+(0.55)^4*(0.45)^4

+(0.55)^5*(0.45)^5

.

.

.

.

}

This is what i solved for in my answer. Could you explain which assumption of mine was false?

[Guest]

i think your math is off adiace as explained below

i think the math should be to win the 4

.55^4

to win in 5

.55^4*.45*4c3

because he has to win the last one otherwise he won in 4 and your double counting

to win in 6

.55^4*.45^2*5c3

same reason he has to win last one

then there is no winning in 7 so then you take the first deuce and in a recursion like problem (i used recursion to solve i dont remember any formula)

.55^3*.45^3*6c3(.55^2+2*.45*.55(.55^2+2*.45*.55(.55^2+...........

first part is self explanitory the rest is win in 2 or tie again then win in two or tie so on so on

only way to win left but as stated above i missed something because i am missing almost 5% sowhere and i dont know where

Umm... i think you have the combinations confused... say he has to win 4 in 5 matches... this can be expressed in two ways:

A winning 4 out of 5 and B winning 1 out of 5

i.e > 5C4 or 5C1 both = 5. Logically this makes sense too.. B wins only once in 5 matches.. he can do this in only 5 different ways.

however 5C3 = 10.

[Guest]

figured it out i dont know where i was losing the .05 in my java recursion but i did but i figured out (i looked at adiace work im not gonna lie) the deuce was (1+a^2+a^3+a^4...)=1/1-a a=.55*.45*2 anyway short answer same old math (i guess the mistake was human somewhere not in my math formula) but new answer

.62315

[Guest]

oh and all i meant adiace is for the win in 5 it should be .55^4*.45*4c1

because it wouldnt make sense for him to win four in a row then play another and lose so there are only 4 places the lose can go otherwise your sorta double counting a win in 6 scenario

same logic applies to win in 6 but thanx for your logic on the deuce it helped, my recursion was flawed somehow

[Guest]

figured it out i dont know where i was losing the .05 in my java recursion but i did but i figured out (i looked at adiace work im not gonna lie) the deuce was (1+a^2+a^3+a^4...)=1/1-a a=.55*.45*2 anyway short answer same old math (i guess the mistake was human somewhere not in my math formula) but new answer

.62315

Hey.. I read finals previous post and modified my equation by multiplying 6C3.

0.69719

(0.55)^4 + (0.55)^4*(0.45)*5C4 + (0.55)^4*(0.45)^2*6C4 + Summation n = 0 to n = infinity {(0.55)^5*(0.45)^3*6C3*[(0.55)*(0.45)]^n}

This is a converging summation there fore S = a/1-r

= 0.6972

69.72% chance of winning thumbsup.gif

Edited May 24, 2009 by adiace

[Guest]

oh and all i meant adiace is for the win in 5 it should be .55^4*.45*4c1

because it wouldnt make sense for him to win four in a row then play another and lose so there are only 4 places the lose can go otherwise your sorta double counting a win in 6 scenario

same logic applies to win in 6 but thanx for your logic on the deuce it helped, my recursion was flawed somehow

dope... thanks for the hint

I have corrected my equations and got the same answer as you

[spoiler='calculations

']

b4 duces

(0.55)^4 + (0.55)^4*(0.45)*(5C4-4C4) + (0.55)^4*(0.45)^2*(6C4-5C4)

infinite duces

+ [(0.55)^5*(0.45)^3*6C3]*Summation n = 0 to n = infinity {2*[(0.55)*(0.45)]^n}

=0.623

[Guest]

0.6231485025 (as accurate as I can give it)

or approx 0.623

[bushindo]

0.6231485025 (as accurate as I can give it)

or approx 0.623

final, adiace, and psychic_mid got it. Well done.