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        riddles >> easy >> Anyone For Tennis?
        
(Message started by: ThudanBlunder on May 9^th, 2007, 6:19pm)

Title: Anyone For Tennis?
Post by ThudanBlunder on May 9^th, 2007, 6:19pm

Two tennis players play a game where the server has probability p > 1/2 of winning any point and probability q of losing it, where p + q = 1

What is the probability that the server will win the game?

Title: Re: Anyone For Tennis?
Post by JiNbOtAk on May 9^th, 2007, 6:30pm

on 05/09/07 at 18:19:39, ThudanBlunder wrote:

What is the probability that the server will the game?

Win or loose ?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on May 9^th, 2007, 6:33pm

on 05/09/07 at 18:30:22, JiNbOtAk wrote:

Win or loose ?

Thanks for pointing out my my omission (now corrected).

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on May 13^th, 2007, 9:55am

Evidently not. ???

Title: Re: Anyone For Tennis?
Post by Eigenray on May 13^th, 2007, 1:39pm

I had gotten confused on the rules of scoring because they weren't what I remembered, but now I realize I was thinking of volleyball.

Let X,Y be the scores of players A,B, respectively, so with probability p, we increase X, and with probability q, we increase Y. Let S = X-Y.

First consider the simplified version where we stop as soon as |S|=2. Let a_i be the probability that player A wins given S=i. We have

a_i = p a_i+1 + q a_i-1,

giving a recurrence with characteristic polynomial pt² - t + q = (t-1)(pt-q). So
a_i = c + d rⁱ,
where r=q/p. The initial condition a_-2=0 gives c = -d/r², and a₂=1 gives d=1/(r²-1/r²), and so

a₀ = (1-r²)/(1-r⁴) = p²/(p²+q²)

is the probability that A wins this simplified game. (This can also be deduced by considering the Martingale r^S.)

It's possible for A to win the simplified game but still lose the real game, but there are only ~~four~~ five ways this can happen. Let (x,y) denote the score X=x, Y=y.

-We can "win" at (2,0) and lose at (2,2) in one way: (0,0)->(2,0)->(2,2). The probability that this happens is p²q²(1-a₀), since the simplified game is the same as the real one starting at (2,2).

-We can "win" at (3,1), and lose at (3,3) in 4 ways: (3,0)->(3,3), or (2,0)->(2,1)->(3,1)->(3,3), or (1,0)->(1,1)->(3,1)->(3,3) or (0,1)->(3,1)->(3,3). The probability that this happens is 4p³q³(1-a₀).

So the probability that A wins the simplified game but loses the real game is

(1-a₀)(p²q²+4p³q³).

And similarly, the probability that A loses the simplified game but wins the real game is

a₀(p²q² + 4p³q³).

So the overall probability is

a₀ + (2a₀-1)(p²q²+4p³q³)
= p²/(p²+q²) [ 1 + (p²-q²)(pq)²(1+4pq) ]

... I think.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on May 13^th, 2007, 4:23pm

Thanks for your reply, Eigenray.

1) In tennis a game is won by the first player to win 4 points, unless the score gets to 3-3 (this is called deuce), in which case the game continues until one of the players gets two points ahead.

2) The solution method and answer, a function of p and/or q, are quite simple.

Title: Re: Anyone For Tennis?
Post by Eigenray on May 13^th, 2007, 7:59pm

Upon further consideration (solving a linear system in 15 variables), I get

[hide]p⁴(1+2q(2-p)+2q²(5-4p))/(1-2pq),[/hide]

which agrees with experiment~~. But I still don't see the error in my previous approach.~~ and my previous solution.

What if they play until one person has at least A points and is ahead by at least B points (tennis: A=4,B=2)?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on May 13^th, 2007, 8:45pm

on 05/13/07 at 19:59:57, Eigenray wrote:

Upon further consideration (solving a linear system in 15 variables), I get

[hide]p⁴(1+2q(2-p)+2q²(5-4p))/(1-2pq),[/hide]?

For normal tennis, employing 1 independent variable, I get p⁴ + 4p⁴q + [(10p⁴q²/(1 - 2pq)]

Title: Re: Anyone For Tennis?
Post by Eigenray on May 13^th, 2007, 10:06pm

It's the same, but written like that it's much more obvious:

p⁴ is the probability of winning at (4,0).

4p⁴q is the probability of winning at (4,1), since you can get to (3,1) in C(3+1,1) ways.

10p⁴q² is the probability of winning at (4,2), since you can get to (3,2) in C(3+2,2) ways.

20p⁵q³ is the probability of winning at (5,3), since you can get to (4,3) in C(4+3,3) ways.

And for nhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 4, #[ways to win at (n+2,n)] = #[ways to get to (n,n)] = 2#[ways to get to (n-1,n-1)] = 2#[ways to win at (n+1,n-1)]. So the probability of winning is

p⁴ + 4p⁴q + 10p⁴q² + 20p⁵q³ + 40p⁶q⁴ + ...
= p⁴ + 4p⁴q + 10p⁴q²/(1-2pq).

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on May 13^th, 2007, 10:29pm

Your generalisation is also interesting and will cover tie-breaks when the server is assumed to have no advantage (which is probably quite true for many occasional players).

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on May 20^th, 2007, 8:16pm

on 05/13/07 at 20:45:59, ThudanBlunder wrote:

For normal tennis, employing 1 independent variable, I get p⁴ + 4p⁴q + [(10p⁴q²/(1 - 2pq)]

For example, when p = 2/3, q = 1/3, the server wins with probability 208/243, roughly a 6 in 7 chance.

on 05/13/07 at 19:59:57, Eigenray wrote:

What if they play until one person has at least A points and is ahead by at least B points (tennis: A=4,B=2)?

Probability of winning without reaching deuce
equals
Probability of winning by A points to r (for r = 0 to A-B)
equals
p^A + ^Ahttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cc.gif₁p^Aq + ^A+1http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cc.gif₂p^Aq² + ^A+2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cc.gif₃p^Aq³+ ...... + ^2A-B-1http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cc.gif_A-Bp^Aq^A-B

When deuces are considered, as the value of B does not affect the common ratio, which remains 2pq,
the required probability equals
p^A + ^Ahttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cc.gif₁p^Aq + ^A+1http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cc.gif₂p^Aq² + ^A+2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cc.gif₃p^Aq³ + ...... + [^2A-B-1http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cc.gif_A-Bp^Aq^A-B/(1-2pq)]

For example, in a tie-break (A = 7, B = 2) this gives probability of winning to be
p⁷ + 7p⁷q + 28p⁷q² + 84p⁷q³ + 210p⁷q⁴ + [462p⁷q⁵/(1 - 2pq)]

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 5^th, 2007, 11:07pm

Clearly it's more than 1/2. And thunderblunder, why answer your own question then quote yourself on it.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 6^th, 2007, 1:58am

on 09/05/07 at 23:07:24, srn347 wrote:

Clearly it's more than 1/2. And thunderblunder, why answer your own question then quote yourself on it.

I answered my own question because my answer appeared to differ from Eigenray's.
I quoted my answer in order to give a specific example with p = 2/3, q = 1/3

You would have realized this if you had listened to Mummy and taken your Ritalin before posting.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 6^th, 2007, 4:30pm

[b]WHAT IS RITALIN?![/b]

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 6^th, 2007, 5:13pm

on 09/06/07 at 16:30:09, srn347 wrote:

[b]WHAT IS RITALIN?![/b]

Ever heard of Google, sonny?

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 6^th, 2007, 8:02pm

Do I dare search?

Title: Re: Anyone For Tennis?
Post by ima1trkpny on Sep 6^th, 2007, 10:59pm

on 09/06/07 at 20:02:06, srn347 wrote:

Do I dare search?

Do I really have to do your work for you so you can understand what was implied?? :o
Ritalin is the brand name of methylphenidate, a drug commonly used to tread ADHD. I'm rather suprised you've never heard of it... as it is pretty common... but even if you've never come across it, it is easy enough to find out about. No excuse for being lazy...

Title: Re: Anyone For Tennis?
Post by mikedagr8 on Sep 7^th, 2007, 3:11am

on 09/06/07 at 22:59:31, ima1trkpny wrote:

Do I really have to do your work for you so you can understand what was implied?? :o
Ritalin is the brand name of methylphenidate, a drug commonly used to tread ADHD... No excuse for being lazy...

Someone with ADHD really has no excuse.

Title: Re: Anyone For Tennis?
Post by mikedagr8 on Sep 7^th, 2007, 4:35am

on 09/06/07 at 17:13:03, ThudanBlunder wrote:

Ever heard of Google, sonny?

Don't think he has ever listened, so it doesn't surprise me that he hasn't, as everyone else is, I am sure.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 7^th, 2007, 6:06pm

Get back on topic. Is that p^4+q^2+10p^4q^2/1-2pq thing provable?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 7^th, 2007, 6:13pm

on 09/07/07 at 18:06:57, srn347 wrote:

Get back on topic. Is that p^4+q^2+10p^4q^2/1-2pq thing provable?

Would you even recognise a proof if you found one in your cornflakes??

Title: Re: Anyone For Tennis?
Post by ima1trkpny on Sep 7^th, 2007, 6:27pm

on 09/07/07 at 18:13:01, ThudanBlunder wrote:

Would you even recognise a proof if you found one in your cornflakes??

T&B, how 'bout we consider giving him another shot... and be a tad kinder... I have a feeling he doesn't really know any better. I think we all understand the vexation, but let's give him one more shot now that it has been made clear what our issues with him are. Fair enough?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 7^th, 2007, 6:30pm

on 09/07/07 at 18:27:33, ima1trkpny wrote:

T&B, how 'bout we consider giving him another shot... and be a tad kinder... I have a feeling he doesn't really know any better.

Yeah, you are right. I shouldn't mock the afflicted.

Title: Re: Anyone For Tennis?
Post by mikedagr8 on Sep 8^th, 2007, 1:51am

on 09/07/07 at 18:30:13, ThudanBlunder wrote:

Yeah, you are right. I shouldn't mock the afflicted.

It's hard to control yourself sometimes, for anyone who has more to give TB for his attitude.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 8^th, 2007, 8:20am

And have you taken your ritalin, thunderblunder(pun intended)? Now were even.

Title: Re: Anyone For Tennis?
Post by Whiskey Tango Foxtrot on Sep 8^th, 2007, 10:14am

on 09/06/07 at 16:30:09, srn347 wrote:

[b]WHAT IS RITALIN?![/b]

If your posts look like the one above I think we're allowed to tease.

Title: Re: Anyone For Tennis?
Post by SWF on Sep 8^th, 2007, 6:36pm

Instead of using the infinite series that Eigenray gives for the win by 2 rule, the way I would do it is let r be probability that the server wins by 2 after the score is tied at 3-3 or more. r = p*p + 2*p*q*r (probability of server winning two straight points plus probability of the score being tied after the next 2 games and server wins after that). Solving for r gives, r=p*p/(1-2*p*q).

Similar to what Eigenray said, add probablity of the four cases: server wins 4-0, 4-1, 4-2, or after score has reached 3-3:
p^4 + 4*p^4*q + 10*p^4*q^2 + 20*p^3*q^3*( p*p/(1-2*p*q) )
The last two terms can be combined to give the 10*p^4*q^2/(1-2pq) appearing in the solutions given previously.

You could continue with a similar analysis of probability of winning a set, and enough sets to win a match, but would need to have two probabilities depending on if a given player is serving or receiving. What is better for determining the better player: the current format of game, set, match or just playing points until one of the players reaches a certain number of points first?

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 8^th, 2007, 6:52pm

Ok.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 8^th, 2007, 7:42pm

on 09/08/07 at 18:36:47, SWF wrote:

What is better for determining the better player: the current format of game, set, match or just playing points until one of the players reaches a certain number of points first?

The rules of table-tennis were changed a few years ago. One change was to increase the diameter of the ball by about 10%. This had the effect of slowing the ball down and allowing more long rallies, thus making the game more attractive for spectators. It also made it easier to follow the fast-moving ball on television. Another change was to the scoring system. Instead of playing 5 sets (3 for ladies) up to 21 points, they now play 7 sets up to 11 points. I think it is intuitively 'obvious' that the second method tends to make the results more unpredictable and therefore the matches more interesting.

on 09/08/07 at 18:36:47, SWF wrote:

You could continue with a similar analysis of probability of winning a set, and enough sets to win a match, but would need to have two probabilities depending on if a given player is serving or receiving.

P(GAME)
Let p = P(POINT), q = 1 - p
p⁴ + 4p⁴q + [10p⁴q²/(1 - 2pq)]

P(TIE-BREAK)
Let p = P(POINT), q = 1 - p
p⁷ + 7p⁷q + 28p⁷q² + 84p⁷q³ + 210p⁷q⁴ + [462p⁷q⁵/(1 - 2pq)]

P(SET WITHOUT TIE-BREAK)
Let p = P(GAME), q = 1 - p
p⁶ + 6p⁶q + 21p⁶q² + 56p⁶q³ + [126q⁶q⁴/(1 - 2pq)]

P(SET WITH TIE-BREAK)
Let p = P(GAME), q = 1 - p
p⁶ + 6p⁶q + 21p⁶q² + 56p⁶q³ + 126q⁶q⁴ + 252p⁷q⁵ + 504p⁶q⁶*P(TIE-BREAK)

5-SET MATCH (MEN, NO TIE-BREAK IN 5th SET)
Let p = P(SET WITH TIE-BREAK), q = 1 - p
p³ + 3p³q + 6p²q²*P(SET WITHOUT TIE-BREAK)

3-SET MATCH (LADIES, ALWAYS TIE-BREAKS)
Let p = P(SET WITH TIE-BREAK), q = 1 - p
p² + 2p²q

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 8^th, 2007, 7:49pm

Doesn't that change the answer? Thunderblunder, quit blundering(no insult intended, just a pun).

Title: Re: Anyone For Tennis?
Post by mikedagr8 on Sep 8^th, 2007, 7:51pm

on 09/08/07 at 19:49:32, srn347 wrote:

Doesn't that change the answer? Thunderblunder, quit blundering(no insult intended, just a pun).

Who is this 'Thunderblunder'? You need to stop making all this sh*t up. Seriously, learn to read.

Ok, I'll stop feeding him.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 8^th, 2007, 7:53pm

^{If you didn't know that i did that intentionally, you need to read!}

Title: Re: Anyone For Tennis?
Post by SWF on Sep 9^th, 2007, 2:27pm

Here is an attempt at comparing the game/set/match approach to a match to just playing until a certain number of points are scored. To keep it simple, ignored difference between serving and receiving probability and also ignored using the tie breakers for sets.

If probability of winning any given point is 0.6, using the formulas ThudanBlunder gives, probability of winning a game is 0.736, probability of winning a set is 0.966, and probability of winning a 5 set match is 0.9996. These values are higher than seen in practice, so some of the assumptions need improvement.

For a fair comparison to a match being a fixed number of points, need to find how many points are in a typical match. For 0.6 probability of winning each point, I come up with an average of 6.48 points per game, 8.07 games per set, and 3.10 sets per match, giving 162.4 points per match. If you make a match be over when the first player scores 98 points, the average number of points played to finish the match will be 162.9 points, assuming one player has 0.6 probability for each point. This results in probability of winning the match being 0.9976, which is less than with the game/set/match approach.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 3:36pm

I had already said I understood it.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 9^th, 2007, 6:54pm

on 09/09/07 at 14:27:01, SWF wrote:

If probability of winning any given point is 0.6, using the formulas ThudanBlunder gives, probability of winning a game is 0.736, probability of winning a set is 0.966, and probability of winning a 5 set match is 0.9996. These values are higher than seen in practice, so some of the assumptions need improvement.

My formulae assume that the same person is always serving. So P(GAME) will be accurate, but not the others.
Similarly, 6.48 points per game ought to be accurate, but not the others.

Still, it is interesting that the first-to-x points probability appears to be less than the 'game, set, and match' one.
This comparison is valid, as we are assuming in each case that the service does not change hands.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 7:05pm

If the players alternate serving or let whoever scores a point serve, the probability would be?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 9^th, 2007, 7:10pm

on 09/09/07 at 19:05:55, srn347 wrote:

If the players alternate serving or let whoever scores a point serve, the probability would be?

f(p,q) of course. ::)

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 7:13pm

Which means?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 9^th, 2007, 7:15pm

on 09/09/07 at 19:13:48, srn347 wrote:

Which means?

A function of p and q, of course.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 7:19pm

That function is?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 9^th, 2007, 7:26pm

on 09/09/07 at 19:19:08, srn347 wrote:

That function is?

A polynomial.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 7:30pm

There are many different kinds of polynominals. Could you just say what it is specifically?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 9^th, 2007, 7:40pm

on 09/09/07 at 19:30:19, srn347 wrote:

Could you just say what it is specifically?

Well, I have worked out the first coefficient.

It is (-84/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif)log_e(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/fraki.gif^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/fraki.gif}) 8)

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 8:26pm

Are you using a calculator of some sort?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 9^th, 2007, 8:30pm

on 09/09/07 at 20:26:11, srn347 wrote:

Are you using a calculator of some sort?

No, I can work out the above expression in my head. My brain is hard-wired for imaginary calculations. 8)

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 8:32pm

When you say he who knows speaks not and vice-verca, does that include typing?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 9^th, 2007, 9:05pm

on 09/09/07 at 20:32:49, srn347 wrote:

When you say he who knows speaks not and vice-verca, does that include typing?

Yes, he who types, speaks not.
He who speaks, types not.

It is written.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 9:11pm

I meant he who types knows not and he who knows types not. Is this also true?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 9^th, 2007, 9:14pm

on 09/09/07 at 21:11:16, srn347 wrote:

I meant he who types knows not and he who knows types not. Is this also true?

I knew what you meant, so typed not.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 9:17pm

Doesn't that mean that since we all type, we all speak not, and we all know?

Title: Re: Anyone For Tennis?
Post by Noke Lieu on Sep 9^th, 2007, 9:21pm

on 09/09/07 at 21:14:03, ThudanBlunder wrote:

I know what you meant, so did not type.

Title: Re: Anyone For Tennis?
Post by SWF on Sep 9^th, 2007, 9:23pm

on 09/09/07 at 18:54:07, ThudanBlunder wrote:

My formulae assume that the same person is always serving. So P(GAME) will be accurate, but not the others.
Similarly, 6.48 points per game ought to be accurate, but not the others.

I meant the assumptions of p=0.6 and ignoring the difference between serving and receiving are probably the main causes of the match set and match probabilities being so high. For comparable players, the better player might have probablity of winning a point of 0.75 when serving and 0.3 when receiving. Also, a good player would account for the game/set/match format in his strategy, and might even intentionally lose some points. So a fixed probability for each point does not tell the full story.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 9:30pm

Intentionally lose points? This is tennis, I don't know what you think it is(golf perhaps).

Title: Re: Anyone For Tennis?
Post by ima1trkpny on Sep 9^th, 2007, 9:42pm

on 09/09/07 at 21:30:10, srn347 wrote:

Intentionally lose points? This is tennis, I don't know what you think it is(golf perhaps).

Sometimes, strategically speaking, taking some sort of loss will set up a situation that creates a net gain... take this situation for example: Let's say you're playing baseball (hopefully you are familiar with the rules, otherwise I will create a new scenario). If you have 2 runners on base and Barry Bonds comes up to bat, the smartest thing strategically to do would be to walk him. Essentially losing in that one encounter in order to avoid him hitting it out of the park and also to "load" the bases setting up double or even triple plays that would allow you to come out ahead in the long run. Think of it in terms of the old adage about losing one battle doesn't necessarily dictate the outcome of the war.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 9:51pm

Walking in baseball I understand, but in tennis what is this "net gain" of losing points?

Title: Re: Anyone For Tennis?
Post by ima1trkpny on Sep 9^th, 2007, 10:20pm

Well the game works of a "best of" either 5 or 7 (depending on male or female players) sets per match, etc
So lets say this is a best of 7. If you have won 3 and your opponent has won 1, and they are currently serving (in which they have a higher probability of winning), by allowing them to win that match, you allow them a victory to gain a point bringing it to 3-2, but you gain the serve. And the idea is that now you have the higher probability and can no go ahead to win with a 4-2 (no need to do the last one since you already have a winning majority), instead of wasting effort fighting the odds against you when it is their serve.
(If I have forgotten or left out any of the rules, etc please correct me as it has been a long time since I have played.)

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 9^th, 2007, 10:25pm

Which is dependant on me already being ahead. Is there a way to use that when it's tied or I'm losing?

Title: Re: Anyone For Tennis?
Post by ima1trkpny on Sep 9^th, 2007, 10:37pm

on 09/09/07 at 22:25:57, srn347 wrote:

Which is dependant on me already being ahead. Is there a way to use that when it's tied or I'm losing?

Working on it... it has been a very long time since I played and my imagination begins to fade when it is late. I just gave the scenario that came to mind considering SWF's post about a good player with advantage... I will work on seeing if there is a way to take advantage of losing some points from a losing or tied relationship... however if someone else can think of anything or can add to my shaky knowledge of possible scenarios I would appreciate it.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 10^th, 2007, 4:11pm

on 09/09/07 at 21:23:41, SWF wrote:

Also, a good player would account for the game/set/match format in his strategy, and might even intentionally lose some points.

I don't see how it can ever be advantageous for a player to lose the next point, but here is an interesting quirk:

Let p = the probabiity that the server wins a point, and q = 1 - p, (0 < p < 1)
Let P(X - Y) be the probability that the server wins the game when the server has X points and the non-server has Y points.

Using your previous method
P(40 - 40) = pP(40 - 30) + qP(30 - 40)
P(40 - 30) = p + qP(40 - 40)
P(30 - 40) = pP(40 - 40)

This leads to
P(40 - 40) = p²/(p² + q²)
and
P(40 - 30) = p + [p²q/(p² + q²)]

In a similar fashion we can derive
P(40 - 15) = p + pq + [p²q²/(p² + q²)]
and
P(30 - 15) = p²(1 + q) + pq + [p²q(pq + 1)/(p² + q²)]
and
P(0 - 0) = p⁴(1 - 16q⁴)/(p⁴ - q⁴)

Btw, this gives the unexpected identity p⁴ + 4p⁴q + [(10p⁴q²/(1 - 2pq)] http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/equiv.gif p⁴(1 - 16q⁴)/(p⁴ - q⁴)

P(0 - 0) > P(40 - 30) simplifies to 8p³ - 4p² - 2p - 1 > 0
which is true for p > 0.9196...
and
P(0 - 0) > P(30 - 15) simplifies to 8p²- 4p - 3 > 0
which is true for p > 0.9114...

So we are led to the inescapable conclusion that it is possible for a player to be ahead during the game and yet have less chance of winning than before (s)he started!

For example, when p = 0.95,
P(0 - 0) = 0.9999077...
and
P(40 - 30) = 0.999862...

And conversely,
P(30 - 40) > P(0 - 0) when p < 0.0804...
and
P(15 - 30) > P(0 - 0) when p < 0.0886...

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 10^th, 2007, 6:59pm

p<1 It also works at p=1. Or at least the original formula(p^4+q^2+10p^4q^2/1-2pq) does.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Sep 10^th, 2007, 7:08pm

on 09/10/07 at 18:59:22, srn347 wrote:

p<1 It also works at p=1. Or at least the original formula(p^4+q^2+10p^4q^2/1-2pq) does.

Obviously if p = 1 there is no more point in playing than there is in reading your posts.

Title: Re: Anyone For Tennis?
Post by SWF on Sep 10^th, 2007, 7:15pm

Although I didn't check the math, that is an interesting conclusion, ThudanBlunder.

Within this probabililty model, I can't see an advantage to intentionally losing points. I was referring to reality. If for example, a player leading the match 2 sets to 1, but is losing the fourth set of the US Open 5 games to zero, and he is worn out from the summer heat, his joints ache, and he doesn't think he can play his best for much longer, even though he is a better player than his spry young opponent. He might intentionally lose the the game to end the set he is likely to lose anyway. Then he can get on with the last set before he is worn out. The other player might similarly try to lose some points to counter this strategy and wear down his opponent before the last set.

It is not strange to see a player lose the fourth set 6-0 or 6-1, and then win the match in the fifth set. Of course this doesn't prove he tried to lose some games, and I bet few pros would admit doing so.

Title: Re: Anyone For Tennis?
Post by mikedagr8 on Sep 11^th, 2007, 3:45am

I agree with that about losing intentionally to gain advantage. SWF makes very good points for tennis, and from my own experiences in soccer and various other sports it is true.
E.g. Soccer; Instead of taking on a team from defence, clear the ball into the other half as far as possible, and then work from there. It is quite common when a team is being pressured.

Title: Re: Anyone For Tennis?
Post by srn347 on Sep 12^th, 2007, 8:23pm

In fact, I've played ayso soccer for a year, I'm a defender, and clearing it to the other half of the field is what I usually do.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Oct 1^st, 2007, 5:41pm

on 09/12/07 at 20:23:15, srn347 wrote:

...I'm a defender, and clearing it to the other half of the field is what I usually do.

You mean you just blindly boot it up the field?

Title: Re: Anyone For Tennis?
Post by mikedagr8 on Oct 1^st, 2007, 5:47pm

on 10/01/07 at 17:41:13, ThudanBlunder wrote:

You mean you just blindly boot it up the field?

Yes. That's the general idea. Of course, there's a lot more of a fine art when you have skill, as to where you kick it. If you are skillful, you can generally set it up in a way to benefit your team, by putting the ball in a position where your attackers can run onto the ball.

Note Before: I've been playing for close to 10 years now, I was a reserve for Australia (don't bother looking me up, you wont find me) in Futsal, and have represented Victoria in soccer as well. This is not coming from an ignorant opinion.

Title: Re: Anyone For Tennis?
Post by sm347 on Oct 1^st, 2007, 6:42pm

what is your position, left outside?

Title: Re: Anyone For Tennis?
Post by mikedagr8 on Oct 1^st, 2007, 6:43pm

on 10/01/07 at 18:42:00, sm347 wrote:

what is your position, left outside?

Depends where I am required in a team. I'm very versatile. My preffered is CDM, but being ambidextrous, I can play anywhere.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Oct 1^st, 2007, 6:50pm

on 10/01/07 at 18:43:27, mikedagr8 wrote:

My preffered is CDM

Yes, we have a CDM in our team and he Cannot Do Much either. :P

Title: Re: Anyone For Tennis?
Post by mikedagr8 on Oct 1^st, 2007, 6:51pm

on 10/01/07 at 18:50:07, ThudanBlunder wrote:

Yes, we have a CDM in our team and he Cannot Do Much either.

LOL. That's a description not a position. That position is Left Back.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Feb 1^st, 2009, 8:25am

on 09/09/07 at 14:27:01, SWF wrote:

Just finished watching Federer - Nadal. There were 347 points (174 -173 to Federer) comprising 51 games.

That is 347/51 = 6.804... points per game. ;)
Not a bad fit, considering these two are more evenly-matched than most opponents.
(No, I didn't count the points.)

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Feb 3^rd, 2009, 4:01am

SMQ, what was your argument that the tie-break system confers no advantage to either server?

Title: Re: Anyone For Tennis?
Post by SMQ on Feb 3^rd, 2009, 10:24am

Hmm, I don't recall making that argument. I'm fairly sure the only real tennis discussion I've had was in this thread (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1211480490), and there I argued mostly from "feel" rather than from mathematical probability since the question I was trying to answer was a "what would you prefer" question.

As to the question of a tiebreak advantage, if we assume that the players are evenly matched (a reasonable assumption since they have played to a tie) and so have the same probability p of winning a point on service (and so a probability of 1 - p of breaking service), the exact probability of winning a tiebreak should be a tractable problem.

I'm working on a program... ;)

--SMQ

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Feb 3^rd, 2009, 11:31am

on 02/03/09 at 10:24:16, SMQ wrote:

Hmm, I don't recall making that argument. --SMQ

Oh, maybe it was Grimbers. Sure I read one somewhere.

Off the top of my head, I would say that the 2nd player has a miniscule advantage as, if every point goes with serve (the most likely scenario, albeit highly improbable), he gets to set point (at 5-6) first. Unfortunately, the absolute expected deviation from this scenario is not zero, as at 5-6 they have not had an equal number of serves.

Title: Re: Anyone For Tennis?
Post by SMQ on Feb 3^rd, 2009, 3:56pm

Assuming I haven't made any programming errors, the tiebreak is mathematically fair.

Call the player serving first in the tiebreak A and the player serving second B. Let A win a point on service with probability p, and B win a point on service with probability q. Clearly, then, A wins a point returning B's serve with probability 1-q, and likewise B wins a point returning A's service with probability 1-p. There are three possible outcomes from the first 12 points: A wins 7-x, B wins x-7, and deuce at 6-6.

P(A wins before deuce) = -462p⁶q⁶ + 1512p⁶q⁵ + 1260p⁵q⁶ - 1890p⁶q⁴ - 4536p⁵q⁵ - 1260p⁴q⁶ + 1120p⁶q³ + 6300p⁵q⁴ + 5040p⁴q⁵ + 560p³q⁶ - 315p⁶q² - 4200p⁵q³ - 7875p⁴q⁴ - 2520p³q⁵ - 105p²q⁶ + 36p⁶q + 1350p⁵q² + 6000p⁴q³ + 4500p³q⁴ + 540p²q⁵ + 6pq⁶ - p⁶ - 180p⁵q - 2250p⁴q² - 4000p³q³ - 1125p²q⁴ - 36pq⁵ + 6p⁵ + 360p⁴q + 1800p³q² + 1200p²q³ + 90pq⁴ - 15p⁴ - 360p³q - 675p²q² - 120pq³ + 20p³ + 180p²q + 90pq² - 15p² - 36pq + 6p

P(B wins before deuce) = -462p⁶q⁶ + 1512p⁵q⁶ + 1260p⁶q⁵ - 1890p⁴q⁶ - 4536p⁵q⁵ - 1260p⁶q⁴ + 1120p³q⁶ + 6300p⁴q⁵ + 5040p⁵q⁴ + 560p⁶q³ - 315p²q⁶ - 4200p³q⁵ - 7875p⁴q⁴ - 2520p⁵q³ - 105p⁶q² + 36pq⁶ + 1350p²q⁵ + 6000p³q⁴ + 4500p⁴q³ + 540p⁵q² + 6p⁶q - q⁶ - 180pq⁵ - 2250p²q⁴ - 4000p³q³ - 1125p⁴q² - 36p⁵q + 6q⁵ + 360pq⁴ + 1800p²q³ + 1200p³q² + 90p⁴q - 15q⁴ - 360pq³ - 675p²q² - 120p³q + 20q³ + 180pq² + 90p²q - 15q² - 36pq + 6q

P(deuce is reached) = 1 - P(A wins before deuce) - P(B wins before deuce)
= 924p⁶q⁶ - 2772p⁶q⁵ - 2772p⁵q⁶ + 3150p⁶q⁴ + 9072p⁵q⁵ + 3150p⁴q⁶ - 1680p⁶q³ - 11340p⁵q⁴ - 11340p⁴q⁵ - 1680p³q⁶ + 420p⁶q² + 6720p⁵q³ + 15750p⁴q⁴ + 6720p³q⁵ + 420p²q⁶ - 42p⁶q - 1890p⁵q² - 10500p⁴q³ - 10500p³q⁴ - 1890p²q⁵ - 42pq⁶ + p⁶ + 216p⁵q + 3375p⁴q² + 8000p³q³ + 3375p²q⁴ + 216pq⁵ + q⁶ - 6p⁵ - 450p⁴q - 3000p³q² - 3000p²q³ - 450pq⁴ - 6q⁵ + 15p⁴ + 480p³q + 1350p²q² + 480pq³ + 15q⁴ - 20p³ - 270p²q - 270pq² - 20q³ + 15p² + 72pq + 15q² - 6p - 6q + 1

(Note that I chose the order of the terms to make it clear that the equations are symmetrical w/rt p and q, i.e. swapping p and q also swaps P(A wins before deuce) and P(B wins before deuce) and has no effect on P(deuce is reached)).

From deuce a player must win two points to win the set. Since deuce can only occur after an even number of points, the next two points will always be served one by each player. To win the set from deuce, then, a player must always win one point serving and one point receiving, and so the probability of winning from deuce is independent of who serves first. Together with the above observation this is sufficient to show that the game is fair.

--SMQ

Title: Re: Anyone For Tennis?
Post by SMQ on Feb 3^rd, 2009, 3:57pm

(edit: gah, unfortunate page break! This is continued from the last post on page 3 (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1178759979;start=50#74)...)

Continuing the analysis for completeness:

P(A wins two consecutive points from deuce) = p(1 - q)
P(B wins two consecutive points from deuce) = (1 - p)q
P(A and B each win one point from deuce) = 2pq - p - q + 1

No matter how many times the score returns to deuce the remaining probability is divided proportionately to the above.

P(A finally wins from deuce) = p(1 - q)/[p(1 - q) + (1 - p)q]
P(B finally wins from deuce) = (1 - p)q/[p(1 - q) + (1 - p)q]

Putting it all together:

P(A wins) = P(A wins before deuce) + P(deuce is reached)*P(A finally wins from deuce)
= (210p⁷q⁶ - 210p⁶q⁷ - 630p⁷q⁵ + 630p⁵q⁷ + 700p⁷q⁴ + 1386p⁶q⁵ - 1386p⁵q⁶ - 700p⁴q⁷ - 350p⁷q³ - 2240p⁶q⁴ + 2240p⁴q⁶ + 350p³q⁷ + 75p⁷q² + 1395p⁶q³ + 2175p⁵q⁴ - 2175p⁴q⁵ - 1395p³q⁶ - 75p²q⁷ - 5p⁷q - 360p⁶q² - 2025p⁵q³ + 2025p³q⁵ + 360p²q⁶ + 5pq⁷ + 29p⁶q + 675p⁵q² + 1150p⁴q³ - 1150p³q⁴ - 675p²q⁵ - 29pq⁶ - 69p⁵q - 600p⁴q² + 600p²q⁴ + 69pq⁵ + 85p⁴q + 225p³q² - 225p²q³ - 85pq⁴ - 55p³q + 55pq³ + 15p²q - 15pq² - pq + p)/(-2pq + p + q)

P(B wins) = P(B wins before deuce) + P(deuce is reached)*P(B finally wins from deuce)
= (210p⁶q⁷ - 210p⁷q⁶ - 630p⁵q⁷ + 630p⁷q⁵ + 700p⁴q⁷ + 1386p⁵q⁶ - 1386p⁶q⁵ - 700p⁷q⁴ - 350p³q⁷ - 2240p⁴q⁶ + 2240p⁶q⁴ + 350p⁷q³ + 75p²q⁷ + 1395p³q⁶ + 2175p⁴q⁵ - 2175p⁵q⁴ - 1395p⁶q³ - 75p⁷q² - 5pq⁷ - 360p²q⁶ - 2025p³q⁵ + 2025p⁵q³ + 360p⁶q² + 5p⁷q + 29pq⁶ + 675p²q⁵ + 1150p³q⁴ - 1150p⁴q³ - 675p⁵q² - 29p⁶q - 69pq⁵ - 600p²q⁴ + 600p⁴q² + 69p⁵q + 85pq⁴ + 225p²q³ - 225p³q² - 85p⁴q - 55pq³ + 55p³q + 15pq² - 15p²q - pq + q)/(-2pq + p + q)

And by the way I've arranged the terms it should again be obvious that the equations are symmetrical w/rt p and q. Thus the overall chance of winning depends only on the players respective chances of winning a point and not on which player serves first.

Q.E.D

--SMQ

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Feb 4^th, 2009, 5:53am

Wow, une veritable tour de force, SMQ! :o Years ago. when this question first occurred to me and I knew no better, I started out trying to calculate those polynomials by hand. ::) But I only had one ream of paper and soon gave up.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Feb 17^th, 2009, 9:26am

on 09/10/07 at 19:15:41, SWF wrote:

Although I didn't check the math, that is an interesting conclusion, ThudanBlunder.

Actually, I can see this anomaly being hijacked by (m)ad men of the future in the following scenario.

After a fortunate case of mistaken identity, somebody's pleasantly surprised-looking Grandpa, who only nipped out for a packet of cigarettes, shuffles onto the Centre Court to face an ageing Nadal over three sets in the first round of the Seniors' Championship. As the crowd hushes, the new courtside Polynomial Probability Processor (PPP), which concurrently monitors all games and which due to time constraints has been carefully installed directly over the dead bodies of the last Luddite holdouts after Wimbledon's version of the Alamo, shows Grandpa with a chance of winning the first game of between 1 in 3 million and 1 in 4 million, depending on whether he is using his new Zimmer frame (http://www.urbandictionary.com/define.php?term=zimmer+frame). Nadal shapes to serve. On cue, Grandpa turns slowly on his heels and waddles off-court to grab a beer. Unfazed, and still 'digging deep' before each serve, Nadal sends down two consecutive aces. However, becoming slightly complacent, he double-faults on the next point due to the umpire overruling a linesman to call the ball out. Stifling a Spanish version of "You can't be serious!", Nadal demands that the decision be referred to Quarkeye, the quantum successor to the primitive, silicon-based Hawkeye, which quickly decides that the ball was out by roughly 173 Angstroms. Instantaneously, PPP's lights start flashing and popping as it proclaims for all to see that, merely by taking an opportune guzzle of the sponsor's beer, Grandpa has improved his chances of winning the first game to 1 in 2 million, even though he is now behind!

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Feb 18^th, 2009, 7:39am

In a previous post I mentioned that Federer won more points than Nadal, in spite of losing the match. We could say his ratio of ‘winning’ defeat was 174/173. In a 2n+1-set final (n > 0) what is the limiting maximum ratio of ‘winning’ defeat possible as n -> http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif?

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Mar 3^rd, 2009, 12:25pm

on 02/18/09 at 07:39:21, ThudanBlunder wrote:

No takers? :o I worked this out in 5-10 minutes. Still wouldn't mind a second opinion.

Title: Re: Anyone For Tennis?
Post by ThudanBlunder on Mar 6^th, 2009, 7:01pm

on 02/18/09 at 07:39:21, ThudanBlunder wrote:

I get [hide]65/31[/hide]. Explanation on request.