200 Points for best (correct) answer to this math problem

[tomcowley]

How exactly? I don't think Justin7 intended this to be, but this thread is turning out to be a great example between the differences of probability and statistics.

Lets say we wanted to look at the the % of times a 1H under hits when the full game under hits (which is basically this problem). We would need to make some "statistical inference" based on data collected to determine what is the relationship between the 1H total and the full game total. We could do this by looking at the 1H scores based on the full game total and model the relationship between them. Then we can use that data to determine the relationship between them.

In this example there's no data or implied relationship between them so we can't just assume one, therefore the way it is stated it is not calculable. You can't even go the route to assume every condition because the chances of all the conditions being equally likely is essentially 0.

If he gave some fixed probabilities or data about the relationship between 1st set and the total game then sure you can calculate it, but without that or data to determine a relationship between them you can't calculate anything unless you assume independence.

It's stipulated that all points are iid.

[rsigley]

Then prob(win game) = prob(win first set)

no need for anything else

[yisman]

good find, durito

[Spektre]

How exactly?

Here is the information given

a) 3 points
b) Prob winning entire game = p, losing = 1-p
c) P is contained in (0,1) for all reals
d) P can't be 0 or 1 which is wrong if c) is true

Nothing about a team that wins the 1st set is 90% to win the entire game, etc. You can't tell anything about set 1 from the game results because you don't know anything about how set 1 affects the final game. For all you know winning set 1 could make you more likely to lose the entire game than win.

The only valid answer without that information is to assume they are independent, in which case you just use the standard uniform distribution which implies their chance to win the entire game is equal to their chance to win the first set.

Although terminology needs to be standardized, you are fundamentally wrong here.

The OP, defines a match as a standard best of 5 setup. First team to win 3 games wins the match.

So, using this terminology, if the probability that Team 1 wins the match is P, what is the probability they will win the first game?

An additional piece of information that was needed, and that I clarified early on in the thread, is that you must make the assumption that the probability of Team 1 winning an individual game is constant throughout the match. Therefore, the probability before the match has begun of Team 1 winning game 1 is the same as winning game 3.

With this assumption made, then the relationship between the probability of Team 1 winning the first game (Or any individual game) and the probability that Team 1 will win the match (defined as P) is:

6x^5 - 15x^4 +10x^3 = P

It has nothing to do with normal distributions.

This is however P=f(x) and the OP wants x=f(P).

I believe this, in closed form, is going to involve divine inspiration as the quintic equation is not solvable in general, and the terms do not seem "easily" factorable.

Spektre

[Spektre]

Then prob(win game) = prob(win first set)

no need for anything else

Incorrect. All points are iid, but the random variables P and X are not.

[rsigley]

Incorrect.

Correct. (Based on the assumptions in the OP)

If it's true that P(winning set 1) = P(winning set 2) = P(winning set 3) which doesn't equal P(winning whole thing)

Just calculate (Assuming P(Winning Set 1) = C

1-P(Win | Lost 1st Set) = 1- [P(Win | Won 2,3,5) + P(Win | Won 2,4,5) + P(Win | Won 3,4,5)]

then just use normal binomial formula to calculate those 3 values. nothing will be ^5, the highest will be ^3

[Spektre]

Correct. (Based on the assumptions in the OP)

Another way to restate the problem...

The chances of flipping an unfair coin and having it land heads up 5 times in a row is 10%

What are the chances the first flip will land heads up?

----
Each flip of the coin is iid.

The chances that the first flip will land heads up is NOT 10%.

Spektre

[rsigley]

See below

[Spektre]

Correct. (Based on the assumptions in the OP)

If it's true that P(winning set 1) = P(winning set 2) = P(winning set 3) which doesn't equal P(winning whole thing)

Just calculate (Assuming P(Winning Set 1) = C

1-P(Win | Lost 1st Set) = 1- [P(Win | Won 2,3,5) + P(Win | Won 2,4,5) + P(Win | Won 3,4,5)]

then just use normal binomial formula to calculate those 3 values. nothing will be ^5, the highest will be ^3

Sorry, I just cannot follow this if you change the wording and interchange them. Is it possible to restate this using the terms "game" to mean an individual contest and "match" to mean a set of games in which one team has one at least 3 games?

[Spektre]

If you want a probability notation what you are looking for is:

P(match_win) = P(123)+P(124)+P(125)+P(134)+P(135)+P(145 )+P(234)+P(235)+P(245)+P(345)

were P(123) = probabilty team 1 wins games 1, 2 and 3

Calculate those probabilities. Perform the addition. You will come up with the fifth order polynomial I gave early on.

Spektre

[rsigley]

Sorry, I just cannot follow this if you change the wording and interchange them. Is it possible to restate this using the terms "game" to mean an individual contest and "match" to mean a set of games in which one team has one at least 3 games?

If it's what you said where P(win match 1) = P(Win match 2) etc then it's just a straight forward application of the binomial distribution like TomG said.

If it's presented as it is in the OP then you can't do anything. I didn't read the thread so I don't know about any additional assumptions. I assumed if there were anymore the OP would be updated.

No need for crazy formulas.

[Spektre]

If it's what you said where P(win match 1) = P(Win match 2) etc then it's just a straight forward application of the binomial distribution like TomG said.

If it's presented as it is in the OP then you can't do anything. I didn't read the thread so I don't know about any additional assumptions. I assumed if there were anymore the OP would be updated.

No need for crazy formulas.

The only additional assumption is that P(win GAME 1)= P(win GAME 2) ... =P(win GAME n) <> p(win MATCH)

Indeed this "crazy formula" is the solution to that problem.

and indeed this "crazy formula" comes from doing exactly what TomG said. (after I gave the formula mind you )

Spektre

[subs]

The only additional assumption is that P(win GAME 1)= P(win GAME 2) ... =P(win GAME n) <> p(win MATCH)

Indeed this "crazy formula" is the solution to that problem.

and indeed this "crazy formula" comes from doing exactly what TomG said. (after I gave the formula mind you )

Spektre

Pay the man?

[ws]

hey all. I should've probably visited players talk first as it's my first post.

this is all beyond me, but could you explain how you guys came up with this equation?
and good luck with the solution.

There are a hundred of ways to come up with the quintic Spektre mentioned.
To avoid potential confusion, let q (rather than P) denote the probability of the 1st player winning the match.

1. We can freely prolong the match till the 1st player wins 3 points -- it can take an arbitrary number of points won by the 2nd player.
If X denotes the total number of points scored in such a prolonged match, then X+3 ~ NegBin(3, p) [or X ~ NegBin(3,p), if you use NB definition that counts both successes and failures].
Solving for p the equation P(X <= 5) = q leads to the quintic above
(if the 1st player won his 3 points during the first 5 points played, then he won the match, otherwise he lost).

2. We can do as TomG and durito suggested, i.e. prolong the match to exactly 5 points -- whichever player won more points, won the match, which leads to X ~ Binom(5, p) and solving (for p) the equation P(X>=3) = q
(this time X is the number of points scored by the 1st player)

3. You can draw a simple Markov model that simulates the problem, similarly to the picture below, and come up with the same quintic:



As to the quintics -- I'm not an algebraist, but general solutions do exist for them (and arbitrary polynomials) -- e.g. using elliptic functions, theta function, Mellin integrals or hipergeometric functions. Whether any of these can be considered to be a "closed form" expression is an entirely different story.
I would be surprised if this particular class of quintics (i.e. Q(p)-q with varying q with Spektre's Q) turned out to be solvable in radicals, but verifying it is not worth the hassle imho.

I agree with rsigley that it's merely an academic exercise.

[Spektre]

Pay the man?

I dunno subs. I think the equation you gave that predates mine by one postis the same in a different format. You thought somehow you could "solve" the equation generically in Excel, but I think yu got me by one post.

Spektre

[Bosseman22]

50%

[subs]

its yours bro, i'm just happy to learn something new and u obviously put in way more skills/time.

Justin maybe do some more of these when u have time, it gets us thinking/talking about different topics?

[Justin7]

I thought Subs and Spektre made the most meaningful contributions to the problem.

good thinking, guys.

[subs]

hehe, v undeserving

how about, for the hundred points, best answer for ideas how u would start to investigate an improvement on this for real tennis betting strategies... never really even looked at live betting because of the need to sit in front of the puter while its happening. so if people don't feel like they're giving up ricebowls might be a good follow on discussion.

[aca]

Yeah, got in a hurry. I am not sure about your reasoning for why it is wrong but it is indeed the wrong formula. Update to be made shortly for the 3 game match situation. Spektre

I got this qubic!

S^2 + 2[(S^2)(1-S)] = P

(S^2)(3-2S) = P


For S= f(P) I got very huge formula but online numerical calculators trow ok numbers (P=0.5 -> S=0.5 etc)!

[Dark Horse]

I first thought this thread was intended as a joke. Especially the (paraphrasing) 'be sure to answer seriously or your answer will be ridiculed' part seemed to set it up like that. Anyway, I'll stick with the simple answer from post 21, which is the same as rsigley has argued. The question did not introduce game elements beyond a coin flip scenario with a weighted coin. The coin remains the same with each flip. The probability of heads or tails doesn't change over 3 or 10,000 flips; although the 3 toss scenario does give the weaker team a better shot, because the smaller sample size allows for luck. (As far as capturing luck in a formula. Good luck.)

[Spektre]

I got this qubic!

S^2 + 2[(S^2)(1-S)] = P

(S^2)(3-2S) = P


For S= f(P) I got very huge formula but online numerical calculators trow ok numbers (P=0.5 -> S=0.5 etc)!

Excellent,

Yes, that's the cubic and it IS solvable but when I started computing it based on the general cubic I started to get a headache!

[Spektre]

I first thought this thread was intended as a joke. Especially the (paraphrasing) 'be sure to answer seriously or your answer will be ridiculed' part seemed to set it up like that. Anyway, I'll stick with the simple answer from post 21, which is the same as rsigley has argued. The question did not introduce game elements beyond a coin flip scenario with a weighted coin. The coin remains the same with each flip. The probability of heads or tails doesn't change over 3 or 10,000 flips; although the 3 toss scenario does give the weaker team a better shot, because the smaller sample size allows for luck. (As far as capturing luck in a formula. Good luck.)

Everything you said is true. The disconnect is that simply because the coin remains the same with each flip, does not mean the probability of winning a best of 3, best of 5, best of n number of flips EQUALS the probability of winning the ONE flip. However there IS a relationship between the probability of winning any given flip and the probability of winning a best of 5. The relationship is the quintic equation given. They are NOT the same.

Spektre

[u21c3f6]

I first thought this thread was intended as a joke. Especially the (paraphrasing) 'be sure to answer seriously or your answer will be ridiculed' part seemed to set it up like that. Anyway, I'll stick with the simple answer from post 21, which is the same as rsigley has argued. The question did not introduce game elements beyond a coin flip scenario with a weighted coin. The coin remains the same with each flip. The probability of heads or tails doesn't change over 3 or 10,000 flips; although the 3 toss scenario does give the weaker team a better shot, because the smaller sample size allows for luck. (As far as capturing luck in a formula. Good luck.)

The only time the chance of winning a point in a 3 out 5 contest is the same as the chance of winning the contest is when either chance is 50%.

This is the kind of thing that I use Excel for. It may be the long way to get there, but once the spreadsheet is set-up I can use it over and over again without worrying about the math. It is the same procedure that I have used to create spreadsheets to deal with the hedges I wager on and/or to compare correlated markets.

The attached spreadsheet actually approaches the problem backwards. You enter the % chance that Team 1 will win a point and the spreadsheet calculates the chance that Team 1 or Team 2 will win the 3 out of 5 contest. By inputting various %'s for the chance of Team 1 winning a point you can easily arrive at the desired % chance of Team 1 winning the contest.

Joe.

[evo34]

I first thought this thread was intended as a joke. Especially the (paraphrasing) 'be sure to answer seriously or your answer will be ridiculed' part seemed to set it up like that. Anyway, I'll stick with the simple answer from post 21, which is the same as rsigley has argued. The question did not introduce game elements beyond a coin flip scenario with a weighted coin. The coin remains the same with each flip. The probability of heads or tails doesn't change over 3 or 10,000 flips; although the 3 toss scenario does give the weaker team a better shot, because the smaller sample size allows for luck. (As far as capturing luck in a formula. Good luck.)

And I had thought your post #21 was a joke. If not, it's probably the worst post in the history of this forum. Do you truly not understand probability? Honest question.

[Dark Horse]

Little smart ass. lol

[mathdotcom]

I thought Subs and Spektre made the most meaningful contributions to the problem.

good thinking, guys.

as opposed to TomG who answered it in a few lines

[TomG]

I originally thought this thread was just a fun brain teaser / puzzle. Subs/Spekre came up with the answer (without showing any work) before me and that's fine that they get the points. I didn't solve it for the points, just for the fun of the puzzle. Anyway, Justin7 doesn't care how the answer is derived. He just wanted to be spoon fed a closed form solution to plug into an Excel spreadsheet so he can use it to lose money betting tennis sets using the full match-up while thinking he is super sharp. Sucks though, there is no close form solution. So now he'll actually have to do a bit of (trivial) work to derive the answer numerically.

[wiffle]

lol justin7ments

[matekus]

First-To-Score

Assuming I have interpreted the problem correctly, there are 10 possible combinations based on whether the favorite or underdog wins the match and on which one of them is the first to score:

Outcomes
0-3 (U)
1-3 (F)
1-3 (U)
2-3 (F)
2-3 (U)
3-2 (F)
3-2 (U)
3-1 (F)
3-1 (U)
3-0 (F)

The sum of the first five combinations equals 1-p (underdog wins) and the sum of the second five combinations equals p. If you could make the working assumption that each combination was equally likely (unrealistic) then the probability of the favorite scoring first reduces to (p+2)/5 = (2*(1-p)/5) + (3*(p/5)) and the underdog to (3-p)/5. In general, the probability of the favorite scoring first in a First-To-Score N points game is (N+P-1)/(2*N-1).

matekus

[Justin7]

I originally thought this thread was just a fun brain teaser / puzzle. Subs/Spekre came up with the answer (without showing any work) before me and that's fine that they get the points. I didn't solve it for the points, just for the fun of the puzzle. Anyway, Justin7 doesn't care how the answer is derived. He just wanted to be spoon fed a closed form solution to plug into an Excel spreadsheet so he can use it to lose money betting tennis sets using the full match-up while thinking he is super sharp. Sucks though, there is no close form solution. So now he'll actually have to do a bit of (trivial) work to derive the answer numerically.

TomG,

If you have ever studied tennis, I'm sure you know why a solution to this problem (odds of winning a set, given odds of winning a match, assuming the odds of winning a point for player #1, or even player 1 and player 2 on on service) is close to useless. It does have some application in figuring out errors in various assumptions, but a perfect solution will not directly help any derivative player.

[subs]

thanks

[Dark Horse]

TomG,

If you have ever studied tennis, I'm sure you know why a solution to this problem (odds of winning a set, given odds of winning a match, assuming the odds of winning a point for player #1, or even player 1 and player 2 on on service) is close to useless. It does have some application in figuring out errors in various assumptions, but a perfect solution will not directly help any derivative player.

There are other best of 3 or 5 games without the tennis problem.

This challenge was a great use of points. But if anything revolutionary had come out of it, on a public forum, the linesmakers would have had the last laugh.

[wantitall4moi]

these kinds of threads sort of add credence to my thoughts on math guys. Answer was given within an hour or so of posting the question, and in the first 10 posts. then it was hashed over and debated for another hundred posts and that first answer never changed. Nor did the formula. Just a bunch of assumptions and what ifs and the usual drivel guys who are supposed to be analytical get caught up in. Which is pretty ironic when you think about it.

[jetsjets1028]

math.. im so good at math