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

**ClimbSomeRocks** · 08-16-11, 10:14 PM

so I shouldn't give my answer for 24 hours? What if I've solved it?

**Justin7** · 08-16-11, 10:24 PM

Originally posted by ClimbSomeRocks

so I shouldn't give my answer for 24 hours? What if I've solved it?

I'm not sure why you think you should wait, if you solved it.

**yisman** · 08-16-11, 10:35 PM

he just said he's taking answers for 24 hours

obviously you can answer now

**DRB** · 08-16-11, 10:43 PM

This is an excellent question but I think you mean Probability of winning is P ( not the same thing as odds ).
So I guess there is a maximum of 5 points played and...............

**Justin7** · 08-16-11, 10:47 PM

Originally posted by DRB

This is an excellent question but I think you mean Probability of winning is P ( not the same thing as odds ).
So I guess there is a maximum of 5 points played and...............

That would be correct, P is a probability between 0 and 1. And, 5 points played is the max.

**Spektre** · 08-16-11, 10:48 PM

I think there is an assumption being made here that each match has the same probability distribution between the two teams. For example, that one team does not fatigue faster than the other and thus has less chance of winning as the match progresses, or any other inter-match change in probability.

**Justin7** · 08-16-11, 10:59 PM

Originally posted by Spektre

I think there is an assumption being made here that each match has the same probability distribution between the two teams. For example, that one team does not fatigue faster than the other and thus has less chance of winning as the match progresses, or any other inter-match change in probability.

You can assume that the probability of team one winning each point is constant throughout the match.

**subs** · 08-16-11, 11:49 PM

i think if P=0.6

then the answer is 0.5537. not trying to claim the points but is that correct?

BTW any1 know of a good tool to run the formula through. the 1 i use times out. thanks

**Spektre** · 08-17-11, 12:03 AM

Yes, if P is 0.6 the answer is 0.5537. Now what if it's any other number

**Pot luck** · 08-17-11, 12:05 AM

There is no general formula for polynomial equations of degree five or higher

**subs** · 08-17-11, 12:06 AM

i was hoping to cheat and plug it all into a maths tool... what would u use?

my clumsy maths is solve [s^3]+3[(s^3)(1-s)]+6[(s^3)((1-s)^2)]=p

where obviously s is your answer

**Spektre** · 08-17-11, 12:12 AM

Originally posted by Pot luck

There is no general formula for polynomial equations of degree five or higher .

We're on the same page.

**Spektre** · 08-17-11, 12:13 AM

Originally posted by subs

i was hoping to cheat and plug it all into a maths tool... what would u use?

my clumsy maths is solve [s^3]+3[(s^3)(1-s)]+6[(s^3)((1-s)^2)]=p

where obviously s is your answer

That is my formula as well, but you need ot solve the quintic equation to get in in general form.

Spektre

**subs** · 08-17-11, 12:20 AM

i always thought that there would b a number cruncher to do this. hhehe shows how little 1 know

**Spektre** · 08-17-11, 12:28 AM

For a given P use numerical methods to solve the equation:

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

Where x is the answer and P is the match win probability given in the question.

Spektre

**TomG** · 08-17-11, 01:35 AM

Let W be Pr(Win Point) and L the Pr(Loss Point). There are 10 ways for a team to win.

WWW
WWLW
WLWW
WLWLW
WWLLW
WLLWW

LWWW
LWWLW
LWLWW
LLWWW

Substitute (1 - W) for L. Sum those up and set them equal to p. Do lots of algebra.

**cappingsports** · 08-17-11, 03:42 AM

Probability of 1st team winning 1st point = P*0.6 + Q*0.4

Where P is probability of 1st team winning and Q is (1-P)

There are 10 ways the 1st team can win, 6 of them result in the 1st team winning the 1st point:

WWW LWWW WLWW WWLW LLWWW LWLWW LWWLW WLLWW WLWLW WWLLW

If 1st team wins, there is a 0.6 probability that 1st team will score 1st point.

There are 10 ways the 2nd team can win, 4 of them result in the 1st team winning the 1st point:

LLL WLLL LWLL LLWL WWLLL WLWLL WLLWL LWWLL LWLWL LLWWL

If 2nd team wins, there is a 0.4 probability that 1st team will score 1st point

Since it's possible a team can both win and lose the match (since neither P or Q is 0 or 1), we need to add the probabilities of 2 scenarios occurring:

- Probability of 1st team scoring 1st point, if they win match = P*0.6
- Probability of 1st team scoring 1st point, if they lose match = Q*0.4

Therefore: Probability of 1st team winning 1st point = P*0.6 + Q*0.4

Example, if P = 0.6 and Q = 0.4

Probability of 1st team winning 1st point = 0.52

That would be my suggestion.

**That Foreign Guy** · 08-17-11, 04:31 AM

The problem is that T (total points) is unknown. For a game with 1 point P obviously = probability of team scoring first but from there is gets a lot messier.

I think the way I'd do it IRL is to convert the moneyline (P) to a pointspread and then use that and the total to work out the expected points scored by each side (and therefore their share of the total points which is the chance of them scoring the first point) and assume there is nothing special about the first point.

Oh, re-reading thread I see T is constrained.

Question - Is 5 the max because you need to get three to win (like men's tennis) or is five the max because the game is naturally low scoring (like soccer where 1-0 is a valid winning result)? I think that affects things a lot, certainly makes brute forcing things harder.

**cappingsports** · 08-17-11, 05:35 AM

I assumed 5 was the max number of points because the winner will always have 3 points and the loser can never have more than 2 points because once one team reaches 3 points, the game is won and the other team can't score any additional points. That is my understanding of the scenario.

**Dark Horse** · 08-17-11, 07:15 AM

With no other known variables than P and Q, the chance of team 1 scoring first is the same as the chance of team 1 winning the game.

P

**aca** · 08-17-11, 08:27 AM

Originally posted by cappingsports

Probability of 1st team winning 1st point = P*0.6 + Q*0.4 Where P is probability of 1st team winning and Q is (1-P)

If P is close to 1 and Q is close to 0
You get only 0.6 for first team winning the first point! too small for me!

You assume that all 10 row cases have same prob but they aren't!

**Spektre** · 08-17-11, 10:16 AM

You are off. If P=0.6, the chance of first team scoring first point is 0.5537.

**Spektre** · 08-17-11, 10:56 AM

It has already been established there is no general solution to a quintic equation.

Here is the solution numerically to a degree of precision. What level of granularity do you want for the points? P is as stated in the problem. x is the probability that Team one will score the first point, given that the probabilities do not change game to game.

P x
0.0001 0.021782
0.01 0.10564
0.02 0.13527
0.03 0.1567
0.04 0.17418
0.05 0.18926
0.06 0.202681
0.07 0.2149
0.08 0.22618
0.09 0.23671
0.1 0.24664
0.2 0.3266
0.3 0.38982
0.4 0.44625
0.5 0.5
0.6 0.55374
0.7 0.610182
0.8 0.6734
0.9 0.75336
0.91 0.76329
0.92 0.773823
0.93 0.7851
0.94 0.79732
0.95 0.81075
0.96 0.82582
0.97 0.8433
0.98 0.86473
0.99 0.89436
0.9999 0.97822

**Justin7** · 08-17-11, 11:17 AM

Originally posted by Spektre

It has already been established there is no general solution to a quintic equation.

There is no generalized solution, but I think a specific solution exists, given that o < p < 1.

**wtt0315** · 08-17-11, 11:19 AM

there is no way of knowing. game and points are different

**wiffle** · 08-17-11, 11:20 AM

50/50

either they do or they don't

**cappingsports** · 08-17-11, 03:06 PM

Originally posted by aca

If P is close to 1 and Q is close to 0 You get only 0.6 for first team winning the first point! too small for me! You assume that all 10 row cases have same prob but they aren't!

Yes, you are correct, just realized my error. I did indeed assume that all 10 ways of winning had the same probability of occurring. That was an incorrect assumption. In actual fact, all 10 ways of winning added together have a probability of P but are not necessarily all equal in probability.

In which case, let P(A), P(B), P(C), P(D), P(E), P(F), P(G), P(H), P(I), P(J) be the different ways of winning for 1st team and Q(A), Q(B), Q(C), Q(D), Q(E), Q(F), Q(G), Q(H), Q(I), Q(J) be the different ways of winning for 2nd team.

Probability of 1st team winning 1st point = (P(A) + P(B) + P(C) + P(D) + P(E) + P(F) + P(G) + P(H) + P(I) + P(J))*0.6 + (Q(A) + Q(B) + Q(C) + Q(D) + Q(E) + Q(F) + Q(G) + Q(H) + Q(I) + Q(J))*0.4

I'm not sure I can take this any further without knowing how win percentage affects the team's chances of winning 3-0, 3-1 or 3-2 or losing 3-0, 3-1 or 3-2. For example, a higher expected win percentage could result in a greater chance of winning 3-0 vs winning 3-2.

**Spektre** · 08-17-11, 03:15 PM

Originally posted by Justin7

There is no generalized solution, but I think a specific solution exists, given that o < p < 1.

There is indeed a specific solution for every p ; 0 < p < 1. In fact, p can be inclusive of 0 and 1 and each would have a specific solution. I believe what you are looking for however is a general solution for any p within the range 0 to 1. I don't think it exists.

**DRB** · 08-17-11, 03:59 PM

I'm impressed Spektre. The sneaky part is the coefficients 3 and 6 in your setup.
[s^3]+3[(s^3)(1-s)]+6[(s^3)((1-s)^2)]=p

**mwgriffin** · 08-17-11, 04:10 PM

**kmdubya** · 08-17-11, 04:11 PM

Isn't it just the cubed root of P?

**Spektre** · 08-17-11, 05:44 PM

Between 0 and 1 ot can be approximated by the formula x=(1/pi) arcsin (2P+1) +0.5

**blanda** · 08-17-11, 06:13 PM

I'm impressed by all of you guys....

**matekus** · 08-17-11, 08:04 PM

Interesting thread!

matekus