[Ganchrow]

A highly related issue came up in a recent thread and has been discussed in similar contexts in the past. I think it's instructive enough to make explicit here.

NBA Team A is playing NBA Team B.

You've handicapped the game and believe that on a neutral court the fair money line for Team A should be +600.

If NBA home teams win 61% of the time, what would be the fair money line when playing on Team A's home court?

What about on Team B's home court?

Please explain your answers.

[InTheHole]

611/589

[SlickFazzer]

Ganch,

I would be interested to see the correct answer, and the math behind this.

Thanks,
Slick

[Ganchrow]

Originally Posted by InTheHole

611/589

No, neither answer is correct.

Out of curiosity, how did you come up with that?

[MoneySportsGuy]

I have no clue but sure would like to know the math behind it

[Panic]

Originally Posted by Ganchrow

No, neither answer is correct?

Out of curiosity, how did you come up with that?

I think he just added and subtracted 11 from the +600. I think he got the number by taking 50% from 61%.

[InTheHole]

Just a guess which I knew was wrong because I calculated it simply in my head as 600 fair value without considering the variable "location". You stated home teams win 61% of the time so an assumption was made that all variables being constant a 11% advantage exist when teams play at home and at the same time doesn't exist when they are playing away.

[SportsLockPicks]

somewhere between +860 and +890

[onlòóker]

Why are NBA teams playing on a field?

[InTheHole]

Originally Posted by Panic

I think he just added and subtracted 11 from the +600.

Yep....not good at math.....like answering one of those "EASY" SAT questions

[InTheHole]

Heah....remember last year with the Celtics...little freaky!

[Panic]

Originally Posted by InTheHole

Yep....not good at math.....like answering one of those "EASY" SAT questions

I was just guessing looking at your numbers. Didnt realize I was close. Only thing I could come up with.

[reno cool]

I'll guess about +350

[MoneySportsGuy]

SAT math questions....ahhh memories of that. WOW

[rjt721]

A. +290
B. +1300

[MoneySportsGuy]

so what is the right answer Ganch?

[Wheell]

Home +384
Away +938

The above is wrong, but if someone got that answer they are on the right track.

Home +480
Away +750

The above is also wrong but is even closer to the truth.

Home +453
Away +795

I'd give full credit to the above answer but it too is wrong due to certain... well, how do I put this... certain issues with how basketball games differ from statistical projections when the home team is a large underdog.

The fact is in real life there are always other factors. A college team and a pro team would have different answers to this problem. Another issue we face is that teams at the edges bend away from normal pythagorian projections. There is no "correct" answer.

[Ganchrow]

Originally Posted by Wheell

Home +384
Away +938

The above is wrong, but if someone got that answer they are on the right track.

Yes, this is what I would judge the correct answer based purely on the information given and the underlying Bayesian probability.

One very important consideration is the degree to which home field advantage (technically, we're talking about advantage expressed in terms of the log of the fair odds ratio) is evenly spread across all ranges of relative home/away team quality, which is the implicit assumption of the original question.

No argument here that this underlying assumption will deviate to varying extents across various sports and odds ranges but the point of this question is to illustrate the underlying math (which I'll do in a later post, unless someone else wants to do so first).

This is analogous to presenting an MLB ML market of +200/-250 and then asking for the "fair" market line. While we know from the historical record that ±214.3 is in reality incorrect, it would still be deemed the "correct" (sport-independent) mathematical answer based on the information given and the implicit and common simplifying assumption that vig is spread equally across all sides of a market.

[InTheHole]

Originally Posted by Ganchrow

Yes, this is what I would judge the correct answer based purely on the information given and the underlying Bayesian probability.

One very important consideration is the degree to which home field advantage (technically, we're talking about advantage expressed in terms of the log of the fair odds ratio) is evenly spread across all ranges of relative home/away team quality, which is the implicit assumption of the original question.

Exactly...thanks for clarifying that for the other posters

[Ganchrow]

Originally Posted by Wheell

Home +480
Away +750

The above is also wrong but is even closer to the truth.

Home +453
Away +795

I'd give full credit to the above answer

Anyway, if you'd care to explain your reasoning behind your above answers I'd love to hear it.

The net effect of what you've done in the former case is simply to multiply (or divide) the neutral court fair ML by 0.8 and then calculate proper Bayesian probability after that for Team A (or B).

The net effect of what you've done in the latter case is simply to divide the home win probability by about 1.41825 (i.e., a probability of 43.01%) and then calculate proper Bayesian probability after that.

Anyway, these both seem quite like data-driven dredges to me. (Not that there's necessarilly anything wrong with that.)

[Wheell]

I would explain but I currently have a cat on my lap.

[donjuan]

I also thought of a "close" way to do it without logs that comes up with the following numbers:

Home: +474
Away: +797

Not perfect but if the purpose was to do the quickest calculation possible that comes up with a reasonably close answer I think it wins.

[Ganchrow]

Originally Posted by donjuan

I also thought of a "close" way to do it without logs that comes up with the following numbers:

Home: +474
Away: +797

Not perfect but if the purpose was to do the quickest calculation possible that comes up with a reasonably close answer I think it wins.

Quite close but no cigar.

Despite what Wheell might claim there really is only one "correct" answer to be found via first principles and can be found by a straightforward of Bayes' theorem.

Btw, you don't actually need to take any logarithms to determine the answer (just like you don't need to take any logs to determine Kelly Optimality) ... I just I personally find the log-based formulation more compact, intuitive, and easier to remember.

So what was your methodology? We can't rule out the fact that my results were simply the product of computational error.

[pico]

Originally Posted by Ganchrow

A highly related issue came up in a recent thread and has been discussed in similar contexts in the past. I think it's instructive enough to make explicit here.

NBA Team A is playing NBA Team B.

You've handicapped the game and believe that on a neutral court the fair money line for Team A should be +600.

If NBA home teams win 61% of the time, what would be the fair money line when playing on Team A's home court?

What about on Team B's home court?

Please explain your answers.

how can you use league home win average for a particular team? let's say thunders fan booing them at home...not much of an advantage.

if you're getting +600 at a neutral field. this team must sucks ass...the spread is probably around 14 pts. the home court will probably be empty and if they play the lakers, they won't be even able to make any baskets at all.

[Ganchrow]

Originally Posted by pico

how can you use league home win average for a particular team? let's say thunders fan booing them at home...not much of an advantage.

if you're getting +600 at a neutral field. this team must sucks ass...the spread is probably around 14 pts. the home court will probably be empty and if they play the lakers, they won't be even able to make any baskets at all.

Home field advantage is a phenomenon which does tend to exist across sports (to varying extents) even for bottom of the barrel teams.

Regardless, you're missing the point. This thread is suuposed to illustrate the fundamentals of probability that underpin asymmetric, non-uniform advantage. Improved selection of model inputs and the tweaking of results (while ultimately extremely important -- as with every model) would best be considered subsequent to the thought process behind deriving these results being sufficiently understood.

And as previously stated:

Originally Posted by Ganchrow

One very important consideration is the degree to which home field advantage (technically, we're talking about advantage expressed in terms of the log of the fair odds ratio) is evenly spread across all ranges of relative home/away team quality, which is the implicit assumption of the original question.

No argument here that this underlying assumption will deviate to varying extents across various sports and odds ranges but the point of this question is to illustrate the underlying math (which I'll do in a later post, unless someone else wants to do so first).

[donjuan]

Quite close but no cigar.

Despite what Wheell might claim there really is only one "correct" answer to be found via first principles and can be found by a straightforward of Bayes' theorem.

Btw, you don't actually need to take any logarithms to determine the answer (just like you don't need to take any logs to determine Kelly Optimality) ... I just I personally find the log-based formulation more compact, intuitive, and easier to remember.

So what was your methodology? We can't rule out the fact that my results were simply the product of computational error.

A bit delirious still but:

Home: 1/7*.61/.5
Away: 1/7*.39/.5

[Wheell]

Here is how to get the answer Ganchrow says is correct:

Log 5 as proposed by Bill James:

(H-H*A)/(H+A-2*H*A)

H is the expected win % of the home team against average league opposition, away is the expected win % of the away team on the road against league opposition.

Let us assume that on a neutral court team B is an average team (.5000), and team a is moderately awful (1/7, or, .1429).

If we hold team B as a constant at home, on a neutral court, and away, we can set team B as the variable and give them strengths of .61 at home, .5 on a neutral court, and .39 away. This method holds up to basic scrutiny as you can take two average teams, hold on as the control, and treat the other as the variable and get what should be the "correct" answer given the information Ganchrow provides.

However, that answer is wrong. Home court advantage is at its largest when the teams are close to even in strength and home court advantage is at its weakest when the game is lopsided. If you are going to win 95% of the team you come to my field (USC @ UCLA), how much can I really gain when you come to my field?

The other two answers I provided use modified log 5 methods that then have a second step to attempt to correct for the problems listed above.

[Ganchrow]

Originally Posted by Wheell

Here is how to get the answer Ganchrow says is correct:

Log 5 as proposed by Bill James:

(H-H*A)/(H+A-2*H*A)

H is the expected win % of the home team against average league opposition, away is the expected win % of the away team on the road against league opposition.

Let us assume that on a neutral court team B is an average team (.5000), and team a is moderately awful (1/7, or, .1429).

If we hold team B as a constant at home, on a neutral court, and away, we can set team B as the variable and give them strengths of .61 at home, .5 on a neutral court, and .39 away. This method holds up to basic scrutiny as you can take two average teams, hold on as the control, and treat the other as the variable and get what should be the "correct" answer given the information Ganchrow provides.

I wrote this question following replying to dcbt in http://forum.sbrforum.com/handicappe...air-line.html/.

The complete derivation from Bayes' Theorem follows:

We define the following 3 events:

Let X_h = Team X wins next game at home
Let X_n = Team X wins next game on neutral field (P(X_n) in this case repreesnts our prior probability estimate of P(X_h))
Let H = any given team wins at home against any given opponent

So we're looking to solve for :

P(X_h | H )

Method 1:
So from Bayes' theorem:

P( X_h | H ) = P( H | X_n ) * P ( X_n ) / P(H)

Directly from Bayes' Theorem:

P(H) = P(X_n) * P( H | X_n ) + (1- P(X_n) * ( 1 - P( H | X_n ))

Substituting in to the P( X_h | H ) equality above gives us:

P( X_h | H ) = P( H | X_n ) * P ( X_n ) / (P(X_n) * P( H | X_n ) + (1- P(X_n) * ( 1 - P( H | X_n ) ) )

So in the first case we have:

P(X_n) = 1/7
P( H | X_n ) = 61%

Which yields:

P( X_h | H ) = 61% * 1/7 / ( 61% * 1/7 + (1- 61%) * ( 1 - 1/7 ) ) ≈ 20.678%

Implying a fair market line of about +383.6 for Team A to win at home.

For Team B to win to win at home we have P(X_n) = 6/7.

P( X_h | H ) = 61% * 6/7 / ( 61% * 6/7 + (1- 61%) * ( 1 - 6/7 ) ) ≈ 90.370%

Implying a fair market line of about +938.5 for Team A to win at on the road.

Method 2:
This is even easier to apply using Bayes' Theorem with logarithmic odds ratios.

If we define the "logit" function lg() as the log of fair fractional odds so lg(x) = ln(x) - ln(1-x), then Bayes theorem gives us:

lg(P(X_h | H )) = lg(P(X_n)) + lg(P(H))

hence for Team A:

lg(P(X_h)) = ln(1/7) - ln(6/7) + ln(61%) - ln(39%) ≈ -1.34445

So because lg^-1(y) = exp(y) / (1+exp(y))

P(X_h) ≈ exp(-1.34445) / (1+exp( -1.34445)) ≈ ≈ 20.678%

Implying (as above) a fair market line of about +383.6 for Team A to win at home.

Similarly for Team B:

lg(P(X_h)) = ln(6/7) - ln(1/7) + ln(61%) - ln(39%) ≈ 2.23907
P(X_h) ≈ exp(2.23907) / (1+exp( 2.23907)) ≈ ≈ 90.370%

Implying (as above) a fair market line of about +938.5 for Team B to win on the road.

Method 3:
This is simply an intuitive way of looking at it.

Imagine you have two sacks of ping pong balls.

In the first sack 1/7 on these balls are marked Team A wins, and the remainder Team B wins.

In the second sack 61% of these balls are marked home team wins, and the remainder away team wins.

The game works as follows ... you pick a ball from each of the two sacks (with replacement) until you select a pair with noncontradictory result (i.e., assuming Team A to be the home team, by picking either Team A Wins+Home Team wins or Team B wins+Away Team Wins).

The probability of picking Team A Wins+Home Team wins would be 1/7 * 61%, while the probability of of picking Team B Wins+Away Team wins would be 6/7 * 39%.

Hence the game terminates after any given set of picks with probability 1/7 * 61% + 6/7 * 39%.

So this gives us for Team A winning at home:

P( X_h | H ) = 61% * 1/7 / ( 61% * 1/7 + (1- 61%) * ( 1 - 1/7 ) ) ≈ 20.678%

which is exactly the same equation fomrlaly derived in Method 1 above.

Similar logic yields the probability of Team A winning on the road.

Originally Posted by Wheell

However, that answer is wrong. Home court advantage is at its largest when the teams are close to even in strength and home court advantage is at its weakest when the game is lopsided. If you are going to win 95% of the team you come to my field (USC @ UCLA), how much can I really gain when you come to my field?

Not sure you really expressed this as you had intended but I'll say that if we home-field advantage yields a randomly selected home team p=61%, then a team winning 5% of the time on neutral ground would win 7.61% at home. That;s how much you could gain.

Originally Posted by Wheell

The other two answers I provided use modified log 5 methods that then have a second step to attempt to correct for the problems listed above.

Yeah, as I mentioned before in the 2^nd solution set you gave it just appears that you multiplied (divided) the neutral court fair ML by 0.8 and then calculated proper Bayesian probability after that for Team A (B).

In the 3^rd solution set you gave it just appears that you divide the home win probability by about 1.41825 (i.e., to yield a probability of 43.01%) in both cases and then calculate proper Bayesian probability after that.

As I said, Wheell, if you'd care to explain the reasoning behind your other 2 answers I'd love to hear it.

[dcbt]

I'm glad you posted this, Ganchrow... I would have chimed in today (since you mentioned this was in response to my question yesterday) but was at work and my spreadsheet was on my home pc. Anyway, it gave me a chance to double check my setup just now, and I got the same answers you did, so it's reassuring to me that I utilized your formulas and explanations correctly. THanks.

[Data]

Originally Posted by Wheell

Here is how to get the answer Ganchrow says is correct:

Log 5 as proposed by Bill James:

(H-H*A)/(H+A-2*H*A)

H is the expected win % of the home team against average league opposition, away is the expected win % of the away team on the road against league opposition.

Let us assume that on a neutral court team B is an average team (.5000), and team a is moderately awful (1/7, or, .1429).

If we hold team B as a constant at home, on a neutral court, and away, we can set team B as the variable and give them strengths of .61 at home, .5 on a neutral court, and .39 away. This method holds up to basic scrutiny as you can take two average teams, hold on as the control, and treat the other as the variable and get what should be the "correct" answer given the information Ganchrow provides.

Or, you can just use an expanded formula that accounts for the league's HCA C:

(H-H*A)*C/((H-H*A)*C+(A-H*A)*(1-C)

However, that answer is wrong. Home court advantage is at its largest when the teams are close to even in strength and home court advantage is at its weakest when the game is lopsided.

It is unclear what you mean here. Since your HCA is not a league-wideconstant your should provide your definition of HCA.

[reno cool]

Originally Posted by Wheell

Here is how to get the answer Ganchrow says is correct:

Log 5 as proposed by Bill James:

(H-H*A)/(H+A-2*H*A)

H is the expected win % of the home team against average league opposition, away is the expected win % of the away team on the road against league opposition.

Let us assume that on a neutral court team B is an average team (.5000), and team a is moderately awful (1/7, or, .1429).

If we hold team B as a constant at home, on a neutral court, and away, we can set team B as the variable and give them strengths of .61 at home, .5 on a neutral court, and .39 away. This method holds up to basic scrutiny as you can take two average teams, hold on as the control, and treat the other as the variable and get what should be the "correct" answer given the information Ganchrow provides.

However, that answer is wrong. Home court advantage is at its largest when the teams are close to even in strength and home court advantage is at its weakest when the game is lopsided. If you are going to win 95% of the team you come to my field (USC @ UCLA), how much can I really gain when you come to my field?

The other two answers I provided use modified log 5 methods that then have a second step to attempt to correct for the problems listed above.

What evidence do you use for this? Would this also be true in case of determining a spread?
I ask this because my gut feeling is the opposite is true if anything, but have not looked into it.

[reno cool]

1/7=.1428*1.56=.2227to.7772

=3.48to1

Why is this wrong?

[Ganchrow]

Originally Posted by reno cool

1/7=.1428*1.56=.2227to.7772

=3.48to1

Why is this wrong?

I'm sorry, but I'm not quite understanding what the above is exactly intended to represent.

Could you perhaps rephrase either in plain English or with clearly defined algebraic variables?

[reno cool]

a. +600 implies 1/7 14.28%
b. 61% vs 39% is 1.564 to 1 ratio

so I'm just multiplying the 2 numbers. get win prob 22.3%

[Ganchrow]

Originally Posted by reno cool

a. +600 implies 1/7 14.28%
b. 61% vs 39% is 1.564 to 1 ratio

so I'm just multiplying the 2 numbers. get win prob 22.3%

OK, I'm with you so far ... but I'm afraid I just don't see your mathematical rationale for multiplying A's expected neutral field probability by the fair home field odds ratio in an attempt to determine A's expected home field probability.

Look at it this way ... if instead of a 61% home win probability we had a 90%, then the quotient would be 9/7 ≈ 1.285714 > 1, which isn't a valid probability.