Behold, the idiocy that is John Hollinger

[endrity]

Originally Posted by chumdawg

D2K is clearly correct here, and I'm amazed anyone would even think to argue otherwise.

The premise is very simple: good teams will win fewer games in a one-possession format than they would in a 48-minute format...and bad teams would also lose fewer.

The net result is, indeed, that the teams would gravitate toward .500.

There is a really good article on this concept on one of the sports websites, but at the moment I can't remember which. The point was that weaker teams improve their chances by introducing more randomness into the contest.

Good teams could just as easily win more, in fact more than they should. There is no reason to assume the bias of a shorter game will have only one direction.

[chumdawg]

Originally Posted by grndmstr_c

You misunderstand me. I'm not claiming that the determinants of winning and losing are the same in a 100 possession game as they are in a one possession game. I'm just saying that skill, as distinct from randomness, is by definition constant.

That said, you make a good point that skill, whatever it is, is surely multidimensional, and there are certainly very complex interactions that go into determining precisely how performance across different dimensions influences the long-run odds of winning in a particular matchup between two specific teams. I just don't view that point as being particularly critical to the discussion at hand.

The point of the article I mentioned, which I will try to find, is that if you could reduce the game down to a small enough size, you would find there is not much difference in skill level. Everybody on the court--or field, as I think the article may have been about football as well--being a professional and all. Or looked at another way, if there *is* a large difference in relative skill levels, much of it is "lost" to the game conditions, as the one-possession format levels the playing field, as it were.

As an illustration, if you were an NFL team that was completely outmatched in terms of talent, would you prefer to play your opponent the full four quarters or instead play a sudden-death overtime? Or do you think there is no difference between the two?

[Murphy3]

Are you talking about each NBA team getting one final possession or are you talking about flipping a coin and whichever team wins gets the ball first and first point wins the game. Or are you going to go with the team with the most points after 5 minutes.. or even 2 minutes..

There's a big difference between that and football OT.

[endrity]

Originally Posted by chumdawg

The point of the article I mentioned, which I will try to find, is that if you could reduce the game down to a small enough size, you would find there is not much difference in skill level. Everybody on the court--or field, as I think the article may have been about football as well--being a professional and all. Or looked at another way, if there *is* a large difference in relative skill levels, much of it is "lost" to the game conditions, as the one-possession format levels the playing field, as it were.

As an illustration, if you were an NFL team that was completely outmatched in terms of talent, would you prefer to play your opponent the full four quarters or instead play a sudden-death overtime? Or do you think there is no difference between the two?

I think in my longer post I explain why this kind of reasoning is wrong. It's not that the skill level differential will be lost, but the size (coefficient, weight) of the error i.e randomness on the final outcome would be much larger.

And yes, a bad team would prefer a shorter game in basketball and football. But the reasoning is that with a much larger impact of the error they have a larger probability of getting a better score. However, that probability would be the same for a getting a worse result than they would normally.

Think of it as a bell shaped distribution. With no randomness the distribution isn't one at all, it's just a vertical line at the mean. As the error gets larger, the tails get fatter and fatter. Therefore the probability of getting results that deviate from the mean gets larger and larger.

[grndmstr_c]

Okay, Ill take the crow. I think I must have been getting myself crossed on winning possessions instead of winning games, but I just worked out a simple proof and dirno and CD are correct on this. Should've taken the time to look at it systematically first, but for those who are interested, here's an asymptotic argument that illustrates why:

Suppose you have Team A and Team B playing a game that proceeds in discrete fashion, i.e., in possessions, and suppose Team A has constant probability of winning a single possession = p. Now, what we're interested in is what if we extend the game so that it has N possessions, and winning the game is defined as winning a majority of the possessions, i.e., winning (N+1)/2 or more possessions. We can use the normal approximation to the binomial if we assume N is large enough, and doing so it's straightforward to show that the probability of winning (N+1)/2 contests is aproximately equal to the cumulative distribution function (cdf) of the standard normal distribution evaluated at the square root of N times a function of p that is positive if p > .5, and negative if p < .5. The standard normal cdf is strictly increasing from 0 to 1 across the real number line, and it follows that the probability of winning (N+1)/2 contests will go to 0 as N goes to infinity if p is even slightly less than .5, and will go to 1 as N goes to infinity if p is even slightly greater than .5. Obviously, the reverse occurs as N goes from large to small.

My apologies gentlemen, and may this serve as a reminder to all that there is real value in striving even to disagree with civility.

[endrity]

Originally Posted by grndmstr_c

Okay, Ill take the crow. I think I must have been getting myself crossed on winning possessions instead of winning games, but I just worked out a simple proof and dirno and CD are correct on this. Should've taken the time to look at it systematically first, but for those who are interested, here's an asymptotic argument that illustrates why:

Suppose you have Team A and Team B playing a game that proceeds in discrete fashion, i.e., in possessions, and suppose Team A has constant probability of winning a single possession = p. Now, what we're interested in is what if we extend the game so that it has N possessions, and winning the game is defined as winning a majority of the possessions, i.e., winning (N+1)/2 or more possessions. We can use the normal approximation to the binomial if we assume N is large enough, and doing so it's straightforward to show that the probability of winning (N+1)/2 contests is aproximately equal to the cumulative distribution function (cdf) of the standard normal distribution evaluated at the square root of N times a function of p that is positive if p > .5, and negative if p < .5. The standard normal cdf is strictly increasing from 0 to 1 across the real number line, and it follows that the probability of winning (N+1)/2 contests will go to 0 as N goes to infinity if p is even slightly less than .5, and will go to 1 as N goes to infinity if p is even slightly greater than .5. Obviously, the reverse occurs as N goes from large to small.

My apologies gentlemen, and may this serve as a reminder to all that there is real value in striving even to disagree with civility.

I guess I look at it differently and make different assumptions. Instead of winning a possession, in basketball you'd have two teams with a possession each trying to score. Let's assume we have good team A and bad team B. And those are the only teams in the league. Now, each team has a probability to score respectively p and q, where p>q. If those probabilities are normally distributed, you'd get various possibilities for winning scores with one single possessions each. If you play the game with an infinite number of possessions, team A would always end up winning, and it's expected score would approach N*p where N is the number of possessions. But the mean would always be p=!q regardless.

[grndmstr_c]

endrity, the catch is this. Winners of multi-possession games can be approximately determined by taking stock of which team had a higher proportion of won possessions. As the number of possessions that go into that per-game sample proportion increases, the difference between that sample proportion and the true probability of winning a given possession will tend to decrease, with the net result being that the better team will win a majority of the possessions more and more often. You're correct about the sample proportion zeroing in on the true mean probability of winning a single possession, but the standard deviation of that sample proportion is shrinking at a root N rate (the same root N that I noted in my last post), and that's what ends up making the difference.

[chumdawg]

Originally Posted by grndmstr_c

Okay, Ill take the crow. I think I must have been getting myself crossed on winning possessions instead of winning games, but I just worked out a simple proof and dirno and CD are correct on this. Should've taken the time to look at it systematically first, but for those who are interested, here's an asymptotic argument that illustrates why:

Suppose you have Team A and Team B playing a game that proceeds in discrete fashion, i.e., in possessions, and suppose Team A has constant probability of winning a single possession = p. Now, what we're interested in is what if we extend the game so that it has N possessions, and winning the game is defined as winning a majority of the possessions, i.e., winning (N+1)/2 or more possessions. We can use the normal approximation to the binomial if we assume N is large enough, and doing so it's straightforward to show that the probability of winning (N+1)/2 contests is aproximately equal to the cumulative distribution function (cdf) of the standard normal distribution evaluated at the square root of N times a function of p that is positive if p > .5, and negative if p < .5. The standard normal cdf is strictly increasing from 0 to 1 across the real number line, and it follows that the probability of winning (N+1)/2 contests will go to 0 as N goes to infinity if p is even slightly less than .5, and will go to 1 as N goes to infinity if p is even slightly greater than .5. Obviously, the reverse occurs as N goes from large to small.

My apologies gentlemen, and may this serve as a reminder to all that there is real value in striving even to disagree with civility.

Yeah, I think that is the way I'm looking at it. For example, let's say that Teams A and B play, and Team A is a 55/45 favorite on each and every possession. If we "pay out" after every single possession, then after enough possessions we would expect the score to be 55/45. But if we wait to "pay out" until we see who is ahead after, say, 100 possessions, then we would expect Team A to win far, far more often than their "per possession" win expectation of 55/45. In fact, after 100 possessions we would expect them to be ahead well better than 90% of the time, right? (I'm just going on gut feel here...100 trials at 55/45 each trial seems like enough to get close to even 100%.)

Am I looking at it right? If so, it would mean that even a 60-win team playing a 20-win team is likely only a small favorite on any one individual possession...hence, the reason that single-possession games would result in all teams moving closer to .500.

[grndmstr_c]

Originally Posted by chumdawg

Yeah, I think that is the way I'm looking at it. For example, let's say that Teams A and B play, and Team A is a 55/45 favorite on each and every possession. If we "pay out" after every single possession, then after enough possessions we would expect the score to be 55/45. But if we wait to "pay out" until we see who is ahead after, say, 100 possessions, then we would expect Team A to win far, far more often than their "per possession" win expectation of 55/45. In fact, after 100 possessions we would expect them to be ahead well better than 90% of the time, right? (I'm just going on gut feel here...100 trials at 55/45 each trial seems like enough to get close to even 100%.)

Am I looking at it right? If so, it would mean that even a 60-win team playing a 20-win team is likely only a small favorite on any one individual possession...hence, the reason that single-possession games would result in all teams moving closer to .500.

The better team in a matchup with 55/45 per-possession odds would actually win in the neighborhood of 97-98% of the time in a 100-possession game, I believe. Then again, it does climb fairly quickly so, for example, even though the normal approximation I'm using is tenuous in this range, the same odds would yield an estimated winning percentage close to 75% over ten possessions. If you buy the last 10 possessions as a reasonable alternative to the last 5 minutes as a definition of crunch time, that at least suggests that a quality closing unit (like Kidd/JET/Marion/Dirk/Chandler) can have a significant positive impact on a team's overall winning percentage in games that are close to tied over the home stretch.

Hollinger has completely missed the boat on that last point if you ask me.

[chumdawg]

Originally Posted by grndmstr_c

If you buy the last 10 possessions as a reasonable alternative to the last 5 minutes as a definition of crunch time, that at least suggests that a quality closing unit (like Kidd/JET/Marion/Dirk/Chandler) can have a significant positive impact on a team's overall winning percentage in games that are close to tied over the home stretch.

Hollinger has completely missed the boat on that last point if you ask me.

Yes! This is the part of our discussion that I am finding most stimulating. I think there are several rich areas to mine. For example, let's say that we are tied with two minutes left, or one minute even. Or to put it in our terms, for two or four or some number of "possessions." What are the thresholds in the respective scenarios where we can say with significant certainty that a team must have been a 55/45 or 60/40 favorite rather than a coin flip, for example.

Since you are crunching numbers, let me make sure I am thinking of "possessions" in the same way that you are. Are you calling a sequence where each team gets the ball one possession or two?

[grndmstr_c]

Originally Posted by chumdawg

Yes! This is the part of our discussion that I am finding most stimulating. I think there are several rich areas to mine. For example, let's say that we are tied with two minutes left, or one minute even. Or to put it in our terms, for two or four or some number of "possessions." What are the thresholds in the respective scenarios where we can say with significant certainty that a team must have been a 55/45 or 60/40 favorite rather than a coin flip, for example.

Since you are crunching numbers, let me make sure I am thinking of "possessions" in the same way that you are. Are you calling a sequence where each team gets the ball one possession or two?

I'm calling a sequence where each team gets the ball one possession, though in my simple examples where there are no ties the I'm just using the word possession to refer to the smallest unit of play from which superiority/inferiority can be decided.

While I'm at it a quick correction. I overstated the probabilities of winning 100 and 10 possessions earlier (forgot to hit the square root key). It'd actually be more like 84% and 63%, respectively, using the 55/45 base odds. The 97-98% and 75% figures should be close to the true values if the base odds are 60/40, though (those larger figures would also be applicable with 55/45 base odds and sample sizes of 400 and 40 possessions, respectively).

The range of things that can occur on a single offensive possession in basketball actually makes specifying the distributions for those hypothesis tests your talking about a fairly laborious affair. You'd need to know the probabilities for each type of shot and turnover on the offensive and defensive end for each team, you'd need to know the shooting efficiencies by type of shot on both ends, you'd need to know rebounding probabilities, and you'd need to have some model for how offensive and defensive proficiencies offset one another.

[chumdawg]

Originally Posted by grndmstr_c

I'm calling a sequence where each team gets the ball one possession, though in my simple examples where there are no ties the I'm just using the word possession to refer to the smallest unit of play from which superiority/inferiority can be decided.

While I'm at it a quick correction. I overstated the probabilities of winning 100 and 10 possessions earlier (forgot to hit the square root key). It'd actually be more like 84% and 63%, respectively, using the 55/45 base odds. The 97-98% and 75% figures should be close to the true values if the base odds are 60/40, though (those larger figures would also be applicable with 55/45 base odds and sample sizes of 400 and 40 possessions, respectively).

The range of things that can occur on a single offensive possession in basketball actually makes specifying the distributions for those hypothesis tests your talking about a fairly laborious affair. You'd need to know the probabilities for each type of shot and turnover on the offensive and defensive end for each team, you'd need to know the shooting efficiencies by type of shot on both ends, you'd need to know rebounding probabilities, and you'd need to have some model for how offensive and defensive proficiencies offset one another.

On the "range of things that can occur" point, I would like to be able to neatly sum all those under one category. Something like "clutch-ness." Meaning, a catch-all for all the possible scenarios, as they impact the game.

In other words, if we are going to make the claim that the Mavs have a closing lineup that is closer to 55/45 or 60/40 than it is to a coin flip, we don't especially care *how* they get there, but *that* they get there.

Ten possessions each, if by that we mean that each team gets the ball ten times, is probably overdoing it if we want to counter Hollinger's claim that tight games are a coin flip. We'd probably need to get it down to five times each, or less, that both teams had the ball.

Given the correction you noted, with regard to the 55/45 favorite, I'm not sure how optimistic I am that the same 55/45 favorite would be expected to be that far removed from chance over the course of five possessions (or "trials").

What would the numbers be for a 55/45 team to win a five-possession-or-less game? (I realize that is a different question than "to win a five-possession game.")

[grndmstr_c]

Easiest to get a sense for it by looking at odd-numbered possession sets (since then there are no ties). Counting down by twos from 11 possessions to 3 with base odds equal to 55/45 the winning probability for the better team would be:

11 -> 63.3%
9 -> 62.1%
7 -> 60.8%
5 -> 59.3%
3 -> 57.5%

[chumdawg]

GMC, I think I found the article I was talking about. It's from Dean Oliver. I don't think it has a static link, so you will have to click on the following link and then click on "Articles" from the menu at the top. There are several interesting--and a couple of related--articles on that page, but the specific one I had in mind is called "The Effect of Controlling Tempo."

http://www.rawbw.com/~deano/

[j0Shi]

I had to dig this thread out, because this is simply amazing:

http://insider.espn.go.com/nba/story...ericks-trouble

From a scientific standpoint it's the most horrible piece I've read in a while. I'm flat out embarrassed that a credible statistician comes up with stupidity like that.

Not saying he won't be right at the end of the season, but his lopsided reasoning is beyond awful.

[LonghornDub]

Originally Posted by j0Shi

I had to dig this thread out, because this is simply amazing:

http://insider.espn.go.com/nba/story...ericks-trouble

From a scientific standpoint it's the most horrible piece I've read in a while. I'm flat out embarrassed that a credible statistician comes up with stupidity like that.

Not saying he won't be right at the end of the season, but his lopsided reasoning is beyond awful.

What a clown, and not because he doesn't like the Mavs.

[dude1394]

Originally Posted by j0Shi

I had to dig this thread out, because this is simply amazing:

http://insider.espn.go.com/nba/story...ericks-trouble

From a scientific standpoint it's the most horrible piece I've read in a while. I'm flat out embarrassed that a credible statistician comes up with stupidity like that.

Not saying he won't be right at the end of the season, but his lopsided reasoning is beyond awful.

I can't tell you definitively who will be winning the championship this year, but I can tell you one team that won't be.

I don't see many statistics in that blurb.

[markus1234]

last season Hollinger told us the same BS. And was wrong of course.

Reading ESPN stuff is like believing in "the end of the world" predictions.

[j0Shi]

Originally Posted by dude1394

I don't see many statistics in that blurb.

Google the headline and append "hoopchina"

[LonghornDub]

Fish is railing on Hollinger big time on twitter right now. Not surprising, but pretty funny.

Course, the guy's probably just a little bit sandy because his power rankings were once again nowhere close to predicting the champion last year.

[41creepingdeath]

Originally Posted by dude1394

I don't see many statistics in that blurb.

I know what that means...

.. easy repeat.

[joemoeschmoe]

What a retarded premise he puts forth.

How many of those seasons he uses for comparisons were there 2 weeks to prepare for the season?

Still a fucking idiot is Hollinger.

[bobatundi]

Originally Posted by j0Shi

I had to dig this thread out, because this is simply amazing:

http://insider.espn.go.com/nba/story...ericks-trouble

From a scientific standpoint it's the most horrible piece I've read in a while. I'm flat out embarrassed that a credible statistician comes up with stupidity like that.

Not saying he won't be right at the end of the season, but his lopsided reasoning is beyond awful.

Hollinger is neither credible nor a statistician. He comes up with his own opinions then screws around with numbers until he "proves" them. Real statisticians seek the truth in numbers and should all be irate that Hollinger sullies their field.

[G-Man]

Interesting that in the so-called analysis, the name Lamar Odum just gets mentioned once, and there is no mention of that fact that Carlisle has yet to figure out how to play him, or how to fit in any of his new pieces... or that guys are not in shape, or that they have yet to even learn the new defensive rotations. Instead, he finds one stat---losing at home by 22---and decides from that tea leaf he can see the entire future.

Can someone find and publish here his last power ranking from the previous season? I know you can't find him predicting the Mavs would beat anybody in the playoffs.... he was probably still predicting Miami and LA comebacks after the Mavs had won 3 games.

Hey John. Did you enjoy watching Portland hoist the banner last week? Since LeBron doesn't have a ring for you to kiss, I guess you'll have to just keep kissing his ass.

[bobbyfg7]

Mavs 4.8% chance to win it all:
http://espn.go.com/nba/hollinger/playoffodds

Mavs 7th best:
http://espn.go.com/nba/hollinger/powerrankings

And the Playoffs:
-Picks Portland in 6
http://espn.go.com/nba/playoffs/2011...zers-mavericks
-Picks Lakers in 5
http://espn.go.com/nba/playoffs/2011...vericks-lakers
-Picks Mavs in 7
http://espn.go.com/nba/playoffs/2011...nder-mavericks
-Picks Heat in 6
http://espn.go.com/nba/playoffs/2011...mavericks-heat

[G-Man]

Originally Posted by bobbyfg7

Mavs 4.8% chance to win it all:
http://espn.go.com/nba/hollinger/playoffodds

Mavs 7th best:
http://espn.go.com/nba/hollinger/powerrankings

And the Playoffs:
-Picks Portland in 6
http://espn.go.com/nba/playoffs/2011...zers-mavericks
-Picks Lakers in 5
http://espn.go.com/nba/playoffs/2011...vericks-lakers
-Picks Mavs in 7
http://espn.go.com/nba/playoffs/2011...nder-mavericks
-Picks Heat in 6
http://espn.go.com/nba/playoffs/2011...mavericks-heat

Thanks for the stats.

Based on his most predictive stat, margin of victory, the Mavs were the 8th best playoff team. And were 1/1000th of a point ahead of OKC in the power rankings.

[mavs777]

Hollinger wrote an article about the west favorites, I don't have insider, let me guess he has Portland as the favorite?