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FrontRange
Joined: 27 Jan 2005
Posts: 131
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 Posted: Fri May 13, 2005 9:46 am Post subject: Series Stats |
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Saw this on ESPN this morning:
In NBA history, when a best-of-7 series is tied 1-1, the Game 3 winner has gone on to win the series 110 of 144 times (76.4 percent).
And is always ticks me off when a stat is given more weight that its worth, so I immdiately wondered what the break-down was for any given game predictating a series winner . . .since by defination winning any game in a seven game series should significantly increase the odds of winner the series. My guess is that since most series don't go 7 games that 65%+ would be about right on any given game (of course, Game 7 would be 100%  ) |
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schtevie
Joined: 18 Apr 2005
Posts: 408
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| Posted: Mon May 16, 2005 8:37 am Post subject: |
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| Assuming the stat was given straight up, it seems to me that it would be a reasonable fact to relate. As a reference, if one assumes that playoff teams are equal in strength, where any team has a 50% chance of winning any remaining game, then one expects a team ahead 2 to 1 to win 11 out of 16 series - or 68.75%. So what needs to be "explained" is the extra 7.65% of series victories. And given the fact that the third game winner is likely the better team (ex ante) hence the 50% estimate is low and the fact that the better team has the homecourt advantage, there ain't that much splainin' to do. (For example, if the 3rd game winner is assumed to be a consistent 55% favorite, all else equal, then it will win 75.9% of series given the 3 first game scenario.) |
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Mike G
Joined: 14 Jan 2005
Posts: 3564
Location: Hendersonville, NC
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| Posted: Mon May 16, 2005 10:26 am Post subject: |
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| schtevie wrote: | | ... if the 3rd game winner is assumed to be a consistent 55% favorite, all else equal, then it will win 75.9% of series given the 3 first game scenario.) |
In fact, the first 2 games are at the home of the better (W-L%) team, so if that better team wins game 3, they've regained home court advantage and are "in control" of the series.
If, however, the underdog team has won one on the road and then wins game 3 at home, they are the clear favorite. (At this point, we may reconsider who was really the favorite to begin with, as there may be some matchup advantage or injury issues, etc.)
If you break down these 3-game standings into "favorite up 2-1" and "underdog up 2-1", I'd guess the former is 78% likely to win the series, and the latter 74%, or some such thing. |
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Mike G
Joined: 14 Jan 2005
Posts: 3564
Location: Hendersonville, NC
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| Posted: Sun May 29, 2005 1:12 pm Post subject: |
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"Miami practically had been written off after the Game 1 loss, and Wade's performance provided a huge lift..."
So wrote Steve Kerr a couple of days ago.
Did anyone actually read such a gloomy future for the Heats, as they went down 0-1 ?
Any series that is 1-0, 1-1, 2-1, 2-2, 3-2, or 3-3 could also be said to be "as close as could possibly be" after X number of games. |
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KD
Joined: 30 Jan 2005
Posts: 163
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| Posted: Sun May 29, 2005 4:39 pm Post subject: |
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It always cracks me up when Barkley and Kenny Smith put their heads together on TNT and try to come up with the "most important" game of a seven-game series.
"See, I say that the third game is the most important game of the series, followed by the fifth -- because usually when you win a fifth, the other team wins the sixth, and you take in it seven -- but the third always has ..."
... and on and on. What bollocks. |
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Ed Küpfer
Joined: 30 Dec 2004
Posts: 785
Location: Toronto
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| Posted: Sun May 29, 2005 11:12 pm Post subject: |
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| KD wrote: | It always cracks me up when Barkley and Kenny Smith put their heads together on TNT and try to come up with the "most important" game of a seven-game series.
"See, I say that the third game is the most important game of the series, followed by the fifth -- because usually when you win a fifth, the other team wins the sixth, and you take in it seven -- but the third always has ..."
... and on and on. What bollocks. |
Hey, I just wrote a usenet message on this very topic, replying to someone who said that the Spurs/Suns Game 3 was the most important game of the year for Steve Nash. I am a big fan of self-quotation:
| I wrote: | You can define the importance of a game by calculating the effect of its result on the probability of the team winning the entire series. If you do this (call the effect of the result its Volatility), then Game 3 down 0-2 has the third lowest volatility of all:
| Code: |
Probability of HCA
team winning series
if game ends in
Series WIN LOSS Volatility
3-3 1.00 0.00 1.00
2-3 0.60 0.00 0.60
2-2 0.76 0.24 0.52
2-1 0.86 0.45 0.41
3-2 1.00 0.60 0.40
1-1 0.61 0.24 0.37
1-2 0.45 0.10 0.35
1-0 0.73 0.39 0.35
0-0 0.59 0.28 0.31
2-0 0.91 0.61 0.30
0-1 0.39 0.12 0.27
3-1 1.00 0.76 0.24
1-3 0.24 0.00 0.24
0-2 0.24 0.04 0.20
3-0 1.00 0.86 0.14
0-3 0.10 0.00 0.10 |
The games at the bottom are the ones whose outcomes are least in doubt. By this measure, the "most important" games are Game 7s, followed by Game 5s with the team with HCA down 2-3, followed by Game 4s with both teams tied. (The games above assume two equally matched teams, shown from the perspective of the team with home court advantage, and HCA is set to 60/40.) |
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KD
Joined: 30 Jan 2005
Posts: 163
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| Posted: Sun May 29, 2005 11:29 pm Post subject: |
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| Nice. Thing is, this thread is bursting with merit and solid info. Kenny and Charles just pull it out of their arse. Like when their top-5 player list always ends up with 13 players on it. |
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Mike G
Joined: 14 Jan 2005
Posts: 3564
Location: Hendersonville, NC
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| Posted: Mon May 30, 2005 10:14 am Post subject: |
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Ed,
Are the odds really 60/40 for the home team? Is that based on playoffs, season, or what? You'd have to know the teams are in fact "evenly matched", wouldn't you?
Then how do the "volatilities" change when you try other variables (say 53/47)? |
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Ed Küpfer
Joined: 30 Dec 2004
Posts: 785
Location: Toronto
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| Posted: Mon May 30, 2005 12:43 pm Post subject: |
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| Mike G wrote: | | Are the odds really 60/40 for the home team? Is that based on playoffs, season, or what? You'd have to know the teams are in fact "evenly matched", wouldn't you? |
The lower bound is 50/50. The upper bound is something like 65/35, which is the highest HCA in any NBA season. A good guess is 60/40. I did a study once on this. By using multiple regression you can estimate the effect of each variable (team 1 strength, team 2 strength, HCA strenght) on the outcome of the game, and I came up with something like 58/42 for HCA strength.
| Mike G wrote: | | Then how do the "volatilities" change when you try other variables (say 53/47)? |
Maybe this will answer that question:
The Volatilities change, but the order of the Important Games stay the same -- except for one: team with HCA up 3-2. As the strength of HCA increases, the Volatility decrease quite a bit. Volatility measures by how much a series is "over." Let's use Detroit and Miami as examples: if the Heat were up 3-0, the series is almost surely "over," and hence Game 4 has a low Volatility. Now, if the Heat were up 3-2, they'd be playing Game 6 in Detroit. If HCA is extreme, Game 6 is almost detetrmined: the Pistons win at home. The series then goes to Miami, where the game is equally determined: Miami wins, and the series is over. If Miami wins Game 6, the series is over. If they lose, they go back to Miami, where HCA assures the Heat of a great chance of winning, and so the series is almost over. If there is no HCA (ie 50/50) then Game 6 in Detroit could just as easily go to the Heat as the Pistons, and if the Pistons manage to win, Game 7 in Miami is just as much of a question mark. With no HCA, if the Heat win Game 6, the series is over. If they lost, the series is up for grabs. Volatility measures the distance betweeen "Series over" and "up for grabs."
HCA team down 2-3 changes a bit too, but in the opposite direction. Using the same teams above, if the Heat are down 2-3, and they win Game 6, the chances of winning Game 7 depend on the strength of HCA. If it is weak, the series is up for grabs. If it is strong, the have a great chance of winning the series. However, if they lose Game 6, they have zero chance of winning the series. Volatility here is higher when HCA is strong because that is the greatest distance between "zero chance" and the chance of the Heat winning Game 7, which is stronger with a higher HCA,
None of the others change all that much.
Here is a table of the numbers used in my graph:
| Code: | HCA 0.50 0.52 0.54 0.56 0.58 0.60 0.62 0.64 0.66
0-0 0.31 0.31 0.31 0.31 0.31 0.32 0.32 0.32 0.33
0-1 0.31 0.31 0.31 0.31 0.31 0.31 0.30 0.30 0.30
0-2 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25
0-3 0.12 0.13 0.13 0.13 0.14 0.14 0.14 0.14 0.15
1-0 0.31 0.31 0.32 0.32 0.32 0.33 0.33 0.34 0.35
1-1 0.37 0.37 0.38 0.37 0.38 0.38 0.39 0.39 0.40
1-2 0.37 0.37 0.38 0.39 0.39 0.40 0.41 0.42 0.42
1-3 0.25 0.25 0.25 0.25 0.25 0.24 0.24 0.24 0.23
2-0 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25
2-1 0.38 0.38 0.37 0.37 0.36 0.35 0.35 0.35 0.34
2-2 0.50 0.50 0.50 0.50 0.51 0.51 0.52 0.53 0.54
2-3 0.50 0.50 0.52 0.54 0.56 0.58 0.60 0.62 0.64
3-0 0.13 0.12 0.12 0.11 0.11 0.10 0.10 0.09 0.08
3-1 0.25 0.25 0.25 0.25 0.25 0.24 0.24 0.23 0.23
3-2 0.50 0.50 0.48 0.46 0.44 0.42 0.40 0.38 0.36
3-3 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 |
I'll note here that those numbers came from simulating the various series -- I didn't calculate the actual conditional probabilities for each possbile pathway. That would be an enormous pain the in ass. What I'm saying is that I might be 1% or 2% off from the actual numbers, but not more than that.
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gabefarkas
Joined: 31 Dec 2004
Posts: 1313
Location: Durham, NC
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| Posted: Mon May 30, 2005 2:12 pm Post subject: |
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Ed -- phenomenal, yet again.
One question though: if the series will end with either one team or another winning, how come the probabilities reading across don't sum to 1? |
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Ed Küpfer
Joined: 30 Dec 2004
Posts: 785
Location: Toronto
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| Posted: Mon May 30, 2005 3:29 pm Post subject: |
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| gabefarkas wrote: | | One question though: if the series will end with either one team or another winning, how come the probabilities reading across don't sum to 1? |
Because I'm showing conditional probablilities, and the conditions aren't independant and exclusive.
Yark. I don't know the terminology. Lemme try again:
Take the 0-0 row.
1 = p(Win series following game win) + p(win series following game loss) + p(Lose series following game win) + p(Lose series following game loss)
The four events on the right side of the equation represent every possible way the series can play out. Either the team wins the first game or it loses. Given that, the Team either wins the series or it loses. That's four different ways:
| Code: |
+--- Win Series [wW]
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+-- Win --+
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| +-- Lose Series [wL]
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Game 1 -+
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| +-- Win Series [lW]
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+-- Lose -+
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+-- Lose Series [lL]
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But in my table, I only show [wW] and [lW]. You'd need [wL] and [lL] to sum to 1.
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Ed Küpfer
Joined: 30 Dec 2004
Posts: 785
Location: Toronto
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| Posted: Tue May 31, 2005 12:50 am Post subject: |
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| Ed Küpfer wrote: | | By using multiple regression you can estimate the effect of each variable (team 1 strength, team 2 strength, HCA strenght) on the outcome of the game, and I came up with something like 58/42 for HCA strength. |
I did this again, just to make sure, using every playoff game since 1974 in my sample (n=2017), and got a HCA of 62%, although it was not statistically significant (itself a surprise, at least to me).
| Code: | Logistic Regression Table
95% CI
Predictor Coef SE Coef Z P Odds Ratio Lower Upper
Constant 0.620857 0.481758 1.29 0.197
HW% 4.66362 0.559507 8.34 0.000 106.02 35.41 317.43
AW% -4.61665 0.561156 -8.23 0.000 0.01 0.00 0.03
Log-Likelihood = -1224.522
Test that all slopes are zero: G = 141.141, DF = 2, P-Value = 0.000 |
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gabefarkas
Joined: 31 Dec 2004
Posts: 1313
Location: Durham, NC
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| Posted: Tue May 31, 2005 4:22 pm Post subject: |
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| gotcha. i didn't catch that they were conditional probabilities. the terminology would be P(A|B) = {probability of event A occuring, given that event B occurs} . |
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