Categories: Sports

Hockey Playoffs: The Game 3 fallacy

Contrary to popular belief, there is no correlation between a Game 3 win and a series victory

by Charlie Gillis on Wednesday, May 6, 2009 5:47pm - 24 Comments

crosbyIt’s right there in the Big Book of Hockey Clichés: “pivotal Game 3.” Type the phrase into Google and you get 25,000 hits—starting with Tuesday night’s contest between the Detroit Red Wings and Anaheim Ducks. Game 3, we’re reminded each year, is the watershed match of a seven-game playoff series. What began as accepted wisdom is now an article of faith.

Too bad it’s wrong.

Truth is, simple math illustrates that a win in Game 3 confers no more edge than you’d expect of any playoff victory—maybe even less. In 300 series over the last 20 NHL playoff seasons, only 191 teams who won Game 3 went on to clinch the set. That’s about 64 per cent, which is not statistically significant. Bear in mind that any team that triumphs in a series has won four times in seven or fewer games. There is a better-than-even chance, then, that one of those victories would have come in Game 3. Bottom line: there is no special correlation between a win in that third outing and a series victory.

Hard to believe, because for some reason Game 3s always feel important. If a team is down by two games, it must—must—win to keep its hopes alive. If it is up by two games, a Game 3 victory puts it on easy street, with a fat, three-game cushion and little likelihood of losing. This much is born out by history: only two NHL teams have ever fought back from a 3-0 deficit to win a series (the New York Islanders did it in 1975, and the Toronto Maple Leafs in 1942). Finally, if the two sides are deadlocked at one win apiece, the third game takes on truly outsized proportions. The loser of Game 3 must then win three out of the remaining four games to prevail, which sounds next to insurmountable.

Yet you’d be surprised how often that happens. Fully 52, or one third, of the teams who found themselves in that jam over the last 20 seasons have clawed their way out. You’d think that more than 66 per cent of teams who won those Game 3s, going ahead 2-1 in a seven-game contest, would win. But there again, the numbers don’t lie. A win in Game 3 is just a win—nothing more, nothing less.

The lesson for coaches and players in all this is obvious: no game counts more than another in the playoffs—short of the one that wins or loses it for you. As for fans and media, there’s no harm in ramping up the hype. When Alex Ovechkin and Sidney Crosby line up tonight for Game 3 in Pittsburgh, it may be nothing more than another playoff game. But in our minds, it will be as pivotal as we care to make it.

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  • sf

    There are a lot of fallacies in sports like this one. Another one is “the dreaded 2-0 lead”. Or the “momentum” fallacy. Or the “leadership in the dressing room” fallacy. Or the “experience” fallacy, that a team needs experienced players in order to win in the playoffs.

    Statistics would debunk every one of these, in my opinion.

  • madeyoulook

    That’s a little over 64 per cent, which is not statistically significant. Bear in mind that any team that triumphs in a series has won four times in seven or fewer games. There is a better-than-even chance, then, that one of those victories would have come in Game 3.

    But Charlie, they are not yet a series winner by Game 3. Since neither team is yet the four-game-winner, you cannot yet retrospectively dismiss the importance of any game, because you don’t yet have the retrospectoscope.

    Oh, and “[t]hat’s a little over 64 per cent, which is not statistically significant” sounded pretty fishy to me. Unless I made an error in Excel, using the binomial distribution, 191 successes (series champions) in 300 trials (game 3 winners trying to win the series) is VERY SIGNIFICANT (probability of 191 or more successes out of 300 even-chance trials is only 1.27313E-06, or 0.00013%). So, before I accuse you of incorrectly declaring non-significance, I would invite you to check your numbers (and have real statisticians wade in here, too).

    I think you’ve blown this one. If I’ve got my math right, the 50-50 contribution of a Game 3 victory has been sufficiently disturbed by the data.

    Anyone?

    • xiv

      Given a 50/50 result of any game, a team with a 2-1 lead in a series should win the series 69% of the time. and a team with a 3-0 lead in the series should win the series 94% of the time. a team down 1-2 would be 31%.

      There is a 25% chance that the team that won the 3rd game is up 3-0. (www)
      There is a 25% chance that the team that won the 3rd game is down 1-2. (llw)
      There is a 50% chance that the team that won the 3rd game is up 2-1. (wlw or lww)

      combining these two sets of statistics:
      (.25)(.94)+(.25)(.31)+(.50)(.69) = 66% chance that the team that won the 3rd game will win the series.

      making a 64% occurrence of that not all that significant

      • madeyoulook

        Thanks, xiv, but I think your 25-25-50 distribution is an unnecessary assumption. Because we actually already have the 300 series in question available for study. Also, you are applying a 50-50 probability of future victories for each game from game 4 on, with no rational basis for that assignment. In effect you appear to be taking many it’s-all-random assumptions to plug into a calculation to demonstrate that it’s all random.

        The question is: does a game 3 victory lead to a better-than-even chance of a series victory or not? Under the null hypothesis (no contribution), the expected series outcome of 300 game-3 victories would be 150-150. Minor variations from 150-150 would not let us reject that null hypothesis with any confidence. Big-deal variations from 150-150 would give us confidence to reject the null hypothesis. The actual result (quoting Charlie) is 191-109. My math says that is a huge variation away from the null hypothesis’ 150-150, and I (hopefully correctly) applied the binomial distribution to so demonstrate.

        Anyone else want to throw themselves at this?

      • xiv

        To clarify, from my last post similar calculations can be run to show that with 50/50 results, a fixed condition of a win in any one of game 1, 2, 3, and 4 all statistically provide a 66% chance to win the series, making game 3 no more special than any of the other first 4. Haven’t run the numbers for optional games.

        • xiv

          I can’t make any predictive values better than 50/50 though. Assigning odds to the winner of a sporting event is more of a witchdoctoring exercise than you’d think. Does a game 3 victory contribute to winning chances of a series, yes. But no more than would statistically expect if each game were a coin flip. You have to bear in mind if say the odds were 70/30 instead of 50/50 the game 3 winner also is more likely to be the 70 than the 30 in the first place. If you had a table of % of series winners who won each of the first 4 games you’d be able to see which if any deviate furthest from the expected 66%. The 25/25/50 distribution rises naturally from the 50/50 assumption. I don’t have the tables the original author did ;) All i can say is if nothing about the actual teams is taken into account, and a 50/50 chance is the best presumption then there will be a 66% chance that the game X winner will win the series. Where X is a number < 4.

  • madeyoulook

    Replying to my own post has led to moderation purgatory, so I will add the major important comment that the 191 championsips out of 300 game-3 winners is actually a hair less than sixty-four per cent.

    And I will repeat my request for someone to check the numbers.

  • Sean Stokholm

    It’s been a while since I’ve watched hockey, but isn’t Game 3 the one where the venue changes from one team’s city to another? I’d be interested to know the breakdown of those 191 wins – how many were for the home team, and how many for the visitors, and if that factor is of greater predictive value.

    • madeyoulook

      Yes, the venue switches for Game 3. But the team with “home ice advantage” (another one statistics might play with), that is also the more favoured team in the standings, hosted the first two.

      FWIW, if the visiting team wins one of the first two, “home ice advantage” is stolen from the team hosting four of the seven, since the series now becomes a best-of-five with three of those remaining five in the other city.

      So, you are combining the weaker team and home ice advantage in Game 3. The data is the data, but good luck interpreting which is contributing what.

  • madeyoulook

    I suppose my binomial distribution fails logically in that we already know that the game three winner has won at least one game, and so is at least “on the way” to four victories out of seven.

    So the better way to go about this, I suppose, would be to compare the “success rate” of 300 Game-1 winners with the success rate of Game-2 winners, with the success rate of Game-n winners (n=1,2,…,6).

    But I don’t have that data. However, I still don’t follow the logic of adding far more random assumptions into any calculations, as xiv has done.

  • Stephen

    I seem to recall that, when factoring sweeps out of the equation, Game 4 winner was actually a more reliable predictor of the eventual series winner then Game 3 winner.

    Admittedly, this becomes less useful as a series goes on (the Game 7 winner correctly predicts the winner of a series an astounding 100% of the time! :) ), but it also kind of holds up from a logical perspective. Winning and losing Game 4 is, again discounting sweeps, the difference between 2-2 and 3-1. Whereas whenever a Game 3 is played in a 1-1 series, losing the game carries little consequence aside from the dreaded “momentum.”

  • Matt

    Agree that the writer’s analysis is faulty. There is a statistical difference. The null hypothesis is that p=0.5 (vs. alternate hypothesis of p>0.5). With 191 wins out of sample size of 300, the numbers with 95% confidence are:

    Test of p = 0.5 vs p > 0.5

    95%
    Lower
    Wins Series Sample p Bound
    191 300 0.637 0.588

    That means, ON AVERAGE, I can say with 95% confidence that the winner of game 3 will win the series AT LEAST 58.8% of the time. Sounds significant to me!

    • Matt

      Format didn’t come out as intended. If I run a 1-Proportion test with 95% confidence it says:
      Wins = 191
      Series = 300
      sample p = 63.7%
      95% lower bound (for entire population) = 58.8%

      • madeyoulook

        Thanks, Matt, except that the 95% confidence interval you calculated (correctly, I will assume) would apply if the observations are a random sample of the entire population. But we have the data of the entire population of 300 series already. I am not sure that particular calculation would therefore apply. No?

        And Charlie, if you’re following this, could I ask you to share the data, if available, on the success rates (i.e. 191/300 for game 3 victors) for the victors of each of the six games? I suspect the definitive answer lies in the proper study of the variance between the 63.6667% rate for Game 3 and the success rates for all the others. Thanks!

  • madeyoulook

    This probably violates all sorts of net-iquette, and so I apologize. But I bring this web page back into awareness this evening given my selfish urge to have a real stats-math geek help clarify the debate over the statistically (dis?)proven hypothesis that a Game 3 victory provides a significant contribution to best-of-seven series success in the last 300 NHL playoff series. It also comes with a repeat request to Charlie to provide, if available, the “success rate” associated with victory in each of the other first six games. Data! we need more data!

    If there is no response to this appeal, I cannot promise not to harangue the Blog Central commentariat in future. Yes, I can be that selfish sometimes.

    Sorry and thank you.

  • madeyoulook

    A propos of nothing, but: Have you ever posted a comment in hopes it (and the associated thread) catches people’s attention, only to watch the notice of it sail through the five-comment list in the window, and nobody bites?

    No? Never happened to you? Oh. I was just wondering. Thanks.

  • madeyoulook

    Psst! Sleepless math geeks! Don’t count sheep! Ponder the little problem above, and share your wisdom!

  • madeyoulook

    Good Friday morning Blog Central folks! Anyone with a penchant for statistics care to take a stab at the discussion above?

    If Charlie Gillis is on duty, any chance the series results associated with victories of the other games are available?

    If only to shut MYL up?

    Anyone?

  • madeyoulook

    Charlie, make me stop! Please! Do you have the data for the other five games?

    Anyone else with a math-y brain! Make me stop! Give us a stats lesson on how to determine significane above. Pretty please?

  • madeyoulook

    OK, the begging continues. That’s alright; Wells hasn’t called me pathetic in a while, so I’d like to give him a legitimate opportunity this time.

    Eleventieth call for any math whiz to apply him or herself to the above discussion, and for Charlie to share more data if available.

    This is a recording…

  • madeyoulook

    OK, last appeal, now that I have monopolized this thread intermittently for over a day now.

    Math geeks, and Charlie, PLEASE indulge my whining with a response to the above discussions.

    Thanks!

    • washedphenom

      I stumbled upon this discussion when looking for hockey scores tonight, and I happen to be a statistician. Its late, but I’ll give it my best shot. I don’t think I”ve made any errors, but feel free to correct. I think the key is asking the right question. It is not whether game #3 winners win the series significantly greater than 50% of the time. Given that they are game #3 winners, they ALREADY are winning more than half the series, because they have one game already counting toward the 4.

      So calculating combinations, the odds of winning 3 or more of the other six games (game 1,2, 4, 5, 6, 7) is equal to .656, extremely close to the .637 figure from the data (191/300). Thus, to test whether it really deviates from chance, you need to test the .637 figure against the chance figure of .656, not .500.

      The article’s conclusion is absolutely correct. Do game #3 winners tend to win series? Yep, because its a win! But winning #3 is no more important than winning any other game. In fact, looking at the data, it seems like winning game #3 gives you a slighty LESS chance of winning the series (.637), versus chance alone (.656)!!! Clearly non-significant, but amusing!

      • madeyoulook

        Thank you, washedphenom. Indeed, I poo-poo’d my own initial binomial distribution test, for the reason you give. As a statistician, though, would you not agree that the BEST test would be to compare the “success rate” associated with Game 1 victories, Game 2 victories, and so on, up to Game 6? Especially if we are to look at the “pivotal” all-important Game 3?

        Although I suppose any Game at 4 and beyond will be inflated, since there is a chance that’s the final game of the series, and the victor automatically becomes the champion.

  • http://www.lookyoungatlanta.com botox Atlanta

    Thanks for sharing those numbers. I really appreciate it. The important thing would be winning all the games of the season.

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