When series end: Game 6
by James Mirtle on May 11, 2009 12:18 PM CDT
in Statistics
6 comments
All four of this year's second round series could potentially end in Game 6, with only one of the higher-seeded teams (Detroit) currently holding the 3-2 series advantage. That means that all three of Carolina, Chicago and Pittsburgh can all wrap up their matchups in Game 6 at home.
Between when the NHL went to solely seven game series in 1986-87 and the 2008 playoffs, Game 6 has been the final game of more of the 315 series than any other option:
The basic numbers here? Fifteen per cent of NHL series end in a sweep, 26 per cent end in five games, 34 per cent end in six and 25 per cent go the distance.
Which teams win?
Checking out whowins.com, when the higher-seeded team (in this case Detroit) leads a series 3-2, they have historically won the series 83.5 per cent of the time and ended the series on the road in Game 6 56 per cent of the time.
When the lower-seeded team (in this case Carolina, Chicago and Pittsburgh) leads a series 3-2, they have historically won the series 75.6 per cent of the time and ended the series at home in Game 6 61 per cent of the time.
Depending on how many Game 7s we end up with, this could go down as a heckuva second round.
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I think Carolina is the only team that closes the deal in 6.
Shut up when you're talking to me!
by Afino on May 11, 2009 12:21 PM CDT reply actions 0 recs
I agree with Carolina, but I think the Wings will do it in 6 as well. The Red Wings have actually gotten better and better in each game this series (despite adding rookies in place of injuries), and it’s apparent Anaheim is woefully unmatched and worn out. I suppose they could somehow squeak out another win at home in game 6, but I just don’t see that as likely.
I do think the Caps and Canucks push their series, but believe Chicago and the Penguins will prevail; however, I could see Caps-Pens going either way, and I’m hoping the Capitals pull it out somehow.
by jameshstephenson on May 11, 2009 1:05 PM CDT up reply actions 0 recs
I suppose they could somehow squeak out another win at home in game 6, but I just don’t see that as likely.
Nothing about Anaheim’s postseason so far has been very likely (including the fact that they qualified). Yeah, they’ll be underdogs in G6, but that’s been true for two rounds.
http://www.battleofcali.com/
by Earl Sleek on May 11, 2009 5:22 PM CDT up reply actions 0 recs
Excellent! I was thinking of compiling this data myself, after a discussion with my boss the other day on the merits of 2-3-2 vs 2-2-1-1-1. I was wondering how many series ended in games 4/5 versus 6/7, because it would have a significant effect on revenue earned for the team with home ice advantage. Obviously, there is a little bit more to this than just what game the series ends in, as a series ending in 6 still got the home team a game 5. Interesting data though.
http://wingsvducks.blogspot.com/
Check out the Wings/Ducks coverage over at CycleLikeSedins!
by IAmJoe on May 11, 2009 1:32 PM CDT reply actions 0 recs
Just for comparative/curiosity purposes
I quickly crunched the numbers (wrote out every binary number from 0 to 63 and counted how many numbers it took for 4 zeroes or ones, whichever comes first, to happen) to see what we would expect the slices to look like under completely neutral circumstances i.e. both are 100% even and have a completely equal chance of winning ever game:
Ends in 4 games = 8 out of 64 possibilities = 12.5%
Ends in 5 games = 16 out of 64 = 25%
Ends in 6 games = 20 of 64 = 31.25%
Ends in 7 games = 20 of 64 = 31.25%
PS: I know there’s a formula I could have plugged into rather than drawing out the whole thing, and I probably learned it in my discreet math class back in school, but damned if I remember it, and writing it all out explicitly would be faster than looking it up and re-learning it =P
by Costa24 on May 11, 2009 3:36 PM CDT reply actions 0 recs
You’re looking for the negative binomial distribution. It gives the probability of having X failures of a binomial variable before the rth success. In this case, we are looking for the number of games a team loses before it gets to four wins. We can stop at seven trials, because four losses also means that the series is over. Then multiply all the probabilities by 2.
The pmf for the negative binomial distribution is the number of combinations of the number of successes minus 1 in the number of trials minus 1, times p^r*(1-p)^x, where p is the probability of success for each trial.
In our case, it’s easy, since p=1-p=0.5. For each length of series, using the math of combinations, it’s {(number of games – 1)!/[(number of wins – 1)!!]}0.5^number of games.
What struck me is how close the probabilities of each length of series are to what we would expect from a random draw with each team having a 50% chance to win each game. I would have expected something more divergent. For one thing, each team does not have a 50% chance to win each game*; some teams are clearly better than others, and have a greater chance of winning. This also implies that each game is a random outcome (in the mathematical sense; I don’t mean that the actual play is random), without there being carryover effects from one game to the next. The only place this breaks down is that it appears that the team up 3 games to 2 wins a disproportionate number of games 6. I’d be curious to know the breakdown of games 6 where the series was 2-2 after 4 games, as opposed to being 3-1.
*The outcomes for four and five game series are almost perfect for a p between .58 and .6.
by J. Michael Neal on May 11, 2009 5:22 PM CDT up reply actions 0 recs
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