The problem isn't necessarily that the calculation is faulty, it's that the wrong thing is being calculated. What's needed is not a ratio of times it's happened against number of years, it's times happened against total number of series played.
Really, the best way to get something resembling relevant data would be to calculate the number of times a team has come back from 2-0 down (having lost both games at home) compared to the total number of series played in the 16 team playoff, 7 game series era, or the 1-8 conference format. The former would provide more data points, while the later would be more representative.
Exactly, his method calculates the odds if there was only one series each postseason. With the current playoff format there are 15 series each post season.
But to get a real accurate number you really couldn't just look at the total number of series. You would have to go through all the series and only look at only ones where the home team lost the first two, and compare the number of times they won the series vs losing it. Any series that didn't start that way is irrelevant data.
Exactly, his method calculates the odds if there was only one series each postseason. With the current playoff format there are 15 series each post season.
But to get a real accurate number you really couldn't just look at the total number of series. You would have to go through all the series and only look at only ones where the home team lost the first two, and compare the number of times they won the series vs losing it. Any series that didn't start that way is irrelevant data.
Yeah this is absolutely correct, don't know why I said total number of series.
The problem isn't necessarily that the calculation is faulty, it's that the wrong thing is being calculated. What's needed is not a ratio of times it's happened against number of years, it's times happened against total number of series played.
Really, the best way to get something resembling relevant data would be to calculate the number of times a team has come back from 2-0 down (having lost both games at home) compared to the total number of series played in the 16 team playoff, 7 game series era, or the 1-8 conference format. The former would provide more data points, while the later would be more representative.
Instead of doing all series, you would actually just take the series that went 2-0 against the home team and how many of those came back to win the series. If you take all the series, you're including the probability that the teams are put in this situation, which doesn't really help as they are already there.
So, going by that. There's a 25% chance we can do it and a 75% chance we won't.
No Shrimper. 18 times in 75 years. I'm not going to do the math as I am far too lazy to account for different team #s and all that but currently there are 15 playoff series a YEAR.
__________________ Shoot me a PM with your concerns.
Instead of doing all series, you would actually just take the series that went 2-0 against the home team and how many of those came back to win the series. If you take all the series, you're including the probability that the teams are put in this situation, which doesn't really help as they are already there.
There's an easy solution to this question. Have pnep pull up a table of all the times a series has gone 0-2 against the home team, divide 18 from that number, and voila we have our probability.
Instead of doing all series, you would actually just take the series that went 2-0 against the home team and how many of those came back to win the series. If you take all the series, you're including the probability that the teams are put in this situation, which doesn't really help as they are already there.
Absolutely, I noticed I had erred after someone pointed it out. The way I set it up (using all series), you would be computing:
"the historical probability that a series, beginning at 0-0, would result in the road team winning the first 2 games, then losing the series"
while the more relevant probability, as you and others have pointed out, is
"the historical conditional probability that a series which is already 2-0 for the team that started on the road results in that team losing the series"
So teams are 40-9 in the 2nd round or later after winning the first 2 games?
That doesn't shocked me. Past the first round one would expect teams to be more evenly matched, so a 2-0 series lead (i.e. needing to only win 2 out of the next 5) is a huge advantage.
So teams are 40-9 in the 2nd round or later after winning the first 2 games?
Road teams are 40-9, yes. 16-6 in the quarterfinals, 15-1 in the semfinals, and 9-2 in the Cup. That kind of makes sense, as you get deeper, the two teams are likely to be more evenly matched.
Ignoring location, teams up 2-0 are:
273-43 in all rounds
79-15 in the 1st round
83-16 in the quarterfinals
69-7 in the semifinals
42-5 in the Cup
Exactly, his method calculates the odds if there was only one series each postseason. With the current playoff format there are 15 series each post season.
But to get a real accurate number you really couldn't just look at the total number of series. You would have to go through all the series and only look at only ones where the home team lost the first two, and compare the number of times they won the series vs losing it. Any series that didn't start that way is irrelevant data.
Coincidentally, if you do the calculation properly it still comes out to 25%.
One "correct" way to do this is to assume some probability 'p' that Vancouver wins each game, and then do a binomial distribution.
Vancouver needs to either win the next 4 straight, or take 3 of the next 4 and then win game 7.
P(Van win) = (p^4) + (4c3)(p^3)(1-p)x(p)
Obviously this is an approximation.
edit: for those interested, it comes out to 18.75% for fair games (ie: p=1/2)
edit2: lol! 18.75% is a really weird coincidence
edit3: Wolfram Alpha simple calcs: p as variable p = 0.5
Last edited by Unaffiliated: 04-14-2012 at 10:50 AM.
One "correct" way to do this is to assume some probability 'p' that Vancouver wins each game, and then do a binomial distribution.
Vancouver needs to either win the next 4 straight, or take 3 of the next 4 and then win game 7.
P(Van win) = (p^4) + (4c3)(p^3)(1-p)x(p)
Obviously this is an approximation.
I'd be interested to see a study done on the degree to which game-to-game results are independent (since of course that's a basic assumption of any binomially distributed process).
I'd be interested to see a study done on the degree to which game-to-game results are independent (since of course that's a basic assumption of any binomially distributed process).
Yeah, good point.
If not, then maybe a Markov Chain would be better. I need to study those for next week
these are the important numbers for the Canucks and Pens.
Also very important is the number of games that the Canucks will ahve Daniel Sedin which is zero. No doubt Duncan Keith will not be a favorite player in his home province for a while.
the Pens have won the SC recently and have a better chance at beating the flyers but it's still a long hill to climb for them.
I'd be interested to see a study done on the degree to which game-to-game results are independent (since of course that's a basic assumption of any binomially distributed process).
With only a few thousand playoff games, that might be hard to determine. I do want to build a database of playoff game results at some point. Maybe I'll start that tonight while I'm watching the "boring" series that aren't Flyers-Pens.
Since I started watching hockey around 1992, I can remember the Habs winning the first 2 games of a series on the road three times. They ended up losing all 3 series.
