Probability of a team going 82-0

[Steerpike]

If we're counting OT and SO losses as losses, then this is tractable as a binomial probability of N choose R, where N is the total number of games and R is the number of wins. The formula for calculating this is:
n! / (n-k)! * p^(n-k) * q^k
Assuming wins and losses are equiprobable (which they must be if every game has a winner and a loser), we obtain:
82! / (0!) * 0.5^(0) * 0.5^(82) = 82! *0.5^82 = 9.830304 * 10^-97

If you're not partial to scientific notation, that's a probability of about 0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001; so, basically, ZERO.

Not only what braniac just said, but under the assumption that every game is a 50-50 coin flip you are overthinking this problem to the point of being dead wrong and not making any sense.

If you are given a probability of a win (pW) all you have to do to find the probability of 82 straight wins is pW^82. If pW = 0.5 then this works out to be 2.0679515 * 10^-25.

If you're not partial to scientific notation, that's a probability of about:

0.0000000000000000000000002; so basically, ZERO, but still what... about 2*10^71 times better than you just waffled out there?


Also if you read what I wrote in my posts, this thread was about the probabilities of winning a game that you would need to have a decent chance at going 82-0 and how you would arrive at that sort of chance through scoring goals at a far higher rate than your opponents.

 

[geofff]

If we're counting OT and SO losses as losses, then this is tractable as a binomial probability of N choose R, where N is the total number of games and R is the number of wins. The formula for calculating this is:
n! / (n-k)! * p^(n-k) * q^k
Assuming wins and losses are equiprobable (which they must be if every game has a winner and a loser), we obtain:
82! / (0!) * 0.5^(0) * 0.5^(82) = 82! *0.5^82 = 9.830304 * 10^-97

If you're not partial to scientific notation, that's a probability of about 0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001; so, basically, ZERO.

I think you are way over-thinking this. There's no need for a factorial in this calculation.

Also, 82! *0.5^82 = 9.830304 * 10^97 (NOT -97)

 

[geofff]

So I did the math and realized it's way easier to change it to "go 82 games without a loss" as I don't want to work out what happens in the event of ties.

All you have to do is score 7.8 goals per game while holding your opponent to 1.5 goals per game and you have a 50% chance of going the entire season without a loss.


The amusing thing is that you end up needing a 0.9963 probability of not losing any particular game to end up with this 50% chance. 82 is a lot of games.

.9963^82 = 0.737888... (not just over 0.5)

0.5^(1/82) = 0.99158261187

so that means you would need a 0.9916 chance of winning/not losing each game to have the 50% of doing it over 82 games

 

[me2]

99% change to win 43.86 82-0
98% chance to win 19.07% 82-0
95% chance to win 1.49% 82-0
90% chance to win 0.017% 82-0

 

[Number54]

Overthinking?

The 50-50 coin-flip isn't overthinking the problem, it's making it simpler. If you had to bet on a team winning a game without knowing the two teams involved, your chances of picking the winner in the dark would be 0.5. Since I have no information about team X's actual odds in each game (and really we'd be equally satisfied with ANY team winning 82, so we don't have to specify a team in advance), the best guess is always the average win rate, which must be 0.5 if every game has a winner and a loser.

I should mention, though, if we decided a-priori that a given team had better than 0.5 chance of winning any given game, it would not make much of a difference; for instance, Chicago had a win ratio of 0.75 in the shortened 2012-13 season. If we granted those as their actual odds in each game, then their chances of an 82 win season would be 0.75^82 which is still 0.0000000000569. It might be 15 orders of magnitude better odds, but still represents less than a 1 in a billion shot of going 82-0.

As for my initial calculated probability, my mistake was in writing the formula; i forgot the k! in the denominator:
n!/ [ (n-k)! *k! ] * p^k * q^n-k

You're you're right that this needs no binomial formula, but only because we've set 82 choose 82 as the conditions, so only one permutation satisfies the desired outcome. But, if you wanted to do this for another number of wins < 82, the binomial formula would be necessary. Moreover. for k wins < n, you'd need to take the sum of all binomial distributions from k:n.

As for the claim that I'm overthinking the problem, I completely disagree. I made my original post in maybe 5 minutes. For example, I should also have accounted for the fact that any one of the 30 teams in the league could equally satisfy these conditions (i.e. we didn't specify that we have to pick the team that would win 82 games in advance). However, that problem's not even tractable with a binomial distribution, because the events are non-independent (i.e. any team that's beaten is taken out of the running). So, I basically constrained myself to a problem I could expound on in 5 minutes. Thanks for noting the error in my calculation, though.

 

[Thenameless]

It would go beyond basic statistics.

If a team was even 20-0 at some point, it would become the ultimate goal of every other team in the league to knock that team down. No team would be able to withstand every team in the league bringing their A game for 62 games afterwards.

Agreed. Teams would even start 72Clarking the 20-0 team, in order to give themselves and other teams a better chance in subsequent games to break the goose-egg.

 

[Copmuter*]

Wouldn't shock me if that lineup went undefeated...

What also has to be factored in is that having those players in one lineup now weakens other teams...

Chicago, for example, won't be as tough to play against without Toews

 

[PumpkinBombX]

Gonna go with a p = 1 x 10^-8
or .8 ^ 82 (80% chance of winning, which would be like 80's oiler or 70's habs territory, but both teams we're too coked out during the regular season to care.)

 

[talkinaway]

I did in the opposite direction. If you take p to be the probability of winning, and you want, say, the team to have a 25% chance of winning every game (it's never going to be 100% unless p = 1, and 25% is a meaty enough number), then you just have to solve p^82 = 0.25. That means p is the 82nd root of 0.25, which is 98.3%.

So the team has to be, on "average", about 98.3% to win EVERY game to have even a 1 in 4 shot at being perfect. And average here isn't arithmetic mean, but it's actually pretty close - if they're 97% in the first half of the season, they need to be 99.7% in the second half. It's actually the geometric mean, but with these extreme cases and keeping the percentages somewhat close to 98.3%, the geometric mean and the arithmetic mean are fairly close.

I don't think any single game NHL matchup in history is ever a 49 to 1 lock. I'd bet the underdog every single time. Not to mention that, of course, teams will target the streaking team.

And, for the record, taking the "overall" percentage down from 25% to 1% only takes p down to 94.5%.

 

[bobholly39]

So I'm super intrigued by this thread:

http://hfboards.mandatory.com/showthread.php?t=1717823

For those who didn't see it, the game is to produce a lineup of players that would win all 82 games in a season. To make it simpler, assume that even if you decide to have Crosby on your team, the Penguins still have him as well or are just as good without him.

For example my fictional team looked like:

Ovechkin—Crosby—Malkin
Benn—Seguin—Stamkos
Hall—Getzlaf—Perry
Toews—Kopitar—Bergeron

Suter—Weber
Karlsson—Pietrangelo
Doughty—Keith

Rask
Lundqvist


You could qualitatively debate the merits of a team but I think it's much more interesting to think of this problem mathematically and try to figure out what Goals for and Goals against rates would result in a team having a decent probability of winning all 82 games.


I think the most logical way to approach this is to look at goals for and goals against per game as Poisson distributions to calculate the probability of winning a game. Multiply that by itself 82 times to get the probability of of winning all 82 games.

I'm a habs fan.

I don't think the habs are the "best" team in the NHL, but I do think we have a strong team.

Put our roster up against your roster for 82 games straight. And you know what? I don't think Habs have a winning record against you by season's end.

But I DO think we win some games - likely at least 10, maybe even 20.

Why? All our guys are NHL level guys. Is Crosby a better center than Plekanec? Yes. But can Plekanec shut down Crosby for a game? yes.

Is Rask a better/as good goalie as Price? Sure, they're comparable. But can Price completely LIGHTS OUT outplay Rask in a game? Yes, it happened in the playoffs this past year.

In a given season, your "all star" team won't play the Habs 82x. But just like it's possible for the Habs to have a few great games/while your team has a few horrible games and it translates into a loss, it wuold be just as possible for other NHL teams. So in a course of 82 games, odds are it happens.

If you want to talk Mathematics. Instead of 82 games, if you were to simulate 10,000 games. This is purely mathematical/hypothetical, so don't take into account that in 10,000 games players would age. So let's imagine you play a game, mark down result, use your time macine to go back in time one day and play it again. If you played that game 10,000x, the odds are your "all star" team might have a run where they reach 82 straight wins. The odds of it happening in ONE season (meaning it would start at from game 1 to 82, or game 83 to 164) instead of midway through a season (so say from game 11 to 92, which is the equivalent of 10 games into first season and 10 games into next season) and stretched out onto the next season are even slimmer.

But I suppose it is mathematically possible.

 

[Hardyvan123]

Short answer : 0%
Long answer : Impossible.

Pretty much this every team runs into a hot goalie and it's too hard to compete at an elite level for 82 games as well, never mind the salary cap problems this team would have...lol

 

[adsfan]

The closest that I remember an NHL coming to a perfect season was the Dryden backed Canadiens. IIRC, they had 8 losses and 12 ties during their 80 game season. A .825 winning percentage.

I think the record for consecutive NHL wins is 17 by the Pens.

 

[Finnish your Czech]

The season is non-linear, in that the probability of winning each individual game varies, so that even if the average probability of winning each game is something ridiculous like 99%, it would still be very unlikely that the team would go 82-0

 

[charlie1]

Assuming each game is independent and the team has the same probability of winning each game, then it is just p^82, where p is the probability of winning a game.

If you have some crazily stacked team where p= 0.95, then the probability of winning all 82 is 0.95^82 = 0.015.

Likewise, for a team with p=0.99 it is 0.99^82=0.44.

So if you can dream up a team that has a 99% chance of winning each game then it's completely plausible that they'll finish the season with a perfect record.

People are invoking the binomial distribution here which isn't really necessary since it just simplifies to the above given the OP's question. If we start asking what is the probability of winning 80/82 games then you would need the binomal distribution.