Chance of repeating 5 goals per game with these underlying stats

[gwh]

We are discussing in the Laine 5 goal game thread about the probability of Laine repeating the 5 goal effort during his career. Nobody has done this since Lemiux in the turn of the 90s.

However long time since high school and can't do this simple math anymore:

This year he has

5 shots in 5 games. (18.6 games per 82 season)
6 shots in 3 games (11.2 games per 82 season)
7 shots in 2 games (7.5 games per 82 season)

Most likely his production drops off after 15 seasons. He has been playing in 95.1% of the available games, making his season length 78 games. His maximum productive games available is 1170 assuming no major injuries.

5 shots in 269 career games
6 shots in 162 career games
7 shots in 108 career games

His career shooting is 18.3% so far.

What is Laine's chance to score 5 or more goals in his career? I am getting 5.5% for 5 shots, 16.9% for 6 shots and a whopping 31.1% for 7 shots.

This totals the chance to repeat during 15 year career at 53%. (which i am quite sure is incorrectly high, due 20 years departure from the math classes)

Anyone with actual math skills willing to give Laine's repeat chance a proper go?

Click to read more...

 

[morehockeystats]

Do you realize that with your assumption of shooting % of 18.3% and the shot volume only in these 539 games he's going to score 562 goals? He'll probably score some 250 in the remaining 631, and he already has what, 99. That means he'll be at 911 goals after what, 1400 career games? With 4-5 years more to go at lesser pace? No wonder 2 5-goal games become a good probability.

The math of the probabilities is correct, the extrapolation of the shp% and shot volume looks fishy.

 

[Ciao]

Yeah, there's a bit more complicated math than taking the total number of games in which he's expected to get five or more shots and multiplying that by his expected shooting percentage.

For example, when calculating the probability of coincident events you use the product of the probability of each event, not the sum of the events.

For example, if the brakes on you car were predicted to fail once in every 100,000 stops and the emergency brakes were also predicted to fail once in every 10,000 stops, then the probability of both sets failing at the same time would be one ten-thousandth of one one-hundred-thousandth, which is a very small number.

Similarly, if a hockey player who scores on 20% of his shots takes exactly five shots in a game, the likelihood of scoring on all five shots in that game would be one fifth of one fifth of one fifth of one fifth of one fifth. I'm not doing the math, but this is a very small number.

It's much more complicated than this, but I think it's pretty unlikely to happen again any time soon.

 

[sr edler]

I'll say there's probably a bigger probability if he doesn't try to round out his game, to a more balanced production. The guy can shoot, obviously.

 

[morehockeystats]

Yeah, there's a bit more complicated math than taking the total number of games in which he's expected to get five or more shots and multiplying that by his expected shooting percentage.

For example, when calculating the probability of coincident events you use the product of the probability of each event, not the sum of the events.

For example, if the brakes on you car were predicted to fail once in every 100,000 stops and the emergency brakes were also predicted to fail once in every 10,000 stops, then the probability of both sets failing at the same time would be one ten-thousandth of one one-hundred-thousandth, which is a very small number.

Similarly, if a hockey player who scores on 20% of his shots takes exactly five shots in a game, the likelihood of scoring on all five shots in that game would be one fifth of one fifth of one fifth of one fifth of one fifth. I'm not doing the math, but this is a very small number.

It's much more complicated than this, but I think it's pretty unlikely to happen again any time soon.

0.2^5 = 0.00032, or 0.032%. Multiply it by 269 games with 5 shots, you get 0.086, or 8.6%
Getting an event to happen with probability of 50% in a game over a span of 500 games requires only 0.1% probability of it to happen in a single game.

 

[Dooman]

0.2^5 = 0.00032, or 0.032%. Multiply it by 269 games with 5 shots, you get 0.086, or 8.6%
Getting an event to happen with probability of 50% in a game over a span of 500 games requires only 0.1% probability of it to happen in a single game.

That math can't be right. That means at a 0.2% probability in a single game would have a 100% probability over 500 games.

 

[DudeWhereIsMakar]

I'll need a VERY... heavy diaper if that's going to happen, if anybody catches my drift.

 

[morehockeystats]

That math can't be right. That means at a 0.2% probability in a single game would have a 100% probability over 500 games.

Yeah, my bad. But not off by much.
0.1% over 500 games means 39.4% it's going to happen. The math is (1-p)^N, i.e. (1-0.001)^500.
0.2% over 500 games would mean 63.3% it's going to happen. The breakeven is around 0.14%.

 

[Dan Kelly]

i'm just glad that Safeway had to pay out the $1 million dollars to a fan after Laine got his 5th goals....

 

[cupface52]

That math can't be right. That means at a 0.2% probability in a single game would have a 100% probability over 500 games.

If my math is right, it'll happen once every 3125 games. Scoring 5 goals on 5 shots with a 20% shooting average.

 

[gwh]

Do you realize that with your assumption of shooting % of 18.3% and the shot volume only in these 539 games he's going to score 562 goals? He'll probably score some 250 in the remaining 631, and he already has what, 99. That means he'll be at 911 goals after what, 1400 career games? With 4-5 years more to go at lesser pace? No wonder 2 5-goal games become a good probability.

The math of the probabilities is correct, the extrapolation of the shp% and shot volume looks fishy.

The shooting percentage is his career average over the last 3 seasons. Crazy number. I think it is high partly because he shoots high often, leading shots to not register as a SOG. In practice he takes more shots than gets recorded and the his true shooting % (including once missing the net) is actually much lower than 18.3%.

I assumed 15 seasons of play with 1170 games. At 18.3% that will be

- 620 goals with 2017-2018 shots per game (2.9). 41 goals per season average. (too low without major injuries)
- 877 goals with 2018-2019 shots per game (4.1) 58 goals per season average. (too high)

If something is wrong in this count, it is the 4.1 shots per game average and using the 5-6-7 shot game numbers from 22 game sample. Career average is 3.0 shots per game.

 

[Artorius Horus T]

What math is to be maketh if Laine scores 6 goals with 6 shots in his next game???. - just an honest question

 

[KingPuckChoo]

I'll need a VERY... heavy diaper if that's going to happen, if anybody catches my drift.

you can't control your bowels? weird flex

 

[morehockeystats]

If my math is right, it'll happen once every 3125 games. Scoring 5 goals on 5 shots with a 20% shooting average.

The point is that author projects there will be at least 270-280 games where Laine would have six or seven times.

As of 5 on 5, I've got:
Probability of 5 on 5 in one game: 0.2^5 = 0.00032
Probability it doesn't happen: 0.99968
Probability it doesn't happen in 500 games: 0.99968^500 = 0.8521.
Probability when it breaks even occurs at log/0.99968/0.5 = ln(0.5)/ln(0.99968) ~= 2166 games.

This is one of the classical Chevalier De Mere's problems I learned some thirty years ago, shame on me for not recognizing it earlier.

 

[Ted Hoffman]

All of this assumes

* Every shot taken is independent of every other. Probably reasonable; not exactly accurate, but we'll ignore that for now.
* Every shot had "equal" probability to go in, represented by whatever shooting % you want to pick. Obviously, that's not true; it's not even true that shot% will be the same from year to year. So, it's likely that "average" shot% overestimates the true probability over time, but in the event Laine is simply firing from between the faceoff circles 10-20' out it might arguably understate his chances. Food for thought, at any rate.
* Situational scoring; Laine had 1 PP goal and that probably increased his chances of getting to 5.
* "Average" goaltending [however you want to define that]. The Blues, ... not average goaltending right now.

And so on and so on. So if all you're wanting is some ad-hoc proxy, the methods used above are probably semi-reasonable. If you're trying to get some semi-precise measurement, you're delving into delusional exactitude quickly.

 

[authentic]

What math is to be maketh if Laine scores 6 goals with 6 shots in his next game???. - just an honest question

If he defies math he would be the second coming.

 

[Barnum]

We are discussing in the Laine 5 goal game thread about the probability of Laine repeating the 5 goal effort during his career. Nobody has done this since Lemiux in the turn of the 90s.

However long time since high school and can't do this simple math anymore:

This year he has

5 shots in 5 games. (18.6 games per 82 season)
6 shots in 3 games (11.2 games per 82 season)
7 shots in 2 games (7.5 games per 82 season)

Most likely his production drops off after 15 seasons. He has been playing in 95.1% of the available games, making his season length 78 games. His maximum productive games available is 1170 assuming no major injuries.

5 shots in 269 career games
6 shots in 162 career games
7 shots in 108 career games

His career shooting is 18.3% so far.

What is Laine's chance to score 5 or more goals in his career? I am getting 5.5% for 5 shots, 16.9% for 6 shots and a whopping 31.1% for 7 shots.

This totals the chance to repeat during 15 year career at 53%. (which i am quite sure is incorrectly high, due 20 years departure from the math classes)

Anyone with actual math skills willing to give Laine's repeat chance a proper go?

All of this is way off. Since Mario did 5 goals in a game the first time, which BTW he did 4 times during his career. All of these players have done it since then.

Nieuwendyk
Sundin
Bondra
Ricci
Zhamnov
Fedorov
Gaborik
Franzen

 

[jmi]

All of this is way off. Since Mario did 5 goals in a game the first time, which BTW he did 4 times during his career. All of these players have done it since then.

Nieuwendyk
Sundin
Bondra
Ricci
Zhamnov
Fedorov
Gaborik
Franzen

And he never claimed no one else than Lemieux had 5 goal night. He said Lemieux is the last one to have multiple five goal nights.

 

[Barnum]

And he never claimed no one else than Lemieux had 5 goal night. He said Lemieux is the last one to have multiple five goal nights.

Ahhh yes. Helps to read throughly. You are correct.