Final Record Probabilites - Final Thoughts

[rluzinski]

I went ahead and estimated the probabilities of many notable final records, most importantly 81-81 (no losing season) and 87-75 (good chance for wildcard):

 

Final Season Probabilities

 

http://i14.photobucket.com/albums/a345/rluzinski/season.gif

 

I found the weighted average of the 1 game probabilites for the rest of the year and used a binomial distribution to calculate overall probabilites. Two important things:

 

* Using a weighted average is technically not correct. Enivsion 4 consecutive games, were 3 games are played against a team that has a .500 w% and 1 game against a team that has a .900 w%. My weighted average method would give the prob. of a sweep at 2.6%, when it's really 1.25%. Since all the 1 game probabilities are pretty close though, I don't think it's THAT off. I'll work on this, however.

 

* The record probabilites are actually for getting that many wins or better. For instance, there is a 57% chance the Brewers will finish at .500 OR BETTER.

 

Let me know what you guys think, or if anyone has any suggestions.

 

EDIT: This is the old, wrong way. See my later post for the correct results.

[homer]

Quote:

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.5% 89 - 73 likely needed for WC

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So yer tellin' me there's a chance. Yeah! I read ya'!

[nsimps]

I think the wildcard winner is going to have 87 wins. For it to be 89, Philly would have to win 60% of its remaining games. The only team I can see getting to 89 would be Florida if their pitching remains strong. (They need to win 61% to reach 89).

[nullbyte]

Good analysis. 43.8% chance for a winning season is a pretty decent prospect. I can't tell you how scared I am of another collapse of some sort...

[huntsvillefan]

Great post rluz, wish there were more like this and the HOF thread.

[Danagger]

I'm sticking by my March prediction...

 

79-83

[bball123h2]

At the risk of a huge reply, what did you do to convert the Pythagorean Win/Loss % to probability, and then probability to probability with home/road? Excellent post, I was just curious on the calculations..

[rluzinski]

"On Pace" Record

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Phili: 86.7

Astros: 86.1

Marlins: 85.5

Mets 84.2

Nats: 84.2

 

One of those 5 teams will overperfom their "on pace" record and get 88 to 89 wins. Baseball Prospectus calculates that the NL W.C. winner will have 88.6 wins, which backs that common-sense approach up.

 

What's the least amount of wins the wildcard has needed? I looked it up and one point.

[nsimps]

Quote:

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One of those 5 teams will overperfom their "on pace" record and get 88 to 89 wins.

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I just don't see real evidence for that (despite what my signature would suggest). None of the competing teams are more than above average and most have difficult schedules for the remainder of the season.

[BrewCityUnit]

I appreciate the efforts of this post, but how can you determine how many wins it is going to take to win the wildcard. What makes your 87 prediction any better than someones 84 win prediction.

 

At the beginning of the year, you would had thrown numbers out there to predict the NL West winner with 90 wins minimun to win the division.

 

Isn't every year different??

[rluzinski]

None of the competing teams are more than above average.

 

My point is simply that they don't have to be to end up with 87-88 wins.

 

Take 10 teams that are all theoretically .500. Ask them all to play 20 games against each other. Who wins each game is a virtual coin flip. You'll see a final record of something like:

 Team Record ---------------- Team 1 13 - 7 Team 2 12 - 8 Team 3 11 - 9 Team 4 11 - 9 Team 5 10 - 10 Team 6 10 - 10 Team 7 9 - 11 Team 8 9 - 11 Team 9 8 - 12 Team 10 7 - 13 

The records will all be grouped around the average record of 10-10, but some teams will win more and some less. With 5 teams, the chances of them ALL playing .500 or worse from here on out isn't very good. If anything, the fact that they play each other makes it even harder for them all to finish .500 or less.

[rluzinski]

I appreciate the efforts of this post, but how can you determine how many wins it is going to take to win the wildcard. What makes your 87 prediction any better than someones 84 win prediction.

 

There are many ways to predict it. The hard way is to look at each game and run a simulation of the rest of the season. Baseball prospectus does this and finds that the average W.C. winner has between 88 and 89 wins.

 

A less scientific way would be just to assume all the teams keep playing at their same pace. Phili would win with about 87 wins. While that doesn't factor many variables in it's a nice ballpark.

 

Finally, looking at prior years is a great way to see what a typical record is needed to win the wildcard. For 84 wins to be enough to win the wildcard the whole NL (minus the 3 divison leaders and the wildcard winner) would have to be at 2 games over .500 or below. Does that sound reasonable?

[nsimps]

Quote:

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Take 10 teams that are all theoretically .500. Ask them all to play 20 games against each other.

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But that assumes an equal probability for all of the games which your first post of the thread discounts.

[rluzinski]

But that assumes an equal probability for all of the games which your first post of the thread discounts.

 

But everyone keeps saying, "none of them are better than .500." I'm just showing that even if they all WERE all .500 teams (they aren't) a team would still win more than half their remaining games. The fact that Phili, Houston and Florida are all obviously better than .500 makes the chances of a 85 win team winning the wildcard even smaller.

 

I was simplying the model to try and prove a point. I have apparently utterly failed.

 

Now if you will excuse me, I need to go buy my playoff tickets. http://forum.brewerfan.net/images/smilies/sick.gif

[nsimps]

Quote:

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Now if you will excuse me, I need to go buy my playoff tickets. :x

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I was just making an estimation that the winner would be more likely to have 87 wins than 89 wins, not that the Brewers were going to win or that your table is horribly wrong.

[rluzinski]

I was just making an estimation that the winner would be more likely to have 87 wins than 89 wins, not that the Brewers were going to win or that your table is horribly wrong.

 

I didn't mean to insinuate that you thought that. Anyway, if you use probabilities higher than .500 for a single game, then there will be a higher variance and it will require more wins to win the wildcard. I used .500 because it's the most conservative (tightest variance).

 

ANYWAY, I was hoping this thread would be more about the Brewers prospects of having a winning record, not making the playoffs.

[bjamesw22]

Good point Russ. And I think seeing as our goal was to finish the year at .500 or better and it's being showed we have a pretty darn good chance. Talk about an improvement from last year!

 

I think another thing you would agree with would be that it's going to be much harder to get to a level where we are a playoff contender/division winner consistently as opposed from starting off mediocre and getting to where we are now.

[splitterpfj]

Thanks rluz, fun thread. I'm hoping for 82 wins, and to finish ahead of the Cubs.

[OldHoss]

Quote:

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At the risk of a huge reply, what did you do to convert the Pythagorean Win/Loss % to probability, and then probability to probability with home/road? Excellent post, I was just curious on the calculations..

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I have a similar or maybe the same question.

 

Does home versus road record matter here, i.e., for Brewer home games should the calculation be based upon the Brewers' Miller Park Pyth W% versus the other team's Miller Park Pyth W% (or overall road Pyth W% if Miller Park is too small of a sample) and conversely for road games?

 

I guess, in other words, I want to know if the Brewers' probabilities change based upon the "favorable" home schedule over the remainder of the season.

[nsimps]

Quote:

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I was hoping this thread would be more about the Brewers prospects of having a winning record, not making the playoffs.

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I understand, but as long as there exists a possibility for a wild card spot, despite actually probability people are going to talk about it. Some people would rather accentuate the romantic possibilities rather than the rationalistic probabilities.

[BrewCityUnit]

Rluz, you always ingnore the part of my post about the NL West this year. How can you explain that? I am confused, just because the average is 87 doesn't mean 87 has to happen every year. I really don't care about averages. I'm sure there will be a year where you will have to win 93 games and likely a year when you will have to win 85. That would average to what, I don't have a calculator, 89?

[Brian the Automator]

[ I'm sure there will be a year where you will have to win 93 games and likely a year when you will have to win 85. That would average to what, I don't have a calculator, 89? ]

 

Only if your sample size is 2.

[rluzinski]

Some people would rather accentuate the romantic possibilities rather than the rationalistic probabilities.

 

I understand that and despite knowing we have a 1 in 50 chance of making the playoffs, I look every day! I will root for them to make it until they are mathematically eliminated. What prompts me to write threads like this is the notion that the Brewers are a modest winning streak away from from having a reasonable chance (say 20%). Some say that me disagreeing is simply a matter of opinion, but I happen to disagree. It's a matter of mathematical probability after looking at the facts. There is a possibility that my models are wrong or that I've used an equation incorrectly. I think I'm pretty close, however, since they agree pretty well with what others have calculated.

 

Rluz, you always ingnore the part of my post about the NL West this year.

 

I've indirectly answered your question by trying to explain how binomial distribution works. There is always a chance that someone will win 162 games or 0 games; that the wildcard winner will need100 games or 80. I'm not here to say it's impossible but rather put a number on it's probability. It's possible that the wildcard winner WILL only neeed 85 wins. It's highly improbable, however.

 

89 wins. That's what I think it will take.

[BrewCityUnit]

Thanks Rluz for the explanation. I agree with your post before about how you look every day and root for them to make it. I can respect that you like to look at percentages as being realistic (because they are), hopefully this is one of those strange years where the statistics don't predict the outcome.

[rluzinski]

I first used the pythagorean theorum to find the "true ability" of a team, in the form of winning %. I then used the log5 equation ti find the probability of the Brewers beating the team on any given day (prob). I adjusted that number by the league average difference in home / splits (home team wins about 4% more games).

 

The second part was a bit more fuzzy math. As I explained, I simply took the weighted average of all those probabilities (take into account how many games the Brewes play in each matchup) to get the average chance the Brewers have to win a game for the rest of the year. It worked to a 52.8% chance. That 2.8% perk represents a combination of home field advantage and strength of schedule.

 

From there I simply use the binomial formula in excel to find the probability of winning so many games. To explain that part quickly. If you flip a coin 10 times, what's the probabily of flipping 6 or more coins?

 

Total Trials = 10

Successful Trials = 6

Prob of success = 50%

 

Answer? There's about a 38% chance of flipping 6 or more heads out of 10.