Final Record Probabilites - Final Thoughts

[LUKE232323]

Isn't that still one game off? It says 1 win and we have 72 wins. If I'm not mistaken, we have 72 wins before play starts today. And we have 18 games left, not 19.

[rluzinski]

That's why I put "Wins in last 19 Games" at the very top of the table. All I did in my spreadsheet was change the probability of winning yeserday's game to 1 (since they already won it). This way, I don't have to change anything else.

 

Sorry for the confusion. Just look at the final record coumn.

[endaround]

Russ, what are you running this in? Just excel? The reason I ask is if you could get a histogram of the final records, that'd be cool.

[rluzinski]

http://i14.photobucket.com/albums/a345/rluzinski/montecarlo18histo.gif

[FVBrewerFan]

Appreciate the effort, but you went to a whole lot of trouble to prove what we already know. The Brewers more than likely will be at or near .500.

 

But there is a short sample left, so even those stats don't mean much. Depends on pitching match-ups, if 3 or 4 guys can get hot at the same time, etc. Of course 13-5 isn't likely, but it could happen.

[rluzinski]

Appreciate the effort, but you went to a whole lot of trouble to prove what we already know. The Brewers more than likely will be at or near .500.

 

You can just always just extrapolate a team's winning % to estimate a team's final record. This whole thread was to calculate the odds of finishing BETTER than expected while incorporating the strength of schedule and home/road aspect of it. With so much talk of what final record the Brewers need to have a shot at the wildcard, I thought this could be of some help.

 

Depends on pitching match-ups, if 3 or 4 guys can get hot at the same time, etc.

 

Most of what is characterised as "getting how" is just normal random variance anyway, so that's pretty much what's being shown above.

[1992casey]

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EDIT: I had made a huge error in my previous first post of my simulation. I was using the probability of beating Arizona for ALL the games, which obviously helped the Brewers ALOT. The probabilities have been updated.

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Has yesterday's chart (the one in the post with the quoted edit) been changed? In other words, does it still have the error?

[LS Brewer]

rluz, thank you for taking the time to post your probability analysis. It is a lot of fun to look at this each morning and find a justification for my wish for the Brewers to finish above .500 and, who knows, win as many as 84-85 games. Please keep up the good work if you have time.

 

Great performance by Capuano last night in getting out of a few jams and cooling off the D'Backs bats.

[rluzinski]

Has yesterday's chart (the one in the post with the quoted edit) been changed? In other words, does it still have the error?

 

It's been fixed. Even one win dramatically changes the odds.

[endaround]

The important thing to note is that the probability weight has shifted from the bad side of .500 to the good side. So the Brewers have better chance at a winning record than a losing one.

[tomwopat]

lets get on the right side of the bell curve!

[rluzinski]

lets get on the right side of the bell curve!

 

Actually, a bell curve is found from data that is normaly distributed. This is actually a binomial distribution (or actually multinomial I think http://forum.brewerfan.net/images/smilies/smile.gif ).

 

Maybe someone who actually knows can clear this up.

[SoCalBrewfan]

(Caveat: I'm not an expert.) A binomial distribution starts to look like a normal (bell) curve for a large enough sample...it also works pretty well for smaller samples if the probability is near .5. (Otherwise the curve is asymmetical, non-bell-like.)

[rluzinski]

Above .500 should still be the goal:

 

http://i14.photobucket.com/albums/a345/rluzinski/montecarlo16.gif

 

http://i14.photobucket.com/albums/a345/rluzinski/montecarlo16histo.gif

[ILBrewerFan]

Maybe someone who actually knows can clear this up.

 

My response may get this thread immediately kicked to the "stats forum". Anyway, I can't say for sure unless I see exactly what you are doing, however, the binomial distribution and the normal distribution are related.

 

As your sample size goes to infinity (large N) the binomial becomes a normal distribution with mean=n*p and variance=n*p*(1-p). So for sufficient N it will be "bell shaped". Most times the binomial is anyway except for extremely polar values of p.

 

A multinomial is an extension of the binomial as well, but I won't bore you with that. Simply that if two random variabls are distributed with a MULT dist their marginal pdf is BIN. Sorry, I like this stuff.

[rluzinski]

No that's great information. Thanks for the clarification. I knew you could approximate a binomial dist. with a normal one: now I know how.

 

I'm still not clear exactly how to use a multinomial, but I suspect it's what I have here. What I am doing is VERY simple. Let's say the Brewers are playing 3, 1 game series in a row and have a probability of beating each team, .5, .6, .55, respectively. I'm just running a crummy simulation, generating a random number for each game and summing up the wins. Repeat 10,000 times and get a decent approximation of the true probabilities of winning 0 games, 1, game, etc... I actually did it for the last 19 games, but as the team plays a game I simply change the probability to 1 or 0, depending on the outcome of the game.

 

I searched high and low for a more elegant way to solve this, but it appears most SABR guys "cheat" by simulating lot's of stuff. Alot of SABR guys (like me) don't have a statistics background, so it's usually just easier to throw a couple random number functions into excel.

 

If you have any advice or questions, I'd love to hear it.

[ILBrewerFan]

It sounds a lot like a multinomial. I posted this in another thread, but it may have been missed. I'm not sure how you calculate the Brewers winning probabilities by team so I will give you an example using a six sided die. First of all the multinomial pdf is as straight forward to calculate as a binomial and it looks like this for a random vector (X1....Xn):

 

f(x1..xn) = m! / (x1!*x2!.......*xn!) * p1^x1 * p2^x2 .......*pn^xn

 

Alright, back to the example. Suppose the die is unbalanced so that the probability of observing a 1 is 1/21, a 2 is 2/21, and so on. (X1, X2, X3, X4, X5, X6) is a random vector that counts the number of times that number comes up in 10 tosses of the die. m=10 trials, n=6 possible outcomes, p1=1/21......p6=6/21. What's the probability that in 10 rolls I get four 6's, three 5's, two 4's, and one 3?

 

10! / (0!0!1!2!3!4!) * (1/21)^0 * (2/21)^0 * (3/21)^1 * (4/21)^2 * (5/21)^3 * (6/21)^4 = .0059

 

So back to baseball and out of the world of colored balls, urns, flips of coins and rolls of die. I would have to think about it some more, but I believe you could treat each opponent as a random variable and the remainder of the season as the number of trials. This is where simulation is handy. You would have to think of all possible outcomes (2 of 3 from Chicago), (2 of 6 from Houston), etc. However, based on your probabilities and plugging the function above you could run through most combinations. (Do you have access to sas, S+, R or any other statistics package? You could whip through this stuff in seconds)

[1992casey]

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My response may get this thread immediately kicked to the "stats forum".

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It's been discussed somewhere. I believe one of the moderators said he'd leave it here for the time being?to reach a larger audience?then move it because of the statistical information it contains.

[ILBrewerFan]

It's been discussed somewhere. I believe one of the moderators said he'd leave it here for the time being?to reach a larger audience?then move it because of the statistical information it contains.

 

Yeah, I was just trying to "make a funny" because I was writing out formulas and stuff. This is what happens when statisticians try to make jokes.

[ja002h]

This is what happens when statisticians try to make jokes.

 

Hey, I laughed at this one http://forum.brewerfan.net/images/smilies/smile.gif

[ILBrewerFan]

Hey, I laughed at this one http://forum.brewerfan.net/images/smilies/smile.gif

 

Sweet! I'm taking this show on the road!

[rluzinski]

But isn't this a bimomial in that only two events can occur per game, a win or a lose. If the team has a 60% chance to win, they have a 40% chance to loss:

 

.6 + .4 = 1

 

Wouldn't a mutlinomial apply for something like soccer, where there are 3 possible events, win, loss or tie? For instance, a team could have a 60% chance to win, 30% chance to tie and 15% to lose:

 

.6 + .3 + .1 = 1

 

You should check out my log5 / series probabilities thread, where I have math similar to some of yours, only with a binomial.

 

LINK

 

With the prob. not all being the same (like they are in a series) you would have to do it all long hand. It's just easier to simulate unless there IS a mathematical way to do it. I just haven't found one.

[endaround]

Right it isn't a multinomial. A multinomial has fixed probabilties and mutliple (more than two) states. Here we have changing probabilties and still just two states. It is series of binomial draws with differing probabilities. You could take an expectations operator through it and firgure out the various moments if you want, but doing it analytically really doesn't gain that much.

 

edit: Finding the mean isn't hard though--its just the mean of the individual binomials added together. The variance becomes more tricky. If the mean is m, and the distribution d, then the variance is:

 

E(d^2)-m^2

[ILBrewerFan]

Right it isn't a multinomial.

 

I stand corrected. Thanks for the info end. Forgot about the changing probabilities thing. Guess I should dust of the counting books.

[Southwest Brewer]

Weren't you kind of describing a binomial, but pi changes for each team they play? pi would be lower on the road against the Astros, and higher against the Pirates?