rluzinski

How depressing. The brewers must be something like 1-4 when 1 game under .500.

This game is crazy. Please don't end bad.

It would be nice to be .500.... again.

Updated 8/24: 81-81: 63.3% 82-80: 50.1% 87-75: 4.4% 89-73: 0.8% Let's get that win this afternoon!

What were the chances of KC losing 18 in a row? Not high, which is why 18+ losing streak have happened, what, 4 times? So I guess once every 25 years http://forum.brewerfan.net/images/smilies/smile.gif

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.

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.

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.

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

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?

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.

"On Pace" Record -------------------------- 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.

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.

I wouldn't attribute it ALL to luck. Anyone watching a game can see when misc. skills can scratch out an extra win or two (playing good 1 run ball). For most teams, however that stuff tends to balance itself out.

is there anyway way to account for a team being at home or on the road? The average home team wins at about a .540 clip. So: WPct(home team) = WPct + 0.040 I was going to add that when I finish the post http://forum.brewerfan.net/images/smilies/smile.gif

I just want a win tomorrow, l so we can get to 1 game under. This team needs to get to .500.

I could care less about Fielder, 1 AB doesn't matter and he can practice all he want's in AAA. JDLR can't be sent down and hasn't had much work this year. He's had problems with control all year and this would have been great time to give him some innings. That's very straight forward logic to me. He needs work BAD and can't get it in a blowout. Great game, let's steal one tomm.

Take a night off from the know-it-all act and enjoy a victory. Right, and when JDLR is put into the game with the bases loaded and he walks the first guy, I'll think of today. If you don't like it, skip over my posts.

This is SO funny. The Brewers have a 9 run lead. GREAT chance to get the young guys some playing time, right? Fielder, JDLR, maybe Capellen? WRONG. Time to get MAGRUDER playing time http://forum.brewerfan.net/images/smilies/smile.gif

The Brewers are up by 10 runs and Cappy throws 115 pitches. Why???

Just when the Pythagorean for the Brewers was fixing itself the Crew has to have a 10 run lead. Owell http://forum.brewerfan.net/images/smilies/smile.gif

Ever wonder how often a .550 team beats a .450 team? I did, and after a little searching I stumbled onto the Log5 equation. Introduced by Bill James in his 1981 Bill James Baseball Abstract, it states: A - A * B WPct = ----------------- A + B - 2 * A * B Where: A = Team A's Winning % B = Team B's Winning % WPct tells us what % of games team A will beat team B, on average. What if we want to calculate the odds of winning 1 game in a 2 game series, however? The first step is to calculate the odds of winning 1 game "one way". The odds of winning the first and losing the second: PROB = WPct * (1 - Wpct) WPct is the probabilty of winning the first game and 1- WPct is the probabilty of losing the second game. The probability of BOTH occuring is simply the product. But that's only one way to win 1 game of a 2 game series, however. The team could lose the first and win the second and still win 1 game in the series. Common sense tells that that the probability of winning a specific number of games in a series is then: PROB(one way) * number of ways So, the probabilty of winning 1 game in a two game series is: WPct * (1-WPct) * 2 If we want to calculate the probability of winning any number of games in any length of a series we need to create a more flexible equation. We'll define the two variables we need as: G = total games in the series W = number of wins in a series The probabilty of winning "one way" is then simply: [WPct ^ W] * [(1-WPt) ^ (G-W)] All that's left to figure out is the "number of ways" part. We could figure it by hand if we wanted to. For a 3 game series, the number of ways to win 1 game is: WLL LWL LLW Three ways to win 1 game in a 3 game series. It just so happens that the Combination function can calculate just how many ways you can win any number of games in any length series: "number of ways" = C(G, W) Putting the final equation together: PROB(G,W) = [WPct ^ W] * [(1-WPt) ^ (G-W)] * C(G, W) A few more points to finish this post. Since the home team wins 4% more games, simply add .04 to their WPct. Also, I use the pythagorean winning percentages for the Log5 formula since I feel it predicts future success better. If anyone is interested in the actual computer code I will be happy to type it in. Let me know if there's any questions or suggestions! Links: Estimating one-game winning percentages Combination Formula Pythagorean Theorum

If I do play-by-play, I try to add things that you definetely wouldn't see on gameday like, "It's nice to see Hart wait for that curveball and drive it to left."

If you are going to have Pierre up, might as well have a slow pitcher at 1B.

Hart was well on his way to a 3rd DP until Delgado bailed him out http://forum.brewerfan.net/images/smilies/smile.gif Good mojo today!