tag:blogger.com,1999:blog-12133335.post113788618686264147..comments2015-01-26T05:43:03.818-05:00Comments on Walk Like a Sabermetrician: Runs Per Winphttp://www.blogger.com/profile/18057215403741682609noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-12133335.post-1139225090113765662006-02-06T06:24:00.000-05:002006-02-06T06:24:00.000-05:00Correction:In the last line of my previous comment...Correction:<BR/><BR/>In the last line of my previous comment I wrote:<BR/><BR/>RPW=2*RPW/x<BR/><BR/>It should be:<BR/><BR/>RPW=2*RPG/x.Ralph Caolanoreply@blogger.comtag:blogger.com,1999:blog-12133335.post-1139177655055085542006-02-05T17:14:00.000-05:002006-02-05T17:14:00.000-05:001. I don't disagree that it is ok to use 2, but I ...1. I don't disagree that it is ok to use 2, but I also don't see any reason why not to use a better estimate if you are inclined to do so.<BR/>3. It was posted on a FanHome thread sometime in the past, but I don't think it is one that is currently on the board. <BR/>4. I didn't think of it that way, that's a good point.<BR/>6. Yes, I was using PW% to abbreviate Pythagorean W%. I differentiated Run Ratio with respect to Run Differential(dRR/dRD) and then PW% with respect to to RR(dPW%/dRR). I multiplied those to get dPW%/dRD, which is RPW. It's good to know that we got the same result.phttp://www.blogger.com/profile/18057215403741682609noreply@blogger.comtag:blogger.com,1999:blog-12133335.post-1139174258586606892006-02-05T16:17:00.000-05:002006-02-05T16:17:00.000-05:00Being a subject close to my heart, I found “Runs p...Being a subject close to my heart, I found “Runs per Win” very interesting. I have the following comments and questions:<BR/><BR/>1. I also enjoyed Dan Fox’s article. However, I thought his use of PythagenPat was overkill. I think PythagenPat always overcomplicates calculations at a team level. A Pythagorean calculation using an exponent of 2 or 1.88 is plenty accurate enough.<BR/>2. I also thought Fox’s use of Palmer’s “square root” formula for RPW was fine, but, being biased, I would have used the formula I derived (By The Numbers, November 2003). It is RPW=2*RPG/x, where x is the Pythagorean exponent. When x=2, RPW is simply RPG.<BR/>3. Can you provide the reference for David Smyth’s formula for RPW?<BR/>4. You mention that Smyth’s formula is undefined when R=Ra and I understand why you think “it’s a shame”. But, when R=Ra, W% is theoretically 0.500, and wins above 0.500 are theoretically zero, also. So, RPW will be 0/0, and, therefore, also undefined (unless you can apply L’Hopital’s Rule).<BR/>5. A little bit of algebra on Smyth’s formula at x=2, yields RPW = 2*(R^2+Ra^2)/(R+Ra). With this formula, you never have the problem of division by zero.<BR/>6. You wrote “However, if we differentiate PW% with respect to RD … we find … RPW = ((2*RPG*(RR^x + 1)^2*(.5 - RD/(2*RPG))^2)/(x*RR^(x-1))” Is PW% Pythagorean winning percentage? If so, what formula did you differentiate to get this? Maybe it doesn’t matter because when I did it, it also distilled down to RPW=2*RPW/x.<BR/><BR/>Ralph CaolaRalph Caolanoreply@blogger.comtag:blogger.com,1999:blog-12133335.post-1138164546704181652006-01-24T23:49:00.000-05:002006-01-24T23:49:00.000-05:00Thanks for the mention, and yes I have made the co...Thanks for the mention, and yes I have made the corrections we emailed about at http://www.hardballtimes.com/main/article/second-look-at-luck/.<BR/><BR/>As to your last question, if I understand you right, I'm using the 10.10 value from Palmer and then applying that value to each team. You're right, that the other would be more accurrate. <BR/><BR/>Thanks for remidning me of the simplified formula. I had seen it somewhere before but forgot all about it.Dan Agonisteshttp://www.blogger.com/profile/17844403476675106059noreply@blogger.com