tag:blogger.com,1999:blog-264226589944705290.post7762512233376031953..comments2023-02-06T08:11:10.397-08:00Comments on God Plays Dice: Some football probabilitiesMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-264226589944705290.post-45860854778237757622010-03-09T22:11:29.027-08:002010-03-09T22:11:29.027-08:00You have to express more your opinion to attract m...You have to express more your opinion to attract more readers, because just a video or plain text without any personal approach is not that valuable. But it is just form my point of viewAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-73152506989485718152008-01-01T15:45:00.000-08:002008-01-01T15:45:00.000-08:00Sebastian, that's the third time you've said the e...Sebastian, that's the third time you've said the exact same thing. As has been pointed out, we're making the simplifying assumption that all teams are evenly matched, since that's the question as it was originally posed to me.John Armstronghttps://www.blogger.com/profile/15177732626660057584noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-47211946423744947462008-01-01T14:56:00.000-08:002008-01-01T14:56:00.000-08:00Dear Isabel,I believe you are oversimplifying, you...Dear Isabel,<BR/><BR/>I believe you are oversimplifying, you are not considering the fact that for this to happen you actually need to have a third team that is very good; I think that considering this, the result is going to be less than 2/11. In fact, for a third team to get into the playoffs it must have played in a division with two better teams, hence it has a worse "environment" that the second team from other divisions, I assume that any team plays more often agains teams from its own division. My point is that for this calculations to be accurate one should consider all the games played by all the teams in the conference. Am I missing something here? I probably am.<BR/>Best greetings from Spain,<BR/>Sebastian.Unknownhttps://www.blogger.com/profile/16647668981553133129noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-26914635912212674282008-01-01T10:50:00.000-08:002008-01-01T10:50:00.000-08:00Dear Isabel,I believe you are oversimplifying, you...Dear Isabel,<BR/><BR/>I believe you are oversimplifying, you are not considering the fact that for this to happen you actually need to have a third team that is very good; I think that considering this, the result is going to be less than 2/11. In fact, for a third team to get into the playoffs it must have played in a division with two better teams, hence it has a worse "environment" that the second team from other divisions, I assume that any team plays more often agains teams from its own division. My point is that for this calculations to be accurate one should consider all the games played by all the teams in the conference. Am I missing something here? I probably am.<BR/>Best greetings from Spain,<BR/>Sebastian.Unknownhttps://www.blogger.com/profile/16647668981553133129noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-18094480469161894442007-12-31T22:25:00.000-08:002007-12-31T22:25:00.000-08:00I meant 8th and 9th ranked teams in my last post a...I meant 8th and 9th ranked teams in my last post as there are only 16 teams in each conference. The probability should be 2/11 that the 8th and 9th ranked teams in a conference come from the same division.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-18326278569311798942007-12-31T22:20:00.000-08:002007-12-31T22:20:00.000-08:00Let's not forget that the original premise is that...Let's not forget that the original premise is that all teams are equal. This means that if a 7-3 teams plays a 3-7 team, the probability of each team winning is still 0.5. I agree an answer of 2/11 does involve having made a simplification regarding the previously mentioned scheduling considerations. However, I expect that this would make only a slight difference since not only must there be more certain losses due to games within the division there must also be more certain wins. If rather than the two wildcards coming from the same division we asked if the teams ranked 16th and 17th (median teams) both came from the same division, I expect the answer would be exactly 2/11.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-25375441953484330072007-12-31T22:08:00.000-08:002007-12-31T22:08:00.000-08:00How good is a team in a division? This can be done...How good is a team in a division? This can be done at any point in the season, and stops evolving at season's end.<BR/><BR/>First, estimate how good the team is by the number of games they've won. Normalize by the number of games they've played. Rank the teams by this normalized 1-goodness.<BR/><BR/>Second, it's better to beat a good team than a bad team. So weight each win by the rank of the team beaten. Normalize. This is the 2-goodness.<BR/><BR/>Iterate this. In the limit, the infinity-goodness is the eigenvalue for that team of the matrix of scores of each team against each other team they've played.<BR/><BR/>This is crudely phrased, but standard in tournaments and weighted digraphs.<BR/><BR/>Now, that's the baseline against which the strange systems of NFL and BCS and MLB are to be compared.<BR/><BR/>I'm close enough to see the blimp over the Rose Bowl right now...<BR/><BR/>May the best team win. Modulo best.<BR/><BR/>Happy New Year.<BR/><BR/>-- Jonathan Vos PostAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-37215546677108201022007-12-31T21:40:00.000-08:002007-12-31T21:40:00.000-08:00A slightly more difficult problem would be to calc...A slightly more difficult problem would be to calculate the probability of both wild cards coming from the same division given the 2001 composition of the league. At that time there were 31 teams, five divisions of five teams and a sixth division with six teams.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-10973426683581553142007-12-31T17:13:00.000-08:002007-12-31T17:13:00.000-08:00Dear Isabel,I believe you are oversimplifying, you...Dear Isabel,<BR/><BR/>I believe you are oversimplifying, you are not considering the fact that for this to happen you actually need to have a third team that is very good; I think that considering this, the result is going to be less than 2/11. In fact, for a third team to get into the playoffs it must have played in a division with two better teams, hence it has a worse "environment" that the second team from other divisions, I assume that any team plays more often agains teams from its own division. My point is that for this calculations to be accurate one should consider all the games played by all the teams in the conference. Am I missing something here? I probably am.<BR/>Best greetings from Spain,<BR/>Sebastian.Unknownhttps://www.blogger.com/profile/16647668981553133129noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-62843353742966754972007-12-31T13:07:00.000-08:002007-12-31T13:07:00.000-08:00"average of 3 losses apiece, just because someone'..."average of 3 losses apiece, just because someone's got to lose those games."<BR/><BR/>There can be a tie game if after 15 minutes of sudden death no one has scored. I'd have to ask my brother in law if there has been a tie as I don't follow football.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-36068489841255985032007-12-31T12:26:00.000-08:002007-12-31T12:26:00.000-08:00Hmm, there's some simplification going here that I...Hmm, there's some simplification going here that I'm not sure is warranted. The NFL schedule is "unbalanced"; each team plays the other teams in its division twice. This should make it harder for multiple teams from a single division to stand out, since the intradivisional games in that division are going to account for 12 losses alone, or an average of 3 losses apiece, just because someone's got to lose those games.<BR/><BR/>I'm sure this would be a big issue if there were three wild card spots and you wanted to know the probability of everyone in a division going to the playoffs. My intuition is that it will still have an effect in the 2-wild-card-team case, although of course not as much.dfanhttps://www.blogger.com/profile/16523251716744122695noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-61544489984320021282007-12-31T10:41:00.000-08:002007-12-31T10:41:00.000-08:00Actually, I'd read the question slightly different...Actually, I'd read the question slightly differently and get 40/121. This is (using your answer) the probability that *a* division sends two wildcards. It could be from one conference (2/11), could be from the other (2/11), or it could happen in both (4/121). Then the probability that it happens at all is 2/11+2/11-4/121 by inclusion-exclusion.John Armstronghttps://www.blogger.com/profile/15177732626660057584noreply@blogger.com