DC Tom Posted December 31, 2016 Posted December 31, 2016 No because that's not factoring in compounding, meaning the chances of missing each of the 17 seasons grows more remote the longer it goes on, as in, just by dumb luck at some point the team SHOULD make it once in 17 years It doesn't factor in a lot of things. It completely ignores that a given team making the playoffs in a year is not a random, independent outcome. But still, Cereal's correct under the assumptions ESPN states they used. That those assumptions are completely incorrect isn't Cereal's fault.
justnzane Posted December 31, 2016 Posted December 31, 2016 (edited) Wouldn't it be? Odds of missing playoffs in division = 3/4 minus odds of being a conference wildcard team = 2/12 Convert to common fractions 3/4 = 9/12 9/12 - 2/12 = 7/12 or 58.333333% chance of missing the playoffs each year calculated over 16 years? which would equal .000179750 which translates into 719/4000000 or 719 chances in 4 million Where is the math wrong? What are the odds of the Bills or any NFL team missing the playoffs 16 years in a row? Final Answer = What is 719 in 4 million Alex? or if you want to approximate it = 1 in 5563 Quick, somebody email Matt Patricia and ask him if I am right. How about the union of probabilities of winning division or earning card= 1/4+2/12= 5/12 This makes use of the assumption that 3 of the 6 AFC playoffs spots can not be earned by the Bills as they can only win 1 division out of the 4. This is opposed to the brute force logic that 6/16 teams make the playoffs in the AFC each year. Thus chances of not getting either spot being 7/12= 58.3% chance of missing the playoffs in a given year. So we agree with slightly different logic to come to the same conclusion. However, you make the mistake of only using 16 years, when we are now on year 17. Thus, the random probability odds are 1 in 1/((7/12)^17), or 1 in 9537. Thus, that 1 in 6 million number is horse crap. When you consider the fact that the probabilities of winning the division and a wild card are not independent and that the team from one year to the next is not random, these odds become more common as shown the Monte Carlo sim that I believe Tom cited earlier. That said, ESPN, if you are looking to hire someone with a passion for sports with a math degree, and understands statistics and probability, please PM me, so I can send in my resume. Edited December 31, 2016 by justnzane
All_Pro_Bills Posted December 31, 2016 Posted December 31, 2016 What are the odds of hiring the wrong coach and GM every time during a 17 year stretch? Or not being able to find a QB? These things aren't random chance or statistical probability equations. You have to be very, very good at being this pitifully bad to pull off missing the playoffs for 17 years. I think the solution is simple. Make a decision on something and then do the opposite. Example - Bills decide to 'move on' from Tyrod. Answer - don't do it. Bills ownership decides to retain Whaley and Brandon. Answer - Fire them. Problem solved, playoffs year 18.
PolishDave Posted December 31, 2016 Posted December 31, 2016 How about the union of probabilities of winning division or earning card= 1/4+2/12= 5/12 This makes use of the assumption that 3 of the 6 AFC playoffs spots can not be earned by the Bills as they can only win 1 division out of the 4. This is opposed to the brute force logic that 6/16 teams make the playoffs in the AFC each year. Thus chances of not getting either spot being 7/12= 58.3% chance of missing the playoffs in a given year. So we agree with slightly different logic to come to the same conclusion. However, you make the mistake of only using 16 years, when we are now on year 17. Thus, the random probability odds are 1 in 1/((7/12)^17), or 1 in 9537. Thus, that 1 in 6 million number is horse crap. When you consider the fact that the probabilities of winning the division and a wild card are not independent and that the team from one year to the next is not random, these odds become more common as shown the Monte Carlo sim that I believe Tom cited earlier. That said, ESPN, if you are looking to hire someone with a passion for sports with a math degree, and understands statistics and probability, please PM me, so I can send in my resume. Yeah man If you kept following the thread, I posted the 17 year math later on. And we got the same answers. Cheers bro.
mead107 Posted December 31, 2016 Posted December 31, 2016 What's the chance we have to wait another 17 years ?
Bleeding Bills Blue Posted December 31, 2016 Posted December 31, 2016 What are the odds that NE wins the division in what - 14 of the last 16 years...? Their success has coincided with our failure. Miami and the Jets have snuck in a few times in that span. But this is Tannehills 5th year, and 1st in the postseason. Rex had 2 years with the Jets. Then there was the year Cassel Started. I get that we should have by now, but many teams would have pretty solid droughts if they had to play a 12-4 team twice a year, as well as only wildcard spots to play for.
BuffaloRebound Posted December 31, 2016 Posted December 31, 2016 What are the odds when you factor in no chance of winning division plus 2 losses to the Pats every year, and only 7 home games in 5 of those years?
DC Tom Posted December 31, 2016 Posted December 31, 2016 How about the union of probabilities of winning division or earning card= 1/4+2/12= 5/12 This makes use of the assumption that 3 of the 6 AFC playoffs spots can not be earned by the Bills as they can only win 1 division out of the 4. This is opposed to the brute force logic that 6/16 teams make the playoffs in the AFC each year. Thus chances of not getting either spot being 7/12= 58.3% chance of missing the playoffs in a given year. So we agree with slightly different logic to come to the same conclusion. However, you make the mistake of only using 16 years, when we are now on year 17. Thus, the random probability odds are 1 in 1/((7/12)^17), or 1 in 9537. Thus, that 1 in 6 million number is horse crap. When you consider the fact that the probabilities of winning the division and a wild card are not independent and that the team from one year to the next is not random, these odds become more common as shown the Monte Carlo sim that I believe Tom cited earlier. That said, ESPN, if you are looking to hire someone with a passion for sports with a math degree, and understands statistics and probability, please PM me, so I can send in my resume. One mistake: the probability of being a wild card isn't 2/12. It's 1/8 (the probability of not winning the division times the probability of being one of the two wildcard teams. (3/4) * (2/12).) To put it in English: you have a 1/4 chance of getting your "division champ" playoff berth, and a 2/12 chance of getting a wild-card berth only if you don't win the division. You need to represent that last condition in your probabilities. That reduces again to a probability of 5/8 of not getting a spot. ESPN, if you're looking for a physicist to explain to your mathematicians how the world works, PM me... What are the odds when you factor in no chance of winning division plus 2 losses to the Pats every year, and only 7 home games in 5 of those years? I am actually modifying my Monte Carlo simulation to take that in to account.
Nanker Posted December 31, 2016 Posted December 31, 2016 I think we have a 100% chance of going without a Playoff appearance in the last 17 years. What do I win?
DC Tom Posted December 31, 2016 Posted December 31, 2016 I think we have a 100% chance of going without a Playoff appearance in the last 17 years. What do I win? [This is an automated response.] You're an idiot. Created by DC Tom-bot, beta version 0.4.
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