-OZ-
Footballguy
At $21, I might not, but let's pretend we forecast Baker to be next year's Mahomes. Do you take him then? Obviously that's best case and unlikely.
Fantasy Football - Footballguys Forums
Welcome to Our Forums. Once you've registered and logged in, you're primed to talk football, among other topics, with the sharpest and most experienced fantasy players on the internet.
Official 2019 FBG Subscriber Contest (5 Viewers)
- Thread starter TheWinz
- Start date
TheWinz
Footballguy
Last year the top 5 most expensive QB's were Rodgers ($27), Watson ($25), Wilson ($24), Newton ($22), and Brady ($21).
Rodgers - 2 of 734 made it to the end, finished as QB6
Watson - 10 of 590 made it, finished as QB4 - the only top 5 QB to have a survival rate (1.69%) higher than the average (1.67%)
Wilson - 2 of 189 made it, finished as QB10
Newton - 1 of 307 made it, finished as QB11
Brady - 4 of 650 made it, finished as QB13
2470 out of 15115 people chose a top 5 QB (16.3%). 19 made it to the end (0.76%). That says quite a bit. I find the expensive QB's to be the worst value in the contest, and I seem to max out around $13/$14 for my top guy.
Drunken Cowboy
Footballguy
SeniorVBDStudent
Footballguy
It will be interesting to see if they shake things up this year and make big shifts in the unit value structure. Historically speaking the contest has typically only featured small tweaks year over year....which is probably wise from the "if it ain't broke, don't fix it" school of thought.
TheWinz
Footballguy
False
- Wilson had the most passing TD's of his career. He also had the least INT's of his career. He was 20th in pass attempts with a mere 427. He had a great season, just not enough volume.
- Watson was also a victim of low volume. He increased his completion % by 6.5 points while cutting his INT ratio in half. Hard to be at the top with only 505 attempts, even if you get 551/5 rushing.
- Rodgers had the 2nd most passing yards of his career. He also only threw 2 INT's in 597 attempts, by far the best attempt/INT ratio in the league. What he lacked was passing TD's, as 9 others threw 30+.
- Brady has had 15 seasons without a missed game. His TD avg for those yrs is 31, so yes, under average. His yardage was above average. His attempts dropped to the lowest since 2010. Underperform or underthrow?
- Newton had his highest completion % as a pro, but his shoulder cost him 2 games. Even with the 2 missed games, he still finished QB11. Every QB ahead of him played in all 15 of their starts thru the first 16 weeks. Prorated, he woulda finished QB8.
I can buy Brady and Newton underperforming, but not Wilson, Watson, or Rodgers.
Gottabesweet
Footballguy
Interseptopus
Footballguy
Drunken Cowboy
Footballguy
SeniorVBDStudent
Footballguy
Statistics wizards, please help if you can with the following question:
Assume a bestball-format team has the following QBs and point per game statistics:
QB1 Avg 26 Sigma 11
QB2 Avg 22 Sigma 7
What is the expected value of QB points for the season for this team?
Assume a bestball-format team has the following QBs and point per game statistics:
QB1 Avg 26 Sigma 11
QB2 Avg 22 Sigma 7
What is the expected value of QB points for the season for this team?
SeniorVBDStudent
Footballguy
Somebody check my work...
Find probability that max (QB1,QB2) > QB0 (say 31)...the real world example might be should I roster Aaron Rodgers or Big Ben + Phillip Rivers...
For QB1 the value is 31-26=5 units above mean or 5/11=.4545 st devs above mean
z table value ~ 0.675
so 32.5% chance value above 31
For QB2 the value is 31-22=9 units above mean or 9/7=1.2857 st devs above mean
z table value ~ 0.9
so 10% chance value is above 31
So there is a combined 32.5+10=42.5% chance max(QB1,QB2) outscores expectation of QB1
So if I can "back into" the QB0 score that has a 50% probability of being exceeded by max(QB1,QB2) score I have the expectation value of max(QB1,QB2)...
Let's try QB0 = 30...
For score 30 vs QB1, 4/11=0.3636 gives z table value .642 so 35.8%
For score 30 vs QB2, 8/7=1.1429 gives z table value .874 so 12.6%
So best ball probability for 30 points expectation is 48.4% or just below 50%
Eyeball interpolation of the two probabilities at 30 and 31 says score 29 would be about 54.5% so 29.67 is approx value that corresponds to 50% expectation (the mean).
So if the pricing of the QBs is $25, $16, $9, then QB0 = QB1 + QB2 = $25, but QB0 has an expectation of 31 points but max(QB1,QB2) is only 29.67 points so QB0 is the better value (in a vacuum which doesn't consider risk and overall roster construction).
Drunken Cowboy
Footballguy
Your 2 QB's will sometimes have their good weeks at the same time. Rounding your numbers a bit. The chance of neither having a big week is 2/3 × .9. So it is about a 40% chance at least one has a good week. You also need to think about floors, injuries, and byes when doing this though. This also assumes perfect knowledge of the actual stars before the year.
SeniorVBDStudent
Footballguy
Drunken Cowboy
Footballguy
You gave the odds of a big week at 32.5%. So the odds of not making that are 1 minus 32.5%
SeniorVBDStudent
Footballguy
So...these two statements don't appear to have anything to do with one another....
Drunken Cowboy
Footballguy
Ignoratio Elenchi
Footballguy
It's been a few years since I've been really invested in this contest (life gets in the way), but I seem to recall finding the opposite, that high-priced QBs were a decent value. Relatively consistent high floor with high ceiling upside as well, seemed to outperform a pair of cheaper QBs for same cost and one less roster spot (e.g. in the kind of calculations VBDStudent's doing). Certainly may have changed over the last 5-10 years though, or I may have been wrong from the start.
TheWinz
Footballguy
I've been reading the last few posts between @SeniorVBDStudent and @Drunken Cowboy. I am just waiting for them to translate to English when they are done
As for the roster spot, I won't be maxing out, so that's a non-issue. As for a top guy outperforming 2 cheaper guys, it hasn't been that way the last few years. Last year, of the top 5 most expensive QB's, only one finished in the top 5. In 2017, of the top 5 most expensive QB's, only one finished in the top 5 (Brady at #3). Only 1 Brady entry finished in the top 50, but he used Brady only 8 times. His other QB was Stafford, who he also used 8 times, and was less than half the price.
SeniorVBDStudent
Footballguy
I've always struggled with the practical application of statistics, so I outlined a statistical procedure for how I THINK you can determine whether two QBs will outperform a single "better" QB in bestball format if you have the projected average and standard deviation for each.I've been reading the last few posts between @SeniorVBDStudent and @Drunken Cowboy. I am just waiting for them to translate to English when they are done
As I noted, I'm not sure if the calculations I used are actually valid for "discrete occurrences of random variables", so I was hoping someone more comfortable than I am with statistics could validate my thought process.
After some back and forth, I realized what Drunken Cowboy was doing. He multiplied the probability of each QB not hitting 31 points, as the chance that neither has a good week (really that neither would hit 31 points) and took 1 minus that as the probability that at least one of the two would hit 31 points. This is different than what I did, which was to add the two probabilities of each meeting 31 points as the cumulative chance that one would exceed 31 points.
Which takes us back to my original point: I've always struggled with the practical application of statistics, so I'm not sure if neither, one or both of us are correct.
Haven’t had time to work on the original question, but based on this Drunken Cowboy is correct. Each QB is a separate event, so probabilities multiply. You only add when combining exclusive outcomes from a single event.
TheWinz
Footballguy
SeniorVBDStudent
Footballguy
Actually looks like neither of us are correct. There is something called the General Addition Rule in statistics which defines the probability of multiple overlapping events (the scores of three QBs for example).
The probability of three events happening is P(A) + P(B) + P(C) - P (A) * P (B) - P(A) * P(C) - P(B) * P(C) + P(A)*P(B)*P(C) which you would have to evaluate multiple times over the range of the elite QB outcomes.
It's all pretty awful.
That isnt what you want here though. That looks like a formula for evaluating the probability of getting one of three outcomes from a single random event. (You are adding the chance of each outcome, then removing the crossover chances of getting both to avoid double counting.)
For combining distinct events, you would always multiply.
ZWK
Footballguy
Assuming normal distributions (which isn't quite right), looks like you'll get about 29.5 points per game when both of those QBs are playing, so about 460 points (28.8 ppg) for weeks 1-16 assuming each guy only misses their bye week (14 games of 29.5 ppg plus 1 game of 26 pts plus 1 game of 22 pts).
TheWinz
Footballguy
All these formulas are giving me a migraine. Wouldn't it just be easier to use the actual facts to prove if the most expensive QB's are worth it?
2018 - Of the 40 QB's offered in the contest, 3 didn't make it to the final cut (Brissett, McCarron, and McCown), but were only owned by 60 entrants, and were all only $3 each. Of the 37 remaining, the lowest survival rate (.27%) was the most expensive QB ($27). The next lowest survival rate (.33%) belonged to the 4th most expensive QB ($22). The most telling stat for me - of the top 5 most expensive QB's, only Cam was in the top 8 after 13 weeks. Let's compare that to the other positions after 13 weeks...
RB - top 6 most expensive were Bell, Gurley, Kamara, Zeke, DJ, and Gordon - Bell was a holdout, so I added Gordon as the 5th. Of these 5, 4 were in the top 7 (along with other expensive guys Barkley, CMC, and Bell's replacement)
WR - top 5 most expensive were AB, Nuke, OBJ, Julio, and MT. Of these 5, all were in the top 8, along with Thielen, Davante, and Tyreek, who were also expensive.
TE - top 5 most expensive were Kelce, Gronk, Ertz, Olsen, and Graham. Of these 5, Kelce and Ertz were the top 2, Gronk and Olsen missed multiple games, and Graham was TE10.
Conclusion - The top WR's are the most reliable, followed by RB's. TE's were about 50/50, given the missed games. QB's were by far the least reliable. Perhaps many years ago when only a few QB's were putting up great stats, the top guys would've been worth it. But in today's NFL, even average QB's can put up worthy numbers. Save your QB bucks and upgrade to a stud RB or WR.
2018 - Of the 40 QB's offered in the contest, 3 didn't make it to the final cut (Brissett, McCarron, and McCown), but were only owned by 60 entrants, and were all only $3 each. Of the 37 remaining, the lowest survival rate (.27%) was the most expensive QB ($27). The next lowest survival rate (.33%) belonged to the 4th most expensive QB ($22). The most telling stat for me - of the top 5 most expensive QB's, only Cam was in the top 8 after 13 weeks. Let's compare that to the other positions after 13 weeks...
RB - top 6 most expensive were Bell, Gurley, Kamara, Zeke, DJ, and Gordon - Bell was a holdout, so I added Gordon as the 5th. Of these 5, 4 were in the top 7 (along with other expensive guys Barkley, CMC, and Bell's replacement)
WR - top 5 most expensive were AB, Nuke, OBJ, Julio, and MT. Of these 5, all were in the top 8, along with Thielen, Davante, and Tyreek, who were also expensive.
TE - top 5 most expensive were Kelce, Gronk, Ertz, Olsen, and Graham. Of these 5, Kelce and Ertz were the top 2, Gronk and Olsen missed multiple games, and Graham was TE10.
Conclusion - The top WR's are the most reliable, followed by RB's. TE's were about 50/50, given the missed games. QB's were by far the least reliable. Perhaps many years ago when only a few QB's were putting up great stats, the top guys would've been worth it. But in today's NFL, even average QB's can put up worthy numbers. Save your QB bucks and upgrade to a stud RB or WR.
Jersey Jammer
Footballguy
What's the expected release of this?
TheWinz
Footballguy
Last year was the latest in 5 years, on 7 Aug. The previous 4 years were 1Aug, 2 Aug, 3 Aug, and 4 Aug. I am going to assume 7 Aug again, plus or minus a few.
SeniorVBDStudent
Footballguy
Its best ball format so I want to evaluate the chance that qb1 OR qb2 will outperform qb3. So for any qb3 score, its that probability that Qb1 > score OR Qb2 > score AND Qb3 < score.
So maybe its multiplying Qb3 probability times the probability of getting one of two outcomes from Qb1 or Qb2 (across the range of possible Qb3 outcomes)
Ignoratio Elenchi
Footballguy
How can QB3 score ever be less than QB3 score? I haven't closely followed this discussion, I'll go back and re-read it at some point because I think what you're trying to do is (relatively) simple. Basically, given the probability distributions of the 2 cheaper QBs you can derive the distribution of max(QB1, QB2), and then compare that to the distribution for QB3. The addition rule you mentioned deals with something different.
rzrback77
Footballguy
Ok, I thnk I have a clearer sense of this, and the answer is not actually as simple as any single formula.
I thought we were still talking about the chance that at least one of two QBs exceed a specified fixed value. In this case, you cannot just add the chances together, and the 1-(chance of neither making the threshold) is the easy way. Longer-winded explanation hidden:
Let’s say the target is 20 fp and that each player has a 45% chance of exceeding it. Obviously the chances that at least one scores 20 or more is not 45+45=90%. This is because adding counts cases where both QBs exceed the target as two successes instead of 1. You can figure out the odds that both QBs exceed the target by multiplying their chances: .45 x .45 = .2025. Subtracting this from the above calculated 90% gives you the real answer of .6975.
Alternatively, you can get there by starting with the chances that the first player makes it (45%). Then consider the chances that the second one makes it WHEN the first one didn’t. This is the product of the two probabilities: .55 for the first missing, and .45 for the second making it: .2475. Add this to the original .45 and you get the correct .6975. (sorry, couldnt get spoiler tags to work.)
The new question, odds that one of QB1 or QB2 outscore a third random event in the score of QB3 is too complex for a simple formula. (Yes, statisticians consider these kinds of formulas “simple” at least by comparison.)
This requires we know, or assume the distribution of scores around their mean. A normal distribution (bell curve) is usually a good assumption for random events, and should be decent for an FF projection that isn’t accounting for opponent, rest days, weather, etc.
Caclulation of the lieklihood of the result of an event with one distribution exceeding an event from another requires the use of normal probability calculations AND we need an estimate of the standard deviation of both distributions.
It is an interesting enough calculation that I am willing to explore it (and subject my students to it this fall) but am presently out of the country and lack a good calculator. Feel free to remind me in a week or two if you haven’t had your draft by then.
I thought we were still talking about the chance that at least one of two QBs exceed a specified fixed value. In this case, you cannot just add the chances together, and the 1-(chance of neither making the threshold) is the easy way. Longer-winded explanation hidden:
Let’s say the target is 20 fp and that each player has a 45% chance of exceeding it. Obviously the chances that at least one scores 20 or more is not 45+45=90%. This is because adding counts cases where both QBs exceed the target as two successes instead of 1. You can figure out the odds that both QBs exceed the target by multiplying their chances: .45 x .45 = .2025. Subtracting this from the above calculated 90% gives you the real answer of .6975.
Alternatively, you can get there by starting with the chances that the first player makes it (45%). Then consider the chances that the second one makes it WHEN the first one didn’t. This is the product of the two probabilities: .55 for the first missing, and .45 for the second making it: .2475. Add this to the original .45 and you get the correct .6975. (sorry, couldnt get spoiler tags to work.)
The new question, odds that one of QB1 or QB2 outscore a third random event in the score of QB3 is too complex for a simple formula. (Yes, statisticians consider these kinds of formulas “simple” at least by comparison.)
This requires we know, or assume the distribution of scores around their mean. A normal distribution (bell curve) is usually a good assumption for random events, and should be decent for an FF projection that isn’t accounting for opponent, rest days, weather, etc.
Caclulation of the lieklihood of the result of an event with one distribution exceeding an event from another requires the use of normal probability calculations AND we need an estimate of the standard deviation of both distributions.
It is an interesting enough calculation that I am willing to explore it (and subject my students to it this fall) but am presently out of the country and lack a good calculator. Feel free to remind me in a week or two if you haven’t had your draft by then.
Last edited by a moderator:
Ignoratio Elenchi
Footballguy
I think there are a few things to point out:TheWinz said:
- As someone noted, these are stats for one season, so we can't infer much from them to begin with.
- Survival rate probably usually has a negative correlation with player price because survival rate is usually correlated with roster size.
- Being in the top X isn't necessarily the stat you want. Aaron Rodgers (most expensive QB last year) finished the season 7th in overall points, but 3rd in terms of lowest-scoring-week. In other words, his floor was 3rd highest among all QBs in a contest where for most of the year the goal is to stay out of the bottom, not be at the top. So "reliable" is probably the wrong word for what you're looking at (or rather, reliable is the right word but you're looking at the wrong stats).
- Comparing individual player ranks isn't really the point anyway. In general higher priced players are going to do better than lower priced players at all positions, because Dodds is pretty good at player pricing in aggregate. The fact that the highest priced WRs ended the season near the top doesn't tell us much about their value in best ball. Antonio Brown was 3rd overall WR but he cost $37 - how many combinations of 2+ cheaper WRs did he outperform? WR is really deep, higher variance, and optimal rosters tend to have 7-8 of them. Is $37 for one of them worth it?
- More generally, you'd want to model weekly scoring by position and price point - last year "pair of cheaper QBs" was a dominant strategy because Mahomes was $11 and Mahomes + literally any other QB likely destroyed any other combo you could come up with. But Mariota was also $11 last year. So what we really want to know is, in general, how well does a pair of $11-ish QBs stack up against $22-ish QBs? Etc.
Last edited by a moderator:
TheWinz
Footballguy
Point 1 - you are right - I only researched 2018 survival rates, so all my conclusions are based on just 2018
Point 2 - I was only pointing out that in 2018, the top QB's had the lowest survival rate of all positions. The average was 1.67% - top 5 QB avg was .79%, top 5 RB was 1.55%, and top 5 WR was 1.32%
Point 3 - Reliable was the wrong word for me to use. Thanx for pointing that out. But even as reliable as Aaron was in terms of having a high floor, he still had the lowest survival rate of all QB's (.27%). I have to assume the reason for this is because people could use the saved money to upgrade from an average guy to a stud at another position.
Point 4 - I did a quick research of the top 50 from last year, and 90% had at least 1 WR $22 or over, and of those, 66% had another WR at least $21. Antonio Brown actually had a higher survival rate (1.83%) than the average (1.67%). No other top postional player met the average (Rodgers - .27%, Gurley - 1.59%, Gronk - .73%, Zuerlein - 1.41%, JAX - 1.01%)
Point 5 - agree 100%, but I can only do this for 2018 using the contest querier. I wish I could find the querier for previous years for a larger sample.
Great discussion
SeniorVBDStudent
Footballguy
I really appreciate the response!Arodin said:
And I encourage those getting a migraine to ignore the math posts.
I agree that 45+45 is wrong (that is only valid for mutually exclusive events). And I agree that .45*.45 gives the chance that both exceed a fixed value.
Part 1
But for bestball, what I want is the chance that either one exceeds the fixed value (i.e. A OR B). The probability for A OR B must deduct the overlap (think overlapping circles...the overlap is counted twice so it must be subtracted once). There are countless google results that explain this with Venn diagrams (here's one: https://bolt.mph.ufl.edu/6050-6052/unit-3/module-6/)
The correct answer is .45 + .45 - (.45*.45). This is the General Addition Rule (for two variables).
That's the first part of the problem (figuring out the probability that two QBs can exceed a fixed score.)
Part 2
But the third QB that you want to compare 2 lower valued QBs has a range of values. Let's use a fictional example: QB3 average = 31 standard deviation = 7. So 95% of the time QB3 will score between 17 and 45 points (avg +/- 2 std dev).
Case 1: What is the chance one of the two lower tier QBs scores > 17 points at the same time the elite QB scores < 17? (I'm starting with Case 1 as the lowest elite QB score expected ignoring everything outside 2 standard deviations from the mean).
This is (A OR B >17) AND C < 17. The stuff in ( ) is solvable with the General Addition Rule and the C < 17 probability is 2.5% (I skipped the z table here because we know the value at "avg - 2 std dev" is half of 100-95%).
Repeat this analysis for 18, 19, ....., 31, ....., 43, 44, 45 points (Cases 2 - 18)
Part 3
Get the probability associated with the elite QB scoring each of the Case points (working left to right on the probability curve, 2.5%, 6%-2.5%, 14%-6%, etc.).
Multiply each case probability times the elite QB scoring probability and sum.
This is the net probability that the two lower tier QBs will outperform the elite QB in best ball format.
Epilog
I'm pretty sure both that (a) this is correct to within the approximations made, and that (b) there are much less brute force ways of doing this.
I started this effort looking for a way to expand the linear maximization model I build last year, but obviously this is hopelessly non-linear so it became more of an academic entertainment exercise.
Ignoratio Elenchi
Footballguy
As I said, I'm guessing there's a generally a slightly negative correlation between player price and survival, which probably contributes a little to this. High-priced players will be more commonly found on small rosters, and small rosters have low survival rates.
And of course, he underperformed relative to his price - he was the highest-priced QB and he was not the top scoring QB. Obviously that will negatively impact survival rate. But that's not an indictment of high-priced QBs as a general strategy, that's just one data point in a year when a popular inexpensive QB went absolutely bananas. I make +EV bets that lose all the time, it doesn't mean it was a bad bet, just means it didn't work out that time. Picking Rodgers last year didn't work out, but that doesn't necessarily mean it was a bad strategy. I'm not saying it was good either, I genuinely don't know, I'm just saying there should be more focus on process, not results.
What percentage of teams in the whole contest met those criteria? $21 was the 20th highest-priced WR last year. It wouldn't surprise me that most people picked at least one of the top 20 WRs, and therefore a large % of finalists have one. I think historically most teams in the final 250 have had 18-person rosters, but that's because every year most teams in the whole contest have 18-person rosters, not because 18-person rosters are good at surviving the regular season.
Again, not surprising, but all that tells us is that Antonio Brown had a good year last year. He scored 15+ points every single week during the regular season (excluding his early-ish bye week, obviously). It doesn't really tell us much at all about whether paying up for a "stud" WR is a good strategy in general, i.e. how likely are you to get that performance by paying for a typical $37 WR, and how does it compare to the kind of performance you'd expect to get on average from a similarly-priced combination of cheaper WRs? (And of course it all has to fit into the larger context as well - what's the right number of WRs to take, and how much is the right amount to spend on WRs in total? Etc.)
SeniorVBDStudent
Footballguy
At the time I wasn't thinking this all the way through.
Now I would say it could be one of two approaches...a "broad brush" approach using tiers to reach "directionally correct" conclusions or (and this is more interesting to me) a rigorous (coded) evaluation of all (reasonable) roster combinations (say QB <=3, TE <=3, RB<=7, WR<=11).
So, as an example, you could evaluate whether ((QBA OR QBB) AND TEA AND (WR1A OR WR1B) AND (WR2A OR WR2B OR WR2C)) is expected to outperform QBC AND TEB AND WR1C AND WR2D. The computations would become quite large as even this singe case would involve 18 x 18 x 18 x 18 cases (by analogy with my prior 18 "cases" of elite QB scoring).
Hence my earlier statement about being pretty sure there is a more elegant and rigorous way of approaching this problem involving some protracted combination of linear algebra, probability integration, discrete distributions and programming but that's beyond me.
TheWinz
Footballguy
It will take a while, but I am willing to explore this one using the contest querier that is still available online, because I think a lot less than 66% spent at least $21 on their second WR. Of course, the results will only reflect 2018, but I think it's worth knowing because it will be a very large sample size. Stay tuned for the results on this one - hopefully I get it done before the contest opens, because once it does, the 2018 querier will be replaced by the 2019 querier.
Ignoratio Elenchi
Footballguy
Well in a sense it's just a sample size of one. The season played out the way it did, but it could have played out a practically infinite number of other ways. Over 4,000 rosters had Mahomes last year, but if we're trying to determine something like "the expected value of an $11 QB" that's not a sample of 4,000+, it's just 1.
https://math.stackexchange.com/questions/2613719/how-do-you-calculate-the-probability-of-the-difference-between-two-normal-distri
Maybe this link helps. If you look down that page to the ‘alternate problem’ it has steps to determine probability between two groups of means/std. (Note it is looking for the difference to be >2 so you are just looking for it to be >0). Also of course I do understand you have two QBs trying to outscore a third qb so you have that general addition rule you mentioned.
I was looking at specific numbers for this from last year ($14 Ben/$13 Rivers vs $27 Rodgers). It’s probably not a great example since Ben had pretty good fantasy year and Rodgers underperformed compared to their prices. The QB combo outscored Aaron 11 of the 15 weeks (ignoring Rodgers bye). I looked for similar priced QB that didn’t do as well as Ben. So even if you picked $15 Stafford / $13 Rivers they outscored Aaron 9 of 15 weeks. And interestingly for this duo QB when you take their weekly high score and average that over the year it is less than Aaron’s average (and has smaller std dev). The other big factor for the contest though would be the last 3 week total. Aaron outscored the Ben duo by a little and the Stafford duo by a lot.
if you pick $25 Watson (instead of Rodgers) he beat the duos 8 of 15 weeks...
Ignoratio Elenchi
Footballguy
Not 100% sure I follow your examples but I think sounds like we're talking about the same thing. Calculating 18 individual "cases" is basically just looking at discrete probability distributions, and while there might be a more "elegant" way to do it, your method may be computationally simplest (e.g. in Excel).
Ultimately if you get too specific it becomes both prohibitively difficult and practically useless. We wouldn't have a lot of confidence in the difference between a $24 QB and a $21 QB, for example. That's why I'd probably just bucket them into broader tiers and do relatively simple math on them, there's probably no incremental benefit to getting too sophisticated with it. Plus there are other complications, like the fact that fantasy scores aren't normally distributed, they're not all truly independent, etc. that will just further complicate your analysis if you try to account for them, or weaken it if you don't. I think the goal should just be to develop a small set of heuristics that guide your roster construction, e.g. "2-3 QBs is optimal," "$35-50 spent on QBs is optimal," etc.
TheWinz
Footballguy
Are you saying that if I go through all 15,115 entries from last year, and find that people who chose 2 top 20 most expensive WR's had a better survival rate than those who didn't, that the data would still be fairly useless, because the 2018 season was what it was, and is in no way predictive of future outcomes? I really would like to know your thought on this before I dig so deep. The data is still there going back many years, but without a querier for each of those seasons, I literally have to sift through every entrant which would take forever.
Ignoratio Elenchi
Footballguy
To a large extent, yes. If you roll a die 1,000 times and it comes up 6 every time, you'd know quite a lot about the probability of rolling a 6. If you roll it once and project that single roll onto 1,000 different screens, you've learned very little. 1,000 entries in last years contest is much more like the latter than the former.
Edit: as an aside, I had records of every contest roster for the last 10 years or so and recently discovered that I lost all that data. Not sure if it can be recovered, but if anyone else has anything similar and wants to send me a data dump shoot me a message.
Last edited by a moderator:
SeniorVBDStudent
Footballguy
Agree with the bolded.
In principle, I would not apply any analysis to historical average and standard deviation. Rather this would be the average and variance of the various 2019 projections for players to be considered. Websites with a statistical slant towards DFS and fantasy in general highlight two relevant points: that 2018 is not a proxy for 2019 and that "wisdom of the crowd" is the most reliably accurate projection source (i.e. while Dodds may be more accurate than Smith or Jones, the consensus of three is likely to be more accurate than any of the three), so its a logical inference to analyze the variance of the projections vs the variance of the history.
Second, I'm pretty sure I won't be launching the comprehensive coding effort I described as being theoretically possible, but I might use this as a "tiebreaker" such as if I'm on the fence to take Mahommes at $32 or Cousins at $21 and Jabroney at $11.
I know this wasn’t the question but using this stats example to find probability qb2 outscores qb1 (using link from my above post) :
26-22 = 4
(11^2+7^2)^0.5 =13
D > 0, standardize D =0
0 - 4 = -4 then divide by 13......... z = -0.31
probability from table: ~38% QB2 outscores QB1 ??
Last edited by a moderator:
Ignoratio Elenchi
Footballguy
I think the question in that example is if you rostered both of those QBs, what's the probability they'd outscore a 3rd QB with, say, mean = 31 and stdev = 7 or whatever.
Basically you have pdfs for QBs A, B and C. And you want to derive the pdf of max(A,B) and compare to C. So in this example, QB1 has mean 26 and stdev 11, and QB2 has mean 22 and stdev 7. If you roster both of them, you'd expect something like an average score of 30 with a stdev of 9 or whatever (just back of envelope math, but effectively you get some distribution for max(QB1,QB2) in a given week). That's what you're then comparing to the distribution for QB3.
You also have to consider that the QB1-QB2 pair will have two weeks with lower expectation (because one of the two will be on bye), and of course QB3 will have one week with 0 expectation (when he's on bye) which you'd have to account for. Plus as I said they're probably not normally distributed, and may not be entirely independent, and you have to take bye week distribution and projected ownership into consideration, and projections are squishy to begin with... all of which means it's probably not worth going too deep into the statistical methods because the results will be very rough guidelines at best, that we could probably approximate in a much simpler manner anyway.
TheWinz
Footballguy
This is what I am trying to do for each position. And to then further break it down by individual dollar amounts. For example, let's say 7 WR's and $84 total is the "sweet spot". How would that $84 best be broken down? 7 guys all around $12? 2 really expensive guys and 5 dart throws? 2 guys around $20 and 5 guys in the $9 range? Of course, every year is different, and there's no telling if a $3 WR like Boyd will come out of nowhere and finish as WR16.
I guess, in the end, each season is it's own season, and any outcome is possible. Who woulda thought that the top finisher didn't roster James Conner, and had to suffer through 13 weeks with Alex Collins and Derrick Henry, or the 2nd place finisher would've had a QB corps of Trubisky, Dalton, and Darnold, with his top RB as Leonard Fournette?
SeniorVBDStudent
Footballguy
Thanks for the link!
Having perused google, there are plenty of writeups for expectation value of max(X,Y) over range [a,b] but its assumed in the references that I looked at that X and Y are identically distributed, which is not the case with two quarterbacks. These references provide the more general answer, but in the form of integrals of pdfs, etc etc. <- not helpful.
Similar threads
