patjams
patjams
Joined: Dec 8, 2013
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December 8th, 2013 at 9:27:33 PM permalink
I was perusing the web tonight looking for some help and came across the site and immediately bookmarked it. Great site and looks to be a great community and I hope that someone here is kind enough to throw me an assist as I am in no way a stats/math guy although I am trying to learn.

My question concerns odds/probability (not sure exactly which). I am in search of some help with the following (forgive me if I am leaving any factors out):

I was playing a fantasy basketball game this weekend with 3 other people (4 total).

There were 2 games (4 teams) on the schedule from which to select a team of 8 players each (all players are available to all 4 teams & selections are not disclosed until the contest locks at which point all teams/players are visible to everyone).

The total player pool from the 4 teams from which to select players is 60 (15 per team).

My question is, what are the odds (or what is the probability) that 2 of the 4 teams entered in the contest would select the exact same 8 players from the pool of 60 players?

Essentially I was having the discussion with another player that there had to be collusion among the two teams who selected the same players as the odds are simply far too great otherwise. I simply could not make my argument as firmly as I wanted to as I do not know the calculations, algorithms, etc... necessary to do so and I would really like to be able to figure it out. Although I know the odds are not as extreme as they would be in one of our "normal" contests (which usually include 10 games or more) I still feel like they are pretty high.

I appreciate in advance any assistance/response as it will be greatly appreciated. Thanks for reading and keep up all the great work at the site!
Mission146
Mission146
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December 8th, 2013 at 9:48:16 PM permalink
Quote: patjams



My question is, what are the odds (or what is the probability) that 2 of the 4 teams entered in the contest would select the exact same 8 players from the pool of 60 players?

Essentially I was having the discussion with another player that there had to be collusion among the two teams who selected the same players as the odds are simply far too great otherwise. I simply could not make my argument as firmly as I wanted to as I do not know the calculations, algorithms, etc... necessary to do so and I would really like to be able to figure it out. Although I know the odds are not as extreme as they would be in one of our "normal" contests (which usually include 10 games or more) I still feel like they are pretty high.

I appreciate in advance any assistance/response as it will be greatly appreciated. Thanks for reading and keep up all the great work at the site!



This is a really difficult question because you do have sixty players, as you mentioned, but of those sixty players, there have to be at least a couple that are going to be completely obvious choices and a greater number of players that nobody is ever going to pick.

Basically, the first thing you'd want to do is look at how many players could conceivably be picked, which is already a subjective measure, and you would eliminate any players that it would be absolutely ridiculous to pick from the sixty.

After that, I'd probably want to look at players from the four teams that would definitely be picked. For example, if one of the teams is the Miami Heat, and LeBron James is healthy, I know just enough about the NBA to assume everyone would pick that player, lest they start at a huge disadvantage.

Essentially, then, you're going to be trimming the eight possible selections down to something less than eight by filling roster spots with players that everyone is absolutely going to take, if applicable, so there are effectively less than eight selections. Those players are also subtracted from the pool of sixty players, leaving less than sixty players. After that, you'll subtract the players that absolutely will not be picked from the pool of sixty, as being very close to impossible.

Finally, a question: What competitive advantage could picking the exact same team possibly offer two different players? If two players were colluding, I would assume that they would pick either slightly different or very different teams and agree to just split the winnings if either team were to win.
Vultures can't be choosers.
Mission146
Mission146
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December 8th, 2013 at 10:09:01 PM permalink
Additionally, after you have eliminated the number of selections/team by determining how many 100%ers there are out there, and reduced the potential number of players that could be selected, (0%ers) then you have to decide whether you want to give every player an equal shot of being picked, or whether or not you want to assign probabilities of certain players being picked as opposed to others.

If every player has an equal shot of being picked, in your view, I could do the math behind it, but you are probably going to get a somewhat less likely result than what is actually the case if the players are weighted. Is this game just between the four of you or are there many others who play the same game under the same parameters? If there are others playing under the same parameters and statistics are available for what % of people picked who, then I would just go with those and that will result in what I would consider about as close as you'd get.

Someone could also come up with Stats and Match-Up based probabilities, I guess, but those would be fairly subjective. Best to know who all players picked, assuming a really strong sample size.
Vultures can't be choosers.
patjams
patjams
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December 8th, 2013 at 10:26:18 PM permalink
Thanks a mil mission as you are absolutely correct in the assumption that there are several players everyone would (and did) take as well as some that nobody would (and therefore none did). The contest was a college basketball event that included 4 teams (Rutgers, Seton Hall, USC, Boston College) and of those 4 teams there were 4 players that were 100% owned and several that were not owned at all.

The advantage that these 2 teams would have if they indeed colluded is that this was what is called a 50/50 game in which the top 50% win and bottom 50% lose. In this case the top 50% (2) were the 2 teams with the exact same roster. The only reason I am so staunch that they colluded is that outside of the 4 obvious guys who were 100% owned, the rest of the players are fairly average and as such "average" about the same amount of points therefore I simply felt it was strange that these 2 teams selected the same 4 "average" guys from a pool of roughly 36.

Taking your response/assistance into account then and trimming the "field" by 24 obvious (4 positive, 20 neg) that would leave me with 36 players for 4 spots which is what I'm having such a hard time with as the other "loser" in the contest and I didn't share a single other player with each other or the other 2. Now that I have examined a bit more closely I don't feel such an absolution about the collusion but I still think it's fishy.

As I said, I really appreciate the response and am thankful to have found the site as I think it will help me towards grasping the subject of statistical analysis so much better than I do. Looking forward to may future exchanges. Be well and have a great evening!
Mission146
Mission146
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December 9th, 2013 at 1:00:06 AM permalink
Quote: patjams

Thanks a mil mission as you are absolutely correct in the assumption that there are several players everyone would (and did) take as well as some that nobody would (and therefore none did). The contest was a college basketball event that included 4 teams (Rutgers, Seton Hall, USC, Boston College) and of those 4 teams there were 4 players that were 100% owned and several that were not owned at all.



You're welcome, good on that, so far.

Quote:

The advantage that these 2 teams would have if they indeed colluded is that this was what is called a 50/50 game in which the top 50% win and bottom 50% lose. In this case the top 50% (2) were the 2 teams with the exact same roster.



I guess I still don't see where the advantage is. It seems that all you would do is either win entirely or lose entirely, which is something that they would have done (individually) in the first place. If they merely wanted to, "Hedge their picks," so to speak, then they would pick differently but agree to share whatever the price is if only one wins.

Quote:

The only reason I am so staunch that they colluded is that outside of the 4 obvious guys who were 100% owned, the rest of the players are fairly average and as such "average" about the same amount of points therefore I simply felt it was strange that these 2 teams selected the same 4 "average" guys from a pool of roughly 36.

Taking your response/assistance into account then and trimming the "field" by 24 obvious (4 positive, 20 neg) that would leave me with 36 players for 4 spots which is what I'm having such a hard time with as the other "loser" in the contest and I didn't share a single other player with each other or the other 2. Now that I have examined a bit more closely I don't feel such an absolution about the collusion but I still think it's fishy.



In this case, we are looking at a pool of 36 possible players to choose from and we are weighting all of them equally. The first thing that needs to be considered is that you have two guys who both must pick players. In that sense, we're going to assign a probability of 100% to one of them because they both have to pick players no matter what happens.

Given that Player A must pick players and we assign Player A 100%, then for Player B to pick the same players means he must pick one of 4/36, then one of 3/35, then one of 2/34 and then one of 1/33.

I do still think thirty-six seems a bit high, though. It seems that only twenty total players would be starters, did everyone pick all starters?

In any case, for Player B to do the same thing as Player A, seemingly at random:

4/36 * 3/35 * 2/34 * 1/33 * 4 = 0.00006790595 or 1/0.00006790595 = 1 in 14,726

I still don't think the odds are that long, though, thirty-six players to have an equal chance seems really high. That's nine players from each team.

Looking at Rutgers game-by-game, they only have anywhere from 6-8 guys they have even played for ten or more minutes in any given game this year.

Seton Hall only has 7-8 guy in double-digits for minutes played.

I could be wrong, but not even delving that deeply into it, I think your 36 needs trimmed a bit more if yourself and these two guys are all knowledgeable about basketball. Just looking at those two teams only yields Max eight players per game to play ten minutes, or more, and there's usually one of the eight that only played 10-12 minutes. I think with a little bit of research, and assuming knowledgeable players, we can probably get the list down closer to 24.

Quote:

As I said, I really appreciate the response and am thankful to have found the site as I think it will help me towards grasping the subject of statistical analysis so much better than I do. Looking forward to may future exchanges. Be well and have a great evening!



Thanks for the compliments, you as well!
Vultures can't be choosers.
beachbumbabs
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beachbumbabs
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December 9th, 2013 at 10:19:17 AM permalink
Quote: patjams

Thanks a mil mission as you are absolutely correct in the assumption that there are several players everyone would (and did) take as well as some that nobody would (and therefore none did). The contest was a college basketball event that included 4 teams (Rutgers, Seton Hall, USC, Boston College) and of those 4 teams there were 4 players that were 100% owned and several that were not owned at all.

The advantage that these 2 teams would have if they indeed colluded is that this was what is called a 50/50 game in which the top 50% win and bottom 50% lose. In this case the top 50% (2) were the 2 teams with the exact same roster. The only reason I am so staunch that they colluded is that outside of the 4 obvious guys who were 100% owned, the rest of the players are fairly average and as such "average" about the same amount of points therefore I simply felt it was strange that these 2 teams selected the same 4 "average" guys from a pool of roughly 36.

Taking your response/assistance into account then and trimming the "field" by 24 obvious (4 positive, 20 neg) that would leave me with 36 players for 4 spots which is what I'm having such a hard time with as the other "loser" in the contest and I didn't share a single other player with each other or the other 2. Now that I have examined a bit more closely I don't feel such an absolution about the collusion but I still think it's fishy.

As I said, I really appreciate the response and am thankful to have found the site as I think it will help me towards grasping the subject of statistical analysis so much better than I do. Looking forward to may future exchanges. Be well and have a great evening!



patjams,

Not negating anything you and Mission have said, but there could be an outside influence that's well short of collusion. There are a lot of fantasy leagues, and so more than a few websites that support creating an ideal fantasy team. If you (or the 2 players you think were in collusion) were to google one of those that allows a "create your own fantasy league", within the same day, it's likely you would receive similar if not identical links. They both go to the same link, they both put in the same parameters because your group isolated 4 specific teams, and voila! They both get the same optimized list that they then put into your draft. Since you didn't do an exclusionary draft (where you go in rounds and a player chosen by one was not available to the rest), it seems to me it's not so much the math as the method, and it's more likely than not that you all would have extremely similar lists. No collusion required, no cheating, just semi-coincidence that they used the same or similar website to do the calculation (there are only so many variables such a site would use, and they're relatively common factors).
If the House lost every hand, they wouldn't deal the game.

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