The Probability that Eric Reid (NFL Player) is being targeted with 6 drug Test in 11 weeks.
$begingroup$
Ok, I thought this was interesting and wanted to see if anyone could figure it out cause the media can't.
Here is the situation,
Eric Reed is an NFL player, He has been in the NFL for 11 weeks making him subject to 11 rounds of "Random" drug test.
-He has been selected for 6 of the 11 possible test.
-There are approximately 72 testable players on a team and each week 10 are selected at random.
-There are 32 teams in the league.
So how would you go about finding the probability that this player was not chosen randomly but was targeted.
Also please feel free to expand on this question in anyway you feel is appropriate as this is not my wheelhouse and I'm sure you could change the question to be more meaningful and statistically significant if you wanted to and that's what I'm really after... this will be linked to in a few forums for people who don't necessarily use math lingo so please feel free to draw conclusions or give simplified perspectives.
Thank You!!
probability binomial-distribution
edited Dec 21 '18 at 1:22
Ethan Bolker
46k553120
asked Dec 21 '18 at 0:15
FrostFrost
141
$endgroup$
|
show 1 more comment
$begingroup$
Ok, I thought this was interesting and wanted to see if anyone could figure it out cause the media can't.
Here is the situation,
Eric Reed is an NFL player, He has been in the NFL for 11 weeks making him subject to 11 rounds of "Random" drug test.
-He has been selected for 6 of the 11 possible test.
-There are approximately 72 testable players on a team and each week 10 are selected at random.
-There are 32 teams in the league.
So how would you go about finding the probability that this player was not chosen randomly but was targeted.
Also please feel free to expand on this question in anyway you feel is appropriate as this is not my wheelhouse and I'm sure you could change the question to be more meaningful and statistically significant if you wanted to and that's what I'm really after... this will be linked to in a few forums for people who don't necessarily use math lingo so please feel free to draw conclusions or give simplified perspectives.
Thank You!!
probability binomial-distribution
edited Dec 21 '18 at 1:22
Ethan Bolker
46k553120
asked Dec 21 '18 at 0:15
FrostFrost
141
$endgroup$
$begingroup$
"I thought this was interesting and wanted to see if anyone could figure it out cause the media can't" - since when is the media correct with mathematics? :)
$endgroup$
– Math1000
Dec 21 '18 at 0:29
$begingroup$
Anything with non zero probability will happen somewhere given enough time.
$endgroup$
– ablmf
Dec 21 '18 at 0:41
1
$begingroup$
Three different things: (1) The probability that Eric Reid is chosen ${} ge 6$ out of 11 times; (2) the probability that there is a player chosen ${} ge 6$ out of 11 times. (3) The probability that somebody somewhere endures an event with probability ${} le 3 cdot 10^{-12}$. The point is: the OP picked one event that seemed unlikely out of all events from recent weeks.
$endgroup$
– GEdgar
Dec 21 '18 at 1:05
1
$begingroup$
I googled about, and found that someone wrote an article about this, including probabilities. They have somewhat different numbers (e.g. they say $53$ active players instead of $72$ players), so get a different answer. (Note that this isn't the probability that the player was/wasn't targeted - that question is beyond the scope of purely mathematical logic)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:12
1
$begingroup$
@MiloBrandt, there are 53 "active" players but there are more testable players, this includes the players on the practice squad. Now... why they care if players on a practice squad are juicing even if they never make it to the active roster is weird but that is the official policy from what I read. And for the record "72" is actually not exactly true, it can be 73 or less but I didn't want to bring that in as a variable.
$endgroup$
– Frost
Dec 21 '18 at 1:53
|
show 1 more comment
$begingroup$
Ok, I thought this was interesting and wanted to see if anyone could figure it out cause the media can't.
Here is the situation,
Eric Reed is an NFL player, He has been in the NFL for 11 weeks making him subject to 11 rounds of "Random" drug test.
-He has been selected for 6 of the 11 possible test.
-There are approximately 72 testable players on a team and each week 10 are selected at random.
-There are 32 teams in the league.
So how would you go about finding the probability that this player was not chosen randomly but was targeted.
Also please feel free to expand on this question in anyway you feel is appropriate as this is not my wheelhouse and I'm sure you could change the question to be more meaningful and statistically significant if you wanted to and that's what I'm really after... this will be linked to in a few forums for people who don't necessarily use math lingo so please feel free to draw conclusions or give simplified perspectives.
Thank You!!
probability binomial-distribution
edited Dec 21 '18 at 1:22
Ethan Bolker
46k553120
asked Dec 21 '18 at 0:15
FrostFrost
141
$endgroup$
Ok, I thought this was interesting and wanted to see if anyone could figure it out cause the media can't.
Here is the situation,
Eric Reed is an NFL player, He has been in the NFL for 11 weeks making him subject to 11 rounds of "Random" drug test.
-He has been selected for 6 of the 11 possible test.
-There are approximately 72 testable players on a team and each week 10 are selected at random.
-There are 32 teams in the league.
So how would you go about finding the probability that this player was not chosen randomly but was targeted.
Also please feel free to expand on this question in anyway you feel is appropriate as this is not my wheelhouse and I'm sure you could change the question to be more meaningful and statistically significant if you wanted to and that's what I'm really after... this will be linked to in a few forums for people who don't necessarily use math lingo so please feel free to draw conclusions or give simplified perspectives.
Thank You!!
probability binomial-distribution
probability binomial-distribution
edited Dec 21 '18 at 1:22
Ethan Bolker
46k553120
asked Dec 21 '18 at 0:15
FrostFrost
141
edited Dec 21 '18 at 1:22
Ethan Bolker
46k553120
asked Dec 21 '18 at 0:15
FrostFrost
141
edited Dec 21 '18 at 1:22
Ethan Bolker
46k553120
edited Dec 21 '18 at 1:22
Ethan Bolker
46k553120
edited Dec 21 '18 at 1:22
Ethan Bolker
46k553120
46k553120
asked Dec 21 '18 at 0:15
FrostFrost
141
asked Dec 21 '18 at 0:15
FrostFrost
141
asked Dec 21 '18 at 0:15
FrostFrost
141
141
$begingroup$
"I thought this was interesting and wanted to see if anyone could figure it out cause the media can't" - since when is the media correct with mathematics? :)
$endgroup$
– Math1000
Dec 21 '18 at 0:29
$begingroup$
Anything with non zero probability will happen somewhere given enough time.
$endgroup$
– ablmf
Dec 21 '18 at 0:41
1
$begingroup$
Three different things: (1) The probability that Eric Reid is chosen ${} ge 6$ out of 11 times; (2) the probability that there is a player chosen ${} ge 6$ out of 11 times. (3) The probability that somebody somewhere endures an event with probability ${} le 3 cdot 10^{-12}$. The point is: the OP picked one event that seemed unlikely out of all events from recent weeks.
$endgroup$
– GEdgar
Dec 21 '18 at 1:05
1
$begingroup$
I googled about, and found that someone wrote an article about this, including probabilities. They have somewhat different numbers (e.g. they say $53$ active players instead of $72$ players), so get a different answer. (Note that this isn't the probability that the player was/wasn't targeted - that question is beyond the scope of purely mathematical logic)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:12
1
$begingroup$
@MiloBrandt, there are 53 "active" players but there are more testable players, this includes the players on the practice squad. Now... why they care if players on a practice squad are juicing even if they never make it to the active roster is weird but that is the official policy from what I read. And for the record "72" is actually not exactly true, it can be 73 or less but I didn't want to bring that in as a variable.
$endgroup$
– Frost
Dec 21 '18 at 1:53
|
show 1 more comment
$begingroup$
"I thought this was interesting and wanted to see if anyone could figure it out cause the media can't" - since when is the media correct with mathematics? :)
$endgroup$
– Math1000
Dec 21 '18 at 0:29
$begingroup$
Anything with non zero probability will happen somewhere given enough time.
$endgroup$
– ablmf
Dec 21 '18 at 0:41
1
$begingroup$
Three different things: (1) The probability that Eric Reid is chosen ${} ge 6$ out of 11 times; (2) the probability that there is a player chosen ${} ge 6$ out of 11 times. (3) The probability that somebody somewhere endures an event with probability ${} le 3 cdot 10^{-12}$. The point is: the OP picked one event that seemed unlikely out of all events from recent weeks.
$endgroup$
– GEdgar
Dec 21 '18 at 1:05
1
$begingroup$
I googled about, and found that someone wrote an article about this, including probabilities. They have somewhat different numbers (e.g. they say $53$ active players instead of $72$ players), so get a different answer. (Note that this isn't the probability that the player was/wasn't targeted - that question is beyond the scope of purely mathematical logic)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:12
1
$begingroup$
@MiloBrandt, there are 53 "active" players but there are more testable players, this includes the players on the practice squad. Now... why they care if players on a practice squad are juicing even if they never make it to the active roster is weird but that is the official policy from what I read. And for the record "72" is actually not exactly true, it can be 73 or less but I didn't want to bring that in as a variable.
$endgroup$
– Frost
Dec 21 '18 at 1:53
$begingroup$
"I thought this was interesting and wanted to see if anyone could figure it out cause the media can't" - since when is the media correct with mathematics? :)
$endgroup$
– Math1000
Dec 21 '18 at 0:29
$begingroup$
"I thought this was interesting and wanted to see if anyone could figure it out cause the media can't" - since when is the media correct with mathematics? :)
$endgroup$
– Math1000
Dec 21 '18 at 0:29
$begingroup$
Anything with non zero probability will happen somewhere given enough time.
$endgroup$
– ablmf
Dec 21 '18 at 0:41
$begingroup$
Anything with non zero probability will happen somewhere given enough time.
$endgroup$
– ablmf
Dec 21 '18 at 0:41
1
1
$begingroup$
Three different things: (1) The probability that Eric Reid is chosen ${} ge 6$ out of 11 times; (2) the probability that there is a player chosen ${} ge 6$ out of 11 times. (3) The probability that somebody somewhere endures an event with probability ${} le 3 cdot 10^{-12}$. The point is: the OP picked one event that seemed unlikely out of all events from recent weeks.
$endgroup$
– GEdgar
Dec 21 '18 at 1:05
$begingroup$
Three different things: (1) The probability that Eric Reid is chosen ${} ge 6$ out of 11 times; (2) the probability that there is a player chosen ${} ge 6$ out of 11 times. (3) The probability that somebody somewhere endures an event with probability ${} le 3 cdot 10^{-12}$. The point is: the OP picked one event that seemed unlikely out of all events from recent weeks.
$endgroup$
– GEdgar
Dec 21 '18 at 1:05
1
1
$begingroup$
I googled about, and found that someone wrote an article about this, including probabilities. They have somewhat different numbers (e.g. they say $53$ active players instead of $72$ players), so get a different answer. (Note that this isn't the probability that the player was/wasn't targeted - that question is beyond the scope of purely mathematical logic)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:12
$begingroup$
I googled about, and found that someone wrote an article about this, including probabilities. They have somewhat different numbers (e.g. they say $53$ active players instead of $72$ players), so get a different answer. (Note that this isn't the probability that the player was/wasn't targeted - that question is beyond the scope of purely mathematical logic)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:12
1
1
$begingroup$
@MiloBrandt, there are 53 "active" players but there are more testable players, this includes the players on the practice squad. Now... why they care if players on a practice squad are juicing even if they never make it to the active roster is weird but that is the official policy from what I read. And for the record "72" is actually not exactly true, it can be 73 or less but I didn't want to bring that in as a variable.
$endgroup$
– Frost
Dec 21 '18 at 1:53
$begingroup$
@MiloBrandt, there are 53 "active" players but there are more testable players, this includes the players on the practice squad. Now... why they care if players on a practice squad are juicing even if they never make it to the active roster is weird but that is the official policy from what I read. And for the record "72" is actually not exactly true, it can be 73 or less but I didn't want to bring that in as a variable.
$endgroup$
– Frost
Dec 21 '18 at 1:53
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
A very rough calculation that needs refinement, and a philosophical observation:
For any one player, this is a Bernoulli trial with chance $10/72 = 0.134$ of success (where "success" means "is tested").
The chance of at least $6$ successes in $11$ trials is about $0.00185$ (according to the Bernoulli trial calculator at https://planetcalc.com/7044/). That's just under $0.2%$; which is pretty small. But there are $32$ teams in the NFL, with $72$ players each. That's more than $2000$ players, so it's not very surprising that one of them was tested more than six times. (I haven't calculated that probability, maybe someone else will.)
Looking back at what happened to Eric Reed, you can wonder whether the fact that he was tested often means something about him. But that's like wondering what someone did that was special that led to his winning the lottery. It doesn't take into account the fact that someone will win. With hindsight you can always single out events that seem highly unlikely and ask for explanations.
See
Super bowl box pool odds
Probability of two people from two different countries meeting in a different country and meeting each other
answered Dec 21 '18 at 1:15
Ethan BolkerEthan Bolker
46k553120
$endgroup$
add a comment |
$begingroup$
If everything is uniformly random (which need not be the case), then there are $32cdot 72 = 2304$ total players in the pool for random testing; since $10$ are chosen each week, the probability that a player is not chosen to be tested is
$$
frac{2303}{2304}cdotfrac{2302}{2303}cdotsfrac{2294}{2295} = frac{2294}{2304}
$$
so the probability that the are chosen is $1-frac{2294}{2304}=frac{5}{1152}$. Call this value $p$.
Now we are in a classic Bernoulli trial setup; we have a probability of success (being called for a drug test) and a number of trials (each of the 11 weeks that have passed), and we wish to know the probability that we succeed a certain number of times. Plugging everything in we find that the probability of being called at least $6$ times would be
$$
sum_{k=6}^{11}{11choose k}p^k(1-p)^{11-k} approx 3cdot 10^{-12}
$$
which is a touch on the low side.
What does this mean? Either nothing, or not that much, or it is the scandal of the century. First, everything that is possible is possible (deep wisdom I know). It could be that this player was (un)lucky enough to win this lottery. Second, the procedure may not be uniformly random. While it might strike you as "unfair" to weight some players as riskier than others, actuaries do this all the time with things like drug use, recidivism rates, life expectency, etc. It might not be a stretch to think that the NFL also has a different distribution than uniform randomness for these drug trials. You can ask why they should deviate from uniform randomness, but that would at least help explain why such a "rare" event has happened. And third, it could be that it isn't really random, or someone messed with the system to circumvent randomness, and we have the next scandal on our hands that we can fret over for 24-72 hours.
EDIT: Based on a comment on this answer, I did some quick research and discovered that the testing is not NFL wide as I have assumed, but that each team tests $10$ of their players each week. This gives you a different $p$ value of $5/36$, and the rest of the numbers change accordingly.
answered Dec 21 '18 at 0:50
DeficientMathDudeDeficientMathDude
1112
$endgroup$
$begingroup$
I think the question was saying that $10/72$ players are tested - that is, $10$ are tested for each team, not that $10$ are tested in the whole NFL. (Also, note that, while your product for the probability of an individual being tested in a given week is correct, it's rather inefficient; the product telescopes to just be $frac{t}n$ where $n$ is the pool of testable players and $t$ is the number tested)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:19
$begingroup$
@MiloBrandt If that is the case the math can easily be adjusted to accommodate the different $p$ value. I don't follow any sportsball games, so I assumed the inclusion of the number of teams was meant to inform us that the testing is not per team, but NFL wide. And while it is inefficient for math people, since this question is directed at non math people, I wrote it out long-form.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:20
$begingroup$
Yea, it is 10 players per team, with 72 testable players on each team. Sorry if that wasn't clear.
$endgroup$
– Frost
Dec 21 '18 at 1:50
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
A very rough calculation that needs refinement, and a philosophical observation:
For any one player, this is a Bernoulli trial with chance $10/72 = 0.134$ of success (where "success" means "is tested").
The chance of at least $6$ successes in $11$ trials is about $0.00185$ (according to the Bernoulli trial calculator at https://planetcalc.com/7044/). That's just under $0.2%$; which is pretty small. But there are $32$ teams in the NFL, with $72$ players each. That's more than $2000$ players, so it's not very surprising that one of them was tested more than six times. (I haven't calculated that probability, maybe someone else will.)
Looking back at what happened to Eric Reed, you can wonder whether the fact that he was tested often means something about him. But that's like wondering what someone did that was special that led to his winning the lottery. It doesn't take into account the fact that someone will win. With hindsight you can always single out events that seem highly unlikely and ask for explanations.
See
Super bowl box pool odds
Probability of two people from two different countries meeting in a different country and meeting each other
answered Dec 21 '18 at 1:15
Ethan BolkerEthan Bolker
46k553120
$endgroup$
add a comment |
$begingroup$
A very rough calculation that needs refinement, and a philosophical observation:
For any one player, this is a Bernoulli trial with chance $10/72 = 0.134$ of success (where "success" means "is tested").
The chance of at least $6$ successes in $11$ trials is about $0.00185$ (according to the Bernoulli trial calculator at https://planetcalc.com/7044/). That's just under $0.2%$; which is pretty small. But there are $32$ teams in the NFL, with $72$ players each. That's more than $2000$ players, so it's not very surprising that one of them was tested more than six times. (I haven't calculated that probability, maybe someone else will.)
Looking back at what happened to Eric Reed, you can wonder whether the fact that he was tested often means something about him. But that's like wondering what someone did that was special that led to his winning the lottery. It doesn't take into account the fact that someone will win. With hindsight you can always single out events that seem highly unlikely and ask for explanations.
See
Super bowl box pool odds
Probability of two people from two different countries meeting in a different country and meeting each other
answered Dec 21 '18 at 1:15
Ethan BolkerEthan Bolker
46k553120
$endgroup$
add a comment |
$begingroup$
A very rough calculation that needs refinement, and a philosophical observation:
For any one player, this is a Bernoulli trial with chance $10/72 = 0.134$ of success (where "success" means "is tested").
The chance of at least $6$ successes in $11$ trials is about $0.00185$ (according to the Bernoulli trial calculator at https://planetcalc.com/7044/). That's just under $0.2%$; which is pretty small. But there are $32$ teams in the NFL, with $72$ players each. That's more than $2000$ players, so it's not very surprising that one of them was tested more than six times. (I haven't calculated that probability, maybe someone else will.)
Looking back at what happened to Eric Reed, you can wonder whether the fact that he was tested often means something about him. But that's like wondering what someone did that was special that led to his winning the lottery. It doesn't take into account the fact that someone will win. With hindsight you can always single out events that seem highly unlikely and ask for explanations.
See
Super bowl box pool odds
Probability of two people from two different countries meeting in a different country and meeting each other
answered Dec 21 '18 at 1:15
Ethan BolkerEthan Bolker
46k553120
$endgroup$
A very rough calculation that needs refinement, and a philosophical observation:
For any one player, this is a Bernoulli trial with chance $10/72 = 0.134$ of success (where "success" means "is tested").
The chance of at least $6$ successes in $11$ trials is about $0.00185$ (according to the Bernoulli trial calculator at https://planetcalc.com/7044/). That's just under $0.2%$; which is pretty small. But there are $32$ teams in the NFL, with $72$ players each. That's more than $2000$ players, so it's not very surprising that one of them was tested more than six times. (I haven't calculated that probability, maybe someone else will.)
Looking back at what happened to Eric Reed, you can wonder whether the fact that he was tested often means something about him. But that's like wondering what someone did that was special that led to his winning the lottery. It doesn't take into account the fact that someone will win. With hindsight you can always single out events that seem highly unlikely and ask for explanations.
See
Super bowl box pool odds
Probability of two people from two different countries meeting in a different country and meeting each other
answered Dec 21 '18 at 1:15
Ethan BolkerEthan Bolker
46k553120
answered Dec 21 '18 at 1:15
Ethan BolkerEthan Bolker
46k553120
answered Dec 21 '18 at 1:15
Ethan BolkerEthan Bolker
46k553120
answered Dec 21 '18 at 1:15
Ethan BolkerEthan Bolker
46k553120
46k553120
add a comment |
add a comment |
$begingroup$
If everything is uniformly random (which need not be the case), then there are $32cdot 72 = 2304$ total players in the pool for random testing; since $10$ are chosen each week, the probability that a player is not chosen to be tested is
$$
frac{2303}{2304}cdotfrac{2302}{2303}cdotsfrac{2294}{2295} = frac{2294}{2304}
$$
so the probability that the are chosen is $1-frac{2294}{2304}=frac{5}{1152}$. Call this value $p$.
Now we are in a classic Bernoulli trial setup; we have a probability of success (being called for a drug test) and a number of trials (each of the 11 weeks that have passed), and we wish to know the probability that we succeed a certain number of times. Plugging everything in we find that the probability of being called at least $6$ times would be
$$
sum_{k=6}^{11}{11choose k}p^k(1-p)^{11-k} approx 3cdot 10^{-12}
$$
which is a touch on the low side.
What does this mean? Either nothing, or not that much, or it is the scandal of the century. First, everything that is possible is possible (deep wisdom I know). It could be that this player was (un)lucky enough to win this lottery. Second, the procedure may not be uniformly random. While it might strike you as "unfair" to weight some players as riskier than others, actuaries do this all the time with things like drug use, recidivism rates, life expectency, etc. It might not be a stretch to think that the NFL also has a different distribution than uniform randomness for these drug trials. You can ask why they should deviate from uniform randomness, but that would at least help explain why such a "rare" event has happened. And third, it could be that it isn't really random, or someone messed with the system to circumvent randomness, and we have the next scandal on our hands that we can fret over for 24-72 hours.
EDIT: Based on a comment on this answer, I did some quick research and discovered that the testing is not NFL wide as I have assumed, but that each team tests $10$ of their players each week. This gives you a different $p$ value of $5/36$, and the rest of the numbers change accordingly.
answered Dec 21 '18 at 0:50
DeficientMathDudeDeficientMathDude
1112
$endgroup$
$begingroup$
I think the question was saying that $10/72$ players are tested - that is, $10$ are tested for each team, not that $10$ are tested in the whole NFL. (Also, note that, while your product for the probability of an individual being tested in a given week is correct, it's rather inefficient; the product telescopes to just be $frac{t}n$ where $n$ is the pool of testable players and $t$ is the number tested)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:19
$begingroup$
@MiloBrandt If that is the case the math can easily be adjusted to accommodate the different $p$ value. I don't follow any sportsball games, so I assumed the inclusion of the number of teams was meant to inform us that the testing is not per team, but NFL wide. And while it is inefficient for math people, since this question is directed at non math people, I wrote it out long-form.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:20
$begingroup$
Yea, it is 10 players per team, with 72 testable players on each team. Sorry if that wasn't clear.
$endgroup$
– Frost
Dec 21 '18 at 1:50
add a comment |
$begingroup$
If everything is uniformly random (which need not be the case), then there are $32cdot 72 = 2304$ total players in the pool for random testing; since $10$ are chosen each week, the probability that a player is not chosen to be tested is
$$
frac{2303}{2304}cdotfrac{2302}{2303}cdotsfrac{2294}{2295} = frac{2294}{2304}
$$
so the probability that the are chosen is $1-frac{2294}{2304}=frac{5}{1152}$. Call this value $p$.
Now we are in a classic Bernoulli trial setup; we have a probability of success (being called for a drug test) and a number of trials (each of the 11 weeks that have passed), and we wish to know the probability that we succeed a certain number of times. Plugging everything in we find that the probability of being called at least $6$ times would be
$$
sum_{k=6}^{11}{11choose k}p^k(1-p)^{11-k} approx 3cdot 10^{-12}
$$
which is a touch on the low side.
What does this mean? Either nothing, or not that much, or it is the scandal of the century. First, everything that is possible is possible (deep wisdom I know). It could be that this player was (un)lucky enough to win this lottery. Second, the procedure may not be uniformly random. While it might strike you as "unfair" to weight some players as riskier than others, actuaries do this all the time with things like drug use, recidivism rates, life expectency, etc. It might not be a stretch to think that the NFL also has a different distribution than uniform randomness for these drug trials. You can ask why they should deviate from uniform randomness, but that would at least help explain why such a "rare" event has happened. And third, it could be that it isn't really random, or someone messed with the system to circumvent randomness, and we have the next scandal on our hands that we can fret over for 24-72 hours.
EDIT: Based on a comment on this answer, I did some quick research and discovered that the testing is not NFL wide as I have assumed, but that each team tests $10$ of their players each week. This gives you a different $p$ value of $5/36$, and the rest of the numbers change accordingly.
answered Dec 21 '18 at 0:50
DeficientMathDudeDeficientMathDude
1112
$endgroup$
$begingroup$
I think the question was saying that $10/72$ players are tested - that is, $10$ are tested for each team, not that $10$ are tested in the whole NFL. (Also, note that, while your product for the probability of an individual being tested in a given week is correct, it's rather inefficient; the product telescopes to just be $frac{t}n$ where $n$ is the pool of testable players and $t$ is the number tested)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:19
$begingroup$
@MiloBrandt If that is the case the math can easily be adjusted to accommodate the different $p$ value. I don't follow any sportsball games, so I assumed the inclusion of the number of teams was meant to inform us that the testing is not per team, but NFL wide. And while it is inefficient for math people, since this question is directed at non math people, I wrote it out long-form.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:20
$begingroup$
Yea, it is 10 players per team, with 72 testable players on each team. Sorry if that wasn't clear.
$endgroup$
– Frost
Dec 21 '18 at 1:50
add a comment |
$begingroup$
If everything is uniformly random (which need not be the case), then there are $32cdot 72 = 2304$ total players in the pool for random testing; since $10$ are chosen each week, the probability that a player is not chosen to be tested is
$$
frac{2303}{2304}cdotfrac{2302}{2303}cdotsfrac{2294}{2295} = frac{2294}{2304}
$$
so the probability that the are chosen is $1-frac{2294}{2304}=frac{5}{1152}$. Call this value $p$.
Now we are in a classic Bernoulli trial setup; we have a probability of success (being called for a drug test) and a number of trials (each of the 11 weeks that have passed), and we wish to know the probability that we succeed a certain number of times. Plugging everything in we find that the probability of being called at least $6$ times would be
$$
sum_{k=6}^{11}{11choose k}p^k(1-p)^{11-k} approx 3cdot 10^{-12}
$$
which is a touch on the low side.
What does this mean? Either nothing, or not that much, or it is the scandal of the century. First, everything that is possible is possible (deep wisdom I know). It could be that this player was (un)lucky enough to win this lottery. Second, the procedure may not be uniformly random. While it might strike you as "unfair" to weight some players as riskier than others, actuaries do this all the time with things like drug use, recidivism rates, life expectency, etc. It might not be a stretch to think that the NFL also has a different distribution than uniform randomness for these drug trials. You can ask why they should deviate from uniform randomness, but that would at least help explain why such a "rare" event has happened. And third, it could be that it isn't really random, or someone messed with the system to circumvent randomness, and we have the next scandal on our hands that we can fret over for 24-72 hours.
EDIT: Based on a comment on this answer, I did some quick research and discovered that the testing is not NFL wide as I have assumed, but that each team tests $10$ of their players each week. This gives you a different $p$ value of $5/36$, and the rest of the numbers change accordingly.
answered Dec 21 '18 at 0:50
DeficientMathDudeDeficientMathDude
1112
$endgroup$
If everything is uniformly random (which need not be the case), then there are $32cdot 72 = 2304$ total players in the pool for random testing; since $10$ are chosen each week, the probability that a player is not chosen to be tested is
$$
frac{2303}{2304}cdotfrac{2302}{2303}cdotsfrac{2294}{2295} = frac{2294}{2304}
$$
so the probability that the are chosen is $1-frac{2294}{2304}=frac{5}{1152}$. Call this value $p$.
Now we are in a classic Bernoulli trial setup; we have a probability of success (being called for a drug test) and a number of trials (each of the 11 weeks that have passed), and we wish to know the probability that we succeed a certain number of times. Plugging everything in we find that the probability of being called at least $6$ times would be
$$
sum_{k=6}^{11}{11choose k}p^k(1-p)^{11-k} approx 3cdot 10^{-12}
$$
which is a touch on the low side.
What does this mean? Either nothing, or not that much, or it is the scandal of the century. First, everything that is possible is possible (deep wisdom I know). It could be that this player was (un)lucky enough to win this lottery. Second, the procedure may not be uniformly random. While it might strike you as "unfair" to weight some players as riskier than others, actuaries do this all the time with things like drug use, recidivism rates, life expectency, etc. It might not be a stretch to think that the NFL also has a different distribution than uniform randomness for these drug trials. You can ask why they should deviate from uniform randomness, but that would at least help explain why such a "rare" event has happened. And third, it could be that it isn't really random, or someone messed with the system to circumvent randomness, and we have the next scandal on our hands that we can fret over for 24-72 hours.
EDIT: Based on a comment on this answer, I did some quick research and discovered that the testing is not NFL wide as I have assumed, but that each team tests $10$ of their players each week. This gives you a different $p$ value of $5/36$, and the rest of the numbers change accordingly.
answered Dec 21 '18 at 0:50
DeficientMathDudeDeficientMathDude
1112
edited Dec 21 '18 at 1:25
answered Dec 21 '18 at 0:50
DeficientMathDudeDeficientMathDude
1112
answered Dec 21 '18 at 0:50
DeficientMathDudeDeficientMathDude
1112
answered Dec 21 '18 at 0:50
DeficientMathDudeDeficientMathDude
1112
1112
$begingroup$
I think the question was saying that $10/72$ players are tested - that is, $10$ are tested for each team, not that $10$ are tested in the whole NFL. (Also, note that, while your product for the probability of an individual being tested in a given week is correct, it's rather inefficient; the product telescopes to just be $frac{t}n$ where $n$ is the pool of testable players and $t$ is the number tested)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:19
$begingroup$
@MiloBrandt If that is the case the math can easily be adjusted to accommodate the different $p$ value. I don't follow any sportsball games, so I assumed the inclusion of the number of teams was meant to inform us that the testing is not per team, but NFL wide. And while it is inefficient for math people, since this question is directed at non math people, I wrote it out long-form.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:20
$begingroup$
Yea, it is 10 players per team, with 72 testable players on each team. Sorry if that wasn't clear.
$endgroup$
– Frost
Dec 21 '18 at 1:50
add a comment |
$begingroup$
I think the question was saying that $10/72$ players are tested - that is, $10$ are tested for each team, not that $10$ are tested in the whole NFL. (Also, note that, while your product for the probability of an individual being tested in a given week is correct, it's rather inefficient; the product telescopes to just be $frac{t}n$ where $n$ is the pool of testable players and $t$ is the number tested)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:19
$begingroup$
@MiloBrandt If that is the case the math can easily be adjusted to accommodate the different $p$ value. I don't follow any sportsball games, so I assumed the inclusion of the number of teams was meant to inform us that the testing is not per team, but NFL wide. And while it is inefficient for math people, since this question is directed at non math people, I wrote it out long-form.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:20
$begingroup$
Yea, it is 10 players per team, with 72 testable players on each team. Sorry if that wasn't clear.
$endgroup$
– Frost
Dec 21 '18 at 1:50
$begingroup$
I think the question was saying that $10/72$ players are tested - that is, $10$ are tested for each team, not that $10$ are tested in the whole NFL. (Also, note that, while your product for the probability of an individual being tested in a given week is correct, it's rather inefficient; the product telescopes to just be $frac{t}n$ where $n$ is the pool of testable players and $t$ is the number tested)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:19
$begingroup$
I think the question was saying that $10/72$ players are tested - that is, $10$ are tested for each team, not that $10$ are tested in the whole NFL. (Also, note that, while your product for the probability of an individual being tested in a given week is correct, it's rather inefficient; the product telescopes to just be $frac{t}n$ where $n$ is the pool of testable players and $t$ is the number tested)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:19
$begingroup$
@MiloBrandt If that is the case the math can easily be adjusted to accommodate the different $p$ value. I don't follow any sportsball games, so I assumed the inclusion of the number of teams was meant to inform us that the testing is not per team, but NFL wide. And while it is inefficient for math people, since this question is directed at non math people, I wrote it out long-form.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:20
$begingroup$
@MiloBrandt If that is the case the math can easily be adjusted to accommodate the different $p$ value. I don't follow any sportsball games, so I assumed the inclusion of the number of teams was meant to inform us that the testing is not per team, but NFL wide. And while it is inefficient for math people, since this question is directed at non math people, I wrote it out long-form.
$endgroup$
– DeficientMathDude
Dec 21 '18 at 1:20
$begingroup$
Yea, it is 10 players per team, with 72 testable players on each team. Sorry if that wasn't clear.
$endgroup$
– Frost
Dec 21 '18 at 1:50
$begingroup$
Yea, it is 10 players per team, with 72 testable players on each team. Sorry if that wasn't clear.
$endgroup$
– Frost
Dec 21 '18 at 1:50
add a comment |
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$begingroup$
"I thought this was interesting and wanted to see if anyone could figure it out cause the media can't" - since when is the media correct with mathematics? :)
$endgroup$
– Math1000
Dec 21 '18 at 0:29
$begingroup$
Anything with non zero probability will happen somewhere given enough time.
$endgroup$
– ablmf
Dec 21 '18 at 0:41
1
$begingroup$
Three different things: (1) The probability that Eric Reid is chosen ${} ge 6$ out of 11 times; (2) the probability that there is a player chosen ${} ge 6$ out of 11 times. (3) The probability that somebody somewhere endures an event with probability ${} le 3 cdot 10^{-12}$. The point is: the OP picked one event that seemed unlikely out of all events from recent weeks.
$endgroup$
– GEdgar
Dec 21 '18 at 1:05
1
$begingroup$
I googled about, and found that someone wrote an article about this, including probabilities. They have somewhat different numbers (e.g. they say $53$ active players instead of $72$ players), so get a different answer. (Note that this isn't the probability that the player was/wasn't targeted - that question is beyond the scope of purely mathematical logic)
$endgroup$
– Milo Brandt
Dec 21 '18 at 1:12
1
$begingroup$
@MiloBrandt, there are 53 "active" players but there are more testable players, this includes the players on the practice squad. Now... why they care if players on a practice squad are juicing even if they never make it to the active roster is weird but that is the official policy from what I read. And for the record "72" is actually not exactly true, it can be 73 or less but I didn't want to bring that in as a variable.
$endgroup$
– Frost
Dec 21 '18 at 1:53