## Answer to Random Drug Testing Question 12/24/11

**Moderators:** mvs_staff, forum_admin, Marilyn

15 posts
• Page

**1**of**1**### Answer to Random Drug Testing Question 12/24/11

In today's Parade there was a question about random drug testing. Every 3 months one quarter of a fixed population is randomly selected for testing. The question was what is the probability that a person would be selected over the course of one year. Marilyn's answer was 25%, but the actual probability is 68%. The probability of not being selected is 0.75x0.75x0.75x0.75 = 0.316, so the probability of being selected at least once is 1 - 0.316 = 0.684.

The binomial probability function defines the probabilities of being selected n times out of N possible times with p being the probability of being selected each time, and q = 1-p:

P(n/N) = [N!/n!(N-n)!] p^n q^(N-n)

The probability of being selected exactly once over the course of the year is:

P(1/4) = [4!/3!1!] [0.25^1][0.75^3] = 0.422

Twice:

P(2/4) = [4!/2!2!] [0.25^2][0.75^2] = 0.211

Three times:

P(3/4) = [4!/1!3!] [0.25^3][0.75^1] = 0.047

Four times:

P(4/4) = [4!/4!0!] [0.25^4][0.75^0] = 0.004

The most likely outcome is being selected once (42.2%). A probability of 25% is nowhere to be found in any of the outcomes.

The binomial probability function defines the probabilities of being selected n times out of N possible times with p being the probability of being selected each time, and q = 1-p:

P(n/N) = [N!/n!(N-n)!] p^n q^(N-n)

The probability of being selected exactly once over the course of the year is:

P(1/4) = [4!/3!1!] [0.25^1][0.75^3] = 0.422

Twice:

P(2/4) = [4!/2!2!] [0.25^2][0.75^2] = 0.211

Three times:

P(3/4) = [4!/1!3!] [0.25^3][0.75^1] = 0.047

Four times:

P(4/4) = [4!/4!0!] [0.25^4][0.75^0] = 0.004

The most likely outcome is being selected once (42.2%). A probability of 25% is nowhere to be found in any of the outcomes.

- Juan Largo
- Thinker
**Posts:**3**Joined:**Sat Sep 06, 2008 4:16 pm

### Re: Answer to Random Drug Testing Question 12/24/11

On 12/25/11, Marilyn wrote:The probability remains 25 percent, despite the repeated testing. One might think that as the number of tests grows, the likelihood of being chosen increases, but as long as the size of the pool remains the same, so does the probability. Goes against your intuition, doesn't it?I manage a drug-testing program for an organization with 400 employees. Every three months, a random-number generator selects 100 names for testing. Afterward, these names go back into the selection pool. Obviously, the probability of an employee being chosen in one quarter is 25 percent. But what is the likelihood of being chosen over the course of a year?

-- Jerry Haskins, Vicksburg, Miss.

Juan Largo wrote:In today's Parade there was a question about random drug testing. Every 3 months one quarter of a fixed population is randomly selected for testing. The question was what is the probability that a person would be selected over the course of one year. Marilyn's answer was 25%, but the actual probability is 68%. The probability of not being selected is 0.75x0.75x0.75x0.75 = 0.316, so the probability of being selected at least once is 1 - 0.316 = 0.684.

The binomial probability function defines the probabilities of being selected n times out of N possible times with p being the probability of being selected each time, and q = 1-p:

P(n/N) = [N!/n!(N-n)!] p^n q^(N-n)

The probability of being selected exactly once over the course of the year is:

P(1/4) = [4!/3!1!] [0.25^1][0.75^3] = 0.422

Twice:

P(2/4) = [4!/2!2!] [0.25^2][0.75^2] = 0.211

Three times:

P(3/4) = [4!/1!3!] [0.25^3][0.75^1] = 0.047

Four times:

P(4/4) = [4!/4!0!] [0.25^4][0.75^0] = 0.004

The most likely outcome is being selected once (42.2%). A probability of 25% is nowhere to be found in any of the outcomes.

Marilyn frequently answers a diferent question then what was seems, to others, to be what was actually asked. This isn't an unpredictable situation, since the questions are short and English can be ambiguous. But it would be helpeful if she could try to indicate what it is she is answering.

It could be that she was answering "Over the course of a year, does the probability that an employee will be chosen in any given quarter change?" Many people intuitively feel that if you aren't chosen in the spring test, the chances you will be selected in the summer will go up. And if you aren't chosen at all in the first three tests of a year, it is almost certain you will be chosen for the fourth since each test picks 1/4 of the employees. She could have easily indicated that by saying "the likelihood of being chosen for any given test...".

We can't know for certain what she was thinking when she gave this blatantly incorrect answer. But I don't think she will understand your math; she has ignored examples in the past where similar math has proven her incorrect. Her application of probability is usually based more on intuition than math, as evidenced by here repeated incorrect (or at least, correct only for a question that isn't what most others see in the text) replies about sequences of dice rolls. She does have a better intuition than most, but she places too much confidence in it.

- JeffJo
- Intellectual
**Posts:**2427**Joined:**Tue Mar 10, 2009 11:01 am

I'd be willing to go along with what you said, but Marilyn did not explain her answer at all. She just threw out a number without offering any reasoning or rationale behind it, which is very unlike her. Maybe she was having a bad day.

- Juan Largo
- Thinker
**Posts:**3**Joined:**Sat Sep 06, 2008 4:16 pm

Juan Largo wrote:I'd be willing to go along with what you said, but Marilyn did not explain her answer at all. She just threw out a number without offering any reasoning or rationale behind it, which is very unlike her. Maybe she was having a bad day.

On the contrary, it is very unusual for her to offer any support for her answers. She usually just states it and expects everybody to see how obvious it is. The best recent example is the question about dice sequences. She answered it three times teh exact same way, by stating that it should seem obvious which sequence was more likely.

- JeffJo
- Intellectual
**Posts:**2427**Joined:**Tue Mar 10, 2009 11:01 am

### Once again Marilyn is ambiguous

Once again Marilyn presents her question in an ambiguous manner. The phrase "being chosen over the course of a year" has several different meanings. It could mean at any time during the year a test occurs, what's the probability of being selected. In that case, Marilyn is correct. But it could also mean the probability of being selected sometime during the year; i.e., there exists a date among the 365/6 in the year such that a drug test is held on that date and you were selected. That is 68%, as our first commenter just computed. (In fact, it is close to 1-1/e).

- jimvb
- Thinker
**Posts:**6**Joined:**Sat Nov 29, 2008 8:43 pm

### Re: Once again Marilyn is ambiguous

jimvb wrote:Once again Marilyn presents her question in an ambiguous manner. The phrase "being chosen over the course of a year" has several different meanings. It could mean at any time during the year a test occurs, what's the probability of being selected.

Only by misinterpretation. It's an easy mistake to make, but clearly in error. A direct comparison was made between "being chosen in one quarter" and "being chosen over the course of a year." That is, comparing one period to another. Your interpretation above requires the comparison to be between "being chosen in one quarter" and "being chosen in another quarter," in which case the word "year" is superfluous.

- JeffJo
- Intellectual
**Posts:**2427**Joined:**Tue Mar 10, 2009 11:01 am

### Juan Largo's Answer to Random Drug Testing Question 12/24/11

Well done, Juan!

In 17 words and a one line of math you have succinctly explained the correct answer to the problem posed in this week's Parade column.

This is a simple problem from freshman statistics, so it's odd that Marilyn got it wrong. One can only hope that readers will be motivated to ponder this problem and seek to understand for themselves how to solve it, rather than accepting the incorrect answer given in the column.

Regards,

Joe

In 17 words and a one line of math you have succinctly explained the correct answer to the problem posed in this week's Parade column.

This is a simple problem from freshman statistics, so it's odd that Marilyn got it wrong. One can only hope that readers will be motivated to ponder this problem and seek to understand for themselves how to solve it, rather than accepting the incorrect answer given in the column.

Regards,

Joe

Juan Largo wrote:In today's Parade there was a question about random drug testing. Every 3 months one quarter of a fixed population is randomly selected for testing. The question was what is the probability that a person would be selected over the course of one year. Marilyn's answer was 25%, but the actual probability is 68%. The probability of not being selected is 0.75x0.75x0.75x0.75 = 0.316, so the probability of being selected at least once is 1 - 0.316 = 0.684.

The binomial probability function defines the probabilities of being selected n times out of N possible times with p being the probability of being selected each time, and q = 1-p:

P(n/N) = [N!/n!(N-n)!] p^n q^(N-n)

The probability of being selected exactly once over the course of the year is:

P(1/4) = [4!/3!1!] [0.25^1][0.75^3] = 0.422

Twice:

P(2/4) = [4!/2!2!] [0.25^2][0.75^2] = 0.211

Three times:

P(3/4) = [4!/1!3!] [0.25^3][0.75^1] = 0.047

Four times:

P(4/4) = [4!/4!0!] [0.25^4][0.75^0] = 0.004

The most likely outcome is being selected once (42.2%). A probability of 25% is nowhere to be found in any of the outcomes.

- jcat
- Thinker
**Posts:**1**Joined:**Sun Oct 23, 2011 12:34 pm

### Re: Answer to Random Drug Testing Question 12/24/11

Parade 12/25/11 wrote:I manage a drug-testing program for an organization with 400 employees. Every three months, a random-number generator selects 100 names for testing. Afterward, these names go back into the selection pool. Obviously, the probability of an employee being chosen in one quarter is 25 percent. But what is the likelihood of being chosen over the course of a year?

-- Jerry Haskins, Vicksburg, Miss.

Marilyn responds:

The probability remains 25 percent, despite the repeated testing. One might think that as the number of tests grows, the likelihood of being chosen increases, but as long as the size of the pool remains the same, so does the probability. Goes against your intuition, doesn't it?

I think Marilyn did not answer the question actually asked which is "what is the likelihood of being chosen over the course of a year?"

This would be one minus the product of not being chosen in any quarter for four consecutive quarters:

1-(1-0.25)^q.

- Code: Select all
`cumulative quarters, cumulative probability of being chosen`

1, 1-0.75 =0.25

2, 1-(0.75)^2=0.4375

3, 1-(0.75)^3=0.5781

4, 1-(0.75)^4=0.6836

So, over the course of four consecutive quarters any employee has a 0.6836 probability of being chosen for testing.

- robert 46
- Intellectual
**Posts:**2526**Joined:**Mon Jun 18, 2007 9:21 am

The probability remains 25 percent, despite the repeated testing. One might think that as the number of tests grows, the likelihood of being chosen increases, but as long as the size of the pool remains the same, so does the probability. Goes against your intuition, doesn't it?

After thinking about this some more, what Marilyn undoubtedly meant was that the likelihood of being selected over the course of a year is unchanged by being selected or not being selected in any previous quarters. It's a common fallacy that if you weren't selected beforehand you are "due" to be selected next time. Maybe Marilyn inferred that the person was committing this fallacy from the way the question was stated and she was just trying to dispel it.

If that's the case, my apologies to Marilyn.

- Juan Largo
- Thinker
**Posts:**3**Joined:**Sat Sep 06, 2008 4:16 pm

### Marilyn didn't answer the question asked

I don't think you need to apologize to Marilyn. Marilyn didn't answer the question that was asked. Someone could have asked a question about what your odds are of being picked in the 4th drawing of the year if you weren't picked in the first 3 (or if you were picked in the first 3). Either way, the answer would be 25%. But that wasn't the question asked.

According to the earlier post, here is what the question said, in part:

"Obviously, the probability of an employee being chosen in one quarter is 25 percent. But what is the likelihood of being chosen over the course of a year?"

I just don't see any reasonable way to interpret the question the way that Marilyn did. The question writer is asking what are the chances of being chosen sometime in the year (presumably in at least 1 of the 4 drawings), not the chances of being chose in any one quarter. It is right there in the question "Obviously, the probability of an employee being chosen in one quarter is 25%." Then Marilyn goes on to explain that the chance of being chosen in any one quarter is 25%.

Marilyn definitely blew this one. I hope she writes a correction.

According to the earlier post, here is what the question said, in part:

"Obviously, the probability of an employee being chosen in one quarter is 25 percent. But what is the likelihood of being chosen over the course of a year?"

I just don't see any reasonable way to interpret the question the way that Marilyn did. The question writer is asking what are the chances of being chosen sometime in the year (presumably in at least 1 of the 4 drawings), not the chances of being chose in any one quarter. It is right there in the question "Obviously, the probability of an employee being chosen in one quarter is 25%." Then Marilyn goes on to explain that the chance of being chosen in any one quarter is 25%.

Marilyn definitely blew this one. I hope she writes a correction.

- svalancius
- Thinker
**Posts:**2**Joined:**Sat Jan 07, 2012 9:47 am

### Re: Marilyn didn't answer the question asked

svalancius wrote:Marilyn definitely blew this one. I hope she writes a correction.

She has already admitted the answer was incorrect, but not a mistake. Instead, she made a tongue-in-cheek comment about being influenced by eggnog, and said she will address it in two weeks (probably the delay caused by the publishing process).

- JeffJo
- Intellectual
**Posts:**2427**Joined:**Tue Mar 10, 2009 11:01 am

### Incorrect, but not a mistake

I'm new to the forum - where did she say she would address this in two weeks?

What Marilyn wrote was true; it just didn't answer the question. I'm glad to hear she is going to address it in a future column.

What Marilyn wrote was true; it just didn't answer the question. I'm glad to hear she is going to address it in a future column.

- svalancius
- Thinker
**Posts:**2**Joined:**Sat Jan 07, 2012 9:47 am

### Re: Incorrect, but not a mistake

svalancius wrote:I'm new to the forum - where did she say she would address this in two weeks?

She answers daily questions on the parade website. Follow the links from the "home" page of this forum, or use this link.

What Marilyn wrote was true; it just didn't answer the question. I'm glad to hear she is going to address it in a future column.

It's also true that frogs have short stumpy legs, but that has no bearing on the question she was asked. An answer can still be wrong, even if true, if it doesn't address the question.

But Marilyn has a very poor record of admitting when she is wrong - as in, almost never, and only when she can find an acceptable excuse. There were at least two other answers in the past couple of months that were outright wrong, and she ignored them. Probably because she can't argue that her answers were right for a different question. There was even one a couple of years ago where she broke the answer down into three separate cases, and each one was wrong for different reasons. I mention that because the proof that one was wrong follows the same reasoning as the drug-test answer, that the chance of a thing happening in at least one of two attempts is less than the twice the chance of it happening in one.

- JeffJo
- Intellectual
**Posts:**2427**Joined:**Tue Mar 10, 2009 11:01 am

### Random Drug Testing Question

I enjoy Marilyn's column, but I was so taken aback at her answer to the question on the morning of Dec. 25 that I registered to participate in this site so that I could comment, only to run into the (reasonable) delays in establishing eligibility to participate. In the meantime, the points I wished to make have been ably and thoroughly made by other commenters.

I will make one observation that simply rephrases the point others have made. I focus on intuition; that is, beyond the conclusive mathematical demonstration presented earlier in the comments, intuition shows the error of Marilyn's answer to the question that was -- unambiguously -- asked. Marilyn's answer was: "One might think that as the number of tests grows, the likelihood of being chosen increases, but as long as the size of the pool remains the same, so does the probability. Goes against your intuition, doesn't it?" Sorry, but as the number of tests grows with a constant pool, the probability of the event occurring at least once in the course of the tests increases. To wit: The odds of rolling a "6" when a six-sided die is tossed once is 1/6. So I throw the die 100 times (an increased number of tests) as the size of the pool remains the same (the die has six sides in each test); do the odds of getting at least one "6" across the 100 throws remain 1 in 6? Of course not. That's why Russian Roulette played 100 times in sequence -- if our intuition means anything --is nearly certain to be fatal.

Looking forward to Marilyn's correction.

I will make one observation that simply rephrases the point others have made. I focus on intuition; that is, beyond the conclusive mathematical demonstration presented earlier in the comments, intuition shows the error of Marilyn's answer to the question that was -- unambiguously -- asked. Marilyn's answer was: "One might think that as the number of tests grows, the likelihood of being chosen increases, but as long as the size of the pool remains the same, so does the probability. Goes against your intuition, doesn't it?" Sorry, but as the number of tests grows with a constant pool, the probability of the event occurring at least once in the course of the tests increases. To wit: The odds of rolling a "6" when a six-sided die is tossed once is 1/6. So I throw the die 100 times (an increased number of tests) as the size of the pool remains the same (the die has six sides in each test); do the odds of getting at least one "6" across the 100 throws remain 1 in 6? Of course not. That's why Russian Roulette played 100 times in sequence -- if our intuition means anything --is nearly certain to be fatal.

Looking forward to Marilyn's correction.

- jerrygf
- Thinker
**Posts:**2**Joined:**Sun Dec 25, 2011 9:21 am**Location:**Gainesville FL

### Re: Random Drug Testing Question

jerrygf wrote:Sorry, but as the number of tests grows with a constant pool, the probability of the event occurring at least once in the course of the tests increases.

We can only speculate on how and/or why Marilyn made this mistake, but the following sequence of inferences does seem plausible:

- Many people do feel - intuitively - that random chance has a memory. That if heads has come up on three consecutive coin flips, somehow tails is more likely on the next one so as to "even things out." A person on this list, who feels he understands probabilty but does not, has incorrectly called this "regression towards the mean" and claimed it was true in a more complex situation.
- I call it "correction toward the mean." Regression is something else entirely, and is true.
- Marilyn knows of this tendency, and has probably answered many questions involving "correction toward the mean." Her first mistake was, without reading the question carefully, she assumed this was another.
- The only way it made sense to be a question about "correction toward the mean," was if the probability asked about was the probability in one quarter.
- So Marilyn's second mistake was that she answered the wrong question, and ...
- Her third was when she admonished the reader about thinking "correction toward the mean" was true.

- JeffJo
- Intellectual
**Posts:**2427**Joined:**Tue Mar 10, 2009 11:01 am

15 posts
• Page

**1**of**1**### Who is online

Users browsing this forum: No registered users and 0 guests