The infamous Ask Marilyn Christmas column

[grapes]

Ask Marilyn Dec. 25, 2011: A drug test is done on employees, with a fourth of them selected at random every three months. "What's the likelihood of being chosen over the course of a year?"

Marilyn's answer: "The probability remains 25 percent, despite the repeated testing. . . . Goes against your intuition, doesn't it?"

Clearly, Marilyn stopped consulting her panel of mathematicians in the early nineties.

[swampyankee]

Wow! grapes can see into the future!

She missed the point, obviously, that the probability of any given employee being tested is 25% on each occasion, but that wasn't the question. Incidentally, if I did my math right, the chance of somebody being tested at least once is about 65%.

[HenrikOlsen]

1 - (0.75^4) = 0.68359375, so the chance of getting picked at least once is ~68.4%.

[grapes]

Wow! grapes can see into the future!

Thousands of years into the future...

My username is Win MegaMillions then.

She missed the point, obviously,

She once wrote a whole book on why Andrew Wiles' proof for Fermat's Last Theorem should be disallowed, because he used hyperbolic functions, and if you can't use hyperbolic functions to square the circle, why should you be allowed to use them for Fermat's Last Theorem.

[HenrikOlsen]

She once wrote a whole book on why Andrew Wiles' proof for Fermat's Last Theorem should be disallowed, because he used hyperbolic functions, and if you can't use hyperbolic functions to square the circle, why should you be allowed to use them for Fermat's Last Theorem.

Not quite, she wrote a book on the history of attempts at proving it and only the last chapter was about Wiles' proof.

But yes, it sounds like she'd definitely not all there all the time when it comes to mathematics.

It sounds like she's at a level where asking what the chances are for a 1 out of 1000 numbers lottery coming up with the same number three times in a row would get the wrong answer. 1/1000,000 is the correct answer.

[Romanus]

Thousands of years into the future...

Kwisatz Haderach?

[orionjim]

I did a quick simulation in Excel and it seems to me she’s right. For 100 employees and running it 20 times I got an average of .256475.

For example for one run I got:
39 were never tested
40 were tested 1 time
18 were tested 2 times
3 were tested 3 times
0 were tested 4 times

The Excel
___Col A___________________Col B __________… to Col CV
1 =rand()
2 =if(a1<=.25,1,0)
3 =rand()
4 =if(a3<=.25,1,0)
5 =rand()
6 =if(a5<=.25,1,0)
7 =rand()
8 =if(a7<=.25,1,0)
9 =(a2+a4+a6+a8)/4

I had columns ‘A’ thru ‘CV’ then I did:
=average(a9:cv9)

With 10 employees there is a lot of variation, with 100 the average is usually from .20 to .30.

Of course I could have done something wrong!

Jim

[Chuck]

Maybe they always pick the ¼ with the highest social security numbers, figuring that would be random enough.

[HenrikOlsen]

I did a quick simulation in Excel and it seems to me she’s right. For 100 employees and running it 20 times I got an average of .256475.

For example for one run I got:
39 were never tested
40 were tested 1 time
18 were tested 2 times
3 were tested 3 times
0 were tested 4 times

<Snip>

Of course I could have done something wrong!

Jim

I don't think you doing anything wrong, but the two of you aren't answering the same question as the rest of us.

With the numbers from your example run 39% were never tested and 40+18+3=61% were chosen at least once, which is rather close to the 68.4% I predicted.

[grapes]

With 10 employees there is a lot of variation, with 100 the average is usually from .20 to .30.

Of course I could have done something wrong!

In addition to Henrik's comment, I would also have 25 employees take the test every single month. Which means that the average should be exactly 25, over any length of time. Apparently, the company facility is set up for 25 candidates every three months...or something like that.

So, yes, each employee should expect to be tested 25% of the time, but I don't think that is the question--I didn't quote the whole column but I quoted the question. Of course it's 25% each and every single time, but--also of course--you'd expect to be tested each year more than 25% of the time.

[Solfe]

On really bad days at work, I used to swear mandatory drug testing would solve most of my problems. My stated percentage was 90%.

[DonM435]

If a company is going to test 25% every quarter year, then they must have the facilities to test everybody annually. However, they'd have to introduce the random element so as not to be predictable with regard to the date, and also so that someone with a successful test wasn't excused from testing for a whole year. With such a systen, some folks might get called in twice, even three times in a year.

I'll be willing to bet they have another rule in place for anyone who is lucky enough to slip through the crack for more than a couple of consecutive years, though.

[orionjim]

I don't think you doing anything wrong, but the two of you aren't answering the same question as the rest of us.

With the numbers from your example run 39% were never tested and 40+18+3=61% were chosen at least once, which is rather close to the 68.4% I predicted.

I see where I went wrong. I was using the words “likelihood” and “probability” as being interchangeable (and they're not); so yes the answer to the question was, as you and Swampyankee said, is ~68 percent.

And Marilyn didn’t answer the question either; however as she said the probability remains 25 percent.

Jim

[Solfe]

Is her answer correct if rephrased to "What are your chances of being tested at the time of each selection process"?

[swampyankee]

Is her answer correct if rephrased to "What are your chances of being tested at the time of each selection process"?

Yes.

[grapes]

We now have a link to the column itself:
http://www.parade.com/askmarilyn/201...-12-25-11.html
Some of the comments are a little rough

Over the course of a year, the company could test everybody, since there are 400 employees, and they test 100 employees every three months. But since it is random, the probability that you will be tested is less than 1, it is more like Henrik's number. But it is greater than 25%.

[orionjim]

To me it looks like she is purposely trying to yank on peoples’ chain.

The definition of Likelihood (from Wolfram MathWorld):

Likelihood is the hypothetical probability that an event that has already occurred would yield a specific outcome. The concept differs from that of a probability in that a probability refers to the occurrence of future events, while a likelihood refers to past events with known outcomes.

She never said that the likelihood wouldn’t increase; and her claim that the probability of being tested remains 25 percent is correct. Of course, as others have said, she never answered the question either.

[SeanF]

My suspicion is that she simply misunderstood the question. Or, perhaps, she didn't understand the question at all, and when she asked her "panel of mathematicians," she incorrectly phrased the question to them.

All this talk of probability reminds me of something, though. I need to see if I can find that old "what are the odds the other child is also a boy" thread...

[orionjim]

My suspicion is that she simply misunderstood the question. Or, perhaps, she didn't understand the question at all, and when she asked her "panel of mathematicians," she incorrectly phrased the question to them.
...

Marilyn says she is going make a correction in the January 22nd issue. See:

http://www.parade.com/askmarilyn/201...revisited.html

[grapes]

Marilyn says she is going make a correction in the January 22nd issue. See:

http://www.parade.com/askmarilyn/201...revisited.html

Funny!

Jerry Grossman of Rochester Hills, Michigan, writes:

Marilyn: You misinterpreted the question. The correct answer is 1 minus [(3/4) raised to the fourth power].Please issue a correction.

Marilyn responds:

Of course! You can look for it in the magazine on January 22nd.

Note to self: Lay off the eggnog next year.

[CDRMSQUARE]

The way I read it, the problem originally posed by Marilyn is about the probability of being selected AT LEAST once over the course of the year, which was corrected by many readers as ~68.6%. To estimate the probability of being selected ONLY once, you need to recognize that entails any number of ways you could be selected only once. For example, the probability of being selected on the first test and not being selected on the 2nd, 3rd, or 4th tests will be 0.25*(0.75**3), or ~10.5%. The probability of being selected on JUST one of the other tests will be the same. Hence there are four ways (combinations) of being selected only once. Each of these combinations is mutually exclusive. For example, being picked for the first test is INDEPENDENT of being picked for the 2nd test, but these two events are NOT mutually exclusive. However, it is not possible to be picked for the first test and not for the last three tests, and to also be picked for the 2nd test but not the others. Because these combinations are mutually exclusive, the probability of being picked ONLY once (be it for the 1st, 2nd, 3rd, or 4th test) is simply the sum of the probabilities of each of the combinations, or 4*10.5% ~ 42%.