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Shammrog
05-29-2007, 07:08 PM
Hello - I believe I am being "set up" at work by my nominal partner.

Without belaboring the details, if:


You have a sample of 462 items.
Assume that this sample is incomplete, and the actual (for arguments sake, a ridiculous figure) missing item rate is 20%, such that the whole population is about 577.
You get a selection from the true population (577) of 12 "missing" items.
What are the odds that you get 12 items that are in fact missing - that is, they should have been in there (in the 462), but were part of the 105 that were missing?

Hopefully this is somewhat clear. Basically, my partner presented me with a request to come up with 12 items this morning. All of which (conveniently) happened to be missing from the 462 actual such items that were prepared. What are the odds, even with an astronomical error rate of 20%, that he would randomly pick 12, all of which were not included and thus "missing?"

Please let me know ASAP. I need statistical help urgently!!!

juise
05-29-2007, 07:39 PM
A rough way of calculating it would be to consider each selection of a data point to be a separate random event. So, you have a 20% chance of choosing a "missing" data point (a 1 in 5 chance). If you were to hit that 1 in 5 chance 12 times in a row, you would calculate P=.2^12=0.0000004096%.

If you wanted to be more exact, you would have to account for the sample taken out of the population each time you select one. If the first time you have a 20% chance (115/577=19.9307%), the next draw, your chance will be slightly lower (114/576=19.7917%). And the next time, even lower (113/575=19.6522%). If you were to follow that logic over 12 draws, multiplying each probability by the next, your total probability falls off a little to 0.0000002434% (to 10 decimal places).

I hope that helped a little (and that I didn't make a gross mistake anywhere).

hurleyfor3
05-29-2007, 09:03 PM
I want to say the probability is

(number of ways to choose 12 among the missing 115 items) divided by (number of ways to choose 12 among all 577 items)

which is

(115 choose 12) divided by (577 choose 12)

which is

((115!)/(12!103!))/((577!)/(12!565!))

or 2.433e-9

or 1 in 410,886,984

hughgs
05-29-2007, 09:14 PM
I want to say the probability is

(number of ways to choose 12 among the missing 115 items) divided by (number of ways to choose 12 among all 577 items)

which is

(115 choose 12) divided by (577 choose 12)

which is

((115!)/(12!103!))/((577!)/(12!565!))

or 2.433e-9

or 1 in 410,886,984

Think about the missing items as red balls and the present items as blue balls. What are the odds of choosing 12 red balls? Hence the low probabilities presented above.

mr. synellinden
05-29-2007, 10:01 PM
I think the simple way to think of it (and I think people have already stated this correctly) is that you were asked for 12 items. Each time you were asked for an item there was a 105/567 (or 115/577 - your math was off in the hypothetical) chance of that item being missing.

The odds of your colleague asking for a missing item twelve straight times is close to 105/567 to the 12th power. (The odds of flipping a coin and it coming up heads 12 times in a row is 1/2 to the 12th power. The odds of a .250 hitter going 5 for 5 is 1/4 to the 5th power.)

However, since each time you are asked for an item that is missing, the pool goes down, the real calculation would be as follows:

105/567 x 104/566 x 103/565 x 102/564 x 101/563 x 100/562 x 99/561 x 98/560 x 97/559 x 96/558 x 95/557 x 94/556

This is really the exact same thing that juise stated, I just put it in different words.

BluBones
05-30-2007, 05:29 AM
In short, the odds are overwhelming that your partner is acting in bad faith. Attack.

Bostondevil
05-30-2007, 07:54 AM
I agree with BluBones.

And everyone else.

(I didn't know there were so many fellow statisticians on this board!)

OZZIE4DUKE
05-30-2007, 07:57 AM
In short, the odds are overwhelming that your partner is acting in bad faith. Attack.

Or, if you use the example from hughgs

Think about the missing items as red balls and the present items as blue balls.
you probably have blue balls... :rolleyes:

hughgs
05-30-2007, 10:21 AM
Or, if you use the example from hughgs

you probably have blue balls... :rolleyes:

Thanks for picking up on that. I almost changed it to green, but held out hope that someone could make the color blue work :).

dukestheheat
05-30-2007, 01:30 PM
hey Bostondevil-

I'm actually not a true statistician, I just play one on TV.

haha!

dth.

hurleyfor3
05-30-2007, 02:12 PM
Yeah, I'm not a statsmatician either, just a dork who remembers Algebra 2.