Comments on: What are the odds of finding 20 needles in 270 haystacks?

Question: What are the odds of finding 20 needles in 270 haystacks?

COBRA! — Wed, 01 Dec 2004 07:40:26 -0800

Suppose there are 20 objects hidden in a collection of 270 haystacks. What are your odds of searching 100 haystacks (assuming that if you choose to search a haystack with an object in it, you'll find that object) without finding an object? My discrete math skillz have eroded to the point where any attempt to work this out myself yields preposterous numbers..

By: Doohickie

Doohickie — Wed, 01 Dec 2004 07:44:08 -0800

I think it would be (80!-60!)/(270!-250!)

Do I pass the test?

By: GrumpyMonkey

GrumpyMonkey — Wed, 01 Dec 2004 07:44:25 -0800

As a follow up to this - what are the chances that you would not find one of those 20 objects after unrolling 100 haystacks over an 8 hour period?

By: andrew cooke

andrew cooke — Wed, 01 Dec 2004 07:47:59 -0800

can you have more than one object in a single haystack?

By: Doohickie

Doohickie — Wed, 01 Dec 2004 07:49:36 -0800

No, those minus signs should be division signs.

If I do my math right, that gives you 4x10-12 (4 times 10 to the minus 12th power).

By: Doohickie

Doohickie — Wed, 01 Dec 2004 07:50:58 -0800

Naw, I'm still screwed up.

By: naxosaxur

naxosaxur — Wed, 01 Dec 2004 07:51:44 -0800

COBRA!, i know what you did last night last night.

By: smackfu

smackfu — Wed, 01 Dec 2004 07:52:49 -0800

At the end there were only 12 objects to find, since 8 of the objects were removed by others and not replaced. That hurts the odds and needs to be reflected.

By: dual_action

dual_action — Wed, 01 Dec 2004 07:53:13 -0800

God, I felt bad for her. I would have given up about hour two.

I can only think there must have been one, she just missed it. I can't imagine how hard it would be to search in the dark.

By: vraxoin

vraxoin — Wed, 01 Dec 2004 07:53:45 -0800

Listen, dude. Stop fucking around with probabilites and just get out there with a pitchfork. You'll find those objects.

By: briank

briank — Wed, 01 Dec 2004 07:54:29 -0800

I was dumbstruck that they could not find a clue packet after eight hours of unrolling the bales.

(For those who don't watch, the question refers to a challenge in this week's episode of "The Amazing Race")

By: golo

golo — Wed, 01 Dec 2004 07:57:29 -0800

If there can be one and only one object per haystack, then the chances of finding an object in any random stack would be 20/270 (0.074). The chances of picking a random stack and NOT finding an object then would be 1-0.074= 0.925 (good chances). The chances of this happening twice would be 0.925 squared (0.857), for three stacks would be 0.925 cubed and so on. If I'm right the chances of choosing 100 haystacks and not finding simngle object would be (1-(20/270))^100=4.54e-4, or something like 0.00045. Not very likely

By: pardonyou?

pardonyou? — Wed, 01 Dec 2004 07:57:43 -0800

I found it distinctly unfair to end an elimination leg with a pure luck challenge. They've done that before, and it just infuriates me.

(still one of the best shows, though)

By: Doohickie

Doohickie — Wed, 01 Dec 2004 07:58:15 -0800

250/270 x 249/269 x 248/268 x ... x 151/171 = 3.8 x 10-9(3.8 times 10 to the minus 9th power)

By: COBRA!

COBRA! — Wed, 01 Dec 2004 07:59:23 -0800

smackfu's right, but I wanted to leave out that complication until the easier version was worked out.

And only one object per haystack, andrew cooke.

It's important to me to work this out, because my wife had ten bucks riding on Lena and Kristy, and I'm trying to console her with the knowledge that she's the victim of a statistical freak accident.

By: Doohickie

Doohickie — Wed, 01 Dec 2004 08:00:59 -0800

golo: With each haystack checked, the denominator decrements by one. For each pick, there are (n-20)/n possible good picks, and n goes from 270 to 171 (one hundred trials)

By: COBRA!

COBRA! — Wed, 01 Dec 2004 08:02:20 -0800

250/270 x 249/269 x 248/268 x ... x 151/171 = 3.8 x 10-9(3.8 times 10 to the minus 9th power)

That's the construction I was using, but that can't be right, can it? That seems outlandish. I guess it was outlandish, though.

By: andrew cooke

andrew cooke — Wed, 01 Dec 2004 08:02:35 -0800

if it's a single object per haystack, then you have 250 haystacks without objects, at first. so the probability of picking an "empty" haystack is 250/270. the probability of picking another is then 249/269. and so on. so the final total probability (for 100 haystacks) is (250x249x...151)/(270x269x...171) which is (250!/150!)*(170!/270!). which is about 1 in 16000. so you are very likely to find an object.

and it doesn't matter how many hours it takes.

By: Doohickie

Doohickie — Wed, 01 Dec 2004 08:06:37 -0800

Yep, I agree with andrew cooke.... 1 chance in 16,223. I ran the numbers wrong the last time.

I think we have an answer.

By: gyc

gyc — Wed, 01 Dec 2004 08:07:17 -0800

pardonyou?: I agree with you completely, but plenty of other teams have been eliminated due to pure luck (such as slow cab drivers and airplane delays). OK, back to the math.

By: golo

golo — Wed, 01 Dec 2004 08:20:22 -0800

Thanks Doohickie. Apparently I was thinking of choosing 100 stacks at a time, not choosing one by one and checking their contents before continuing.

By: box

box — Wed, 01 Dec 2004 08:25:19 -0800

In terms of probability, it's my understanding that it doesn't matter whether you choose 100 stacks at a time or if you choose one and check before continuing. (Well, assuming you don't accidentally pick a stack you already chose.)

By: Divine_Wino

Divine_Wino — Wed, 01 Dec 2004 08:28:09 -0800

However, it's still a real world problem and I think she might have missed one or two of the clues in bales she unrolled (at one point her sister told her to look around in the already unrolled stacks which she did listlessly) what with the fatigue and the panic and the running around, so that fuzzies the odds. (as Dual_Action points out)

As long as we're on the topic that screamy dude with the died blue hair at the back of his head needs to be fired out of a cannon into a pit of rats with herpes. I want to find his wife and pay for her divorce lawyer even though I'm poor.

By: Galvatron

Galvatron — Wed, 01 Dec 2004 08:35:47 -0800

andrew cooke's got it. Another way of arriving at the solution: there are C(250, 100) unique selections of 100 haystacks that have no hidden item (where C(., .) is a binomial coefficient). There are C(270, 100) total unique selections of 100 haystacks. So the probability of not finding an item is C(250, 100) / C(270, 100) = (250!/100!/150!) / (270!/100!/170!) = (250!/150!)*(170!/270!)

By: saintsguy

saintsguy — Wed, 01 Dec 2004 09:17:52 -0800

There is an additional assumption being made that if you search a haystack, then you are guaranteed to find the object hidden within it.

Of course, this is completely untrue, hence "needle in a haystack".

This complication shortens the odds of not finding a single clue.

By: yerfatma

yerfatma — Wed, 01 Dec 2004 09:17:57 -0800

This was on in the background last night while we were applying for mortgage refinancing. My contribution to the financial pursuit went something like this:

"She's still fucking looking . . . still looking . . . oh my God, the old people caught up to them . . . it's dark now."

By: Mayor Curley

Mayor Curley — Wed, 01 Dec 2004 09:37:35 -0800

As long as we're on the topic that screamy dude with the died blue hair at the back of his head needs to be fired out of a cannon into a pit of rats with herpes.

Send the guy with the two little bundles of hair on his forehead with them. And I'm very worried that the old couple and the father/daughter team will get eliminated and there will be no one to root for.

By: COBRA!

COBRA! — Wed, 01 Dec 2004 09:58:38 -0800

You can always watch and root for Bolo to spontaneously combust. It's going to happen.

By: drewbeck

drewbeck — Wed, 01 Dec 2004 11:16:54 -0800

GRANDPARENTS!!!

By: spilon

spilon — Wed, 01 Dec 2004 14:42:42 -0800

I dunno the answer to the math problem, but who knew wathching pro wrestlers trying to count stuffed animals and non-stick bakeware could be so damn entertaining!

By: amarynth

amarynth — Tue, 07 Dec 2004 12:57:50 -0800

Here's a game that let's you relive Lena's search from the comfort of your own home.