# Probability Friday Fun

January 22, 2010 1:55 PM Subscribe

Hi everyone. Say I have a list of 50 items. I pick 10 of them. I put them back.
What is the probability that, the SECOND time I pick 10 items, I pick an item I already picked the first time?
How about the THIRD time, from the first or second?
How about the probability of picking 3 of the same item, or 5? Thanks!

*How about the THIRD time, from the first or second?*

If I haven't screwed this up, the probability should be about 99.3%.

The k term indicates the number of items of overlap between the first and second picking rounds. This term should be the probability that the overlap in the first two rounds is k and this term should be the probability of avoiding the first two picking rounds in the third when the overlap in the first two rounds is k.

posted by mhum at 2:26 PM on January 22, 2010

*What is the probability that, the SECOND time I pick 10 items, I pick an item I already picked the first time*

Exactly one or at least one?

assuming that you mean at least one then I think Perplexity is correct, but a more economical notation is 1-40P10/50P10

posted by Neiltupper at 2:36 PM on January 22, 2010

Thanks all. @Neiltupper, for the first question, I just want to know at least one. For the last portion, "3 of the same item, or 5" would be where I am asking about the calculation for EXACTLY 3, or 5.

Appreciate the help!

posted by EduTek at 2:41 PM on January 22, 2010

Appreciate the help!

posted by EduTek at 2:41 PM on January 22, 2010

*How about the probability of picking 3 of the same item,*

There's some ambiguity here. I'll assume that this means that at least one item is picked in all three picking rounds. In which case, I think the probability is around 34.3%. Here, the second term (50-k, 10)/(50,10) represents the probability of avoiding the intersection of the first two picking rounds in the third round.

posted by mhum at 2:42 PM on January 22, 2010

*For the last portion, "3 of the same item, or 5" would be where I am asking about the calculation for EXACTLY 3, or 5."*

Oh wait. I guess I misunderstood. Can you give an example of five of the same item? Do you mean that an item can be selected more than once in a given round? If that's the case, I'm not sure that the previous analyses hold up.

posted by mhum at 2:44 PM on January 22, 2010

For ease of terminology, let's say the 10 you picked the first time are marked red, the others black.

The probability of picking n red items in the next ten is

40! * 10! * 40!

----------------------------- * 10!

(30+n)! * (10-n)! * 50!

So:

n=1, p(n red) = 0.2662

n=3, p(n red) = 0.0181

n=5, p(n red) = 0.0006

I'm having trouble parsing "How about the THIRD time, from the first or second?" Could you clarify what you mean by that?

posted by Salvor Hardin at 2:46 PM on January 22, 2010

The probability of picking n red items in the next ten is

40! * 10! * 40!

----------------------------- * 10!

(30+n)! * (10-n)! * 50!

So:

n=1, p(n red) = 0.2662

n=3, p(n red) = 0.0181

n=5, p(n red) = 0.0006

I'm having trouble parsing "How about the THIRD time, from the first or second?" Could you clarify what you mean by that?

posted by Salvor Hardin at 2:46 PM on January 22, 2010

Yeah, I was reading this as "no item can be selected more than once in a round, i mean that I select the 10 items in round 1, then start over and select 10 items in round 2, what is the probability that 3 are the same?"

Which way are you asking?

posted by jeb at 2:46 PM on January 22, 2010

Which way are you asking?

posted by jeb at 2:46 PM on January 22, 2010

Sure. I have 50 numbers, 1 through 50. I cannot pick the same number once. During the second round, I have all of my 50 numbers back in the pool.

As for the how about the third time, I mean:

Round 1: I have 50 numbers, I pick 10. I put them back.

Round 2: I have 50 numbers, I pick 10. I put them back.

Round 3: I have 50 numbers, I pick 10. What's the probability that I picked at least one of these numbers during round 1 OR round 2.

posted by EduTek at 2:55 PM on January 22, 2010

As for the how about the third time, I mean:

Round 1: I have 50 numbers, I pick 10. I put them back.

Round 2: I have 50 numbers, I pick 10. I put them back.

Round 3: I have 50 numbers, I pick 10. What's the probability that I picked at least one of these numbers during round 1 OR round 2.

posted by EduTek at 2:55 PM on January 22, 2010

*Round 3: I have 50 numbers, I pick 10. What's the probability that I picked at least one of these numbers during round 1 OR round 2.*

In this case, my previous answer should hold up.

posted by mhum at 2:57 PM on January 22, 2010

@mhum: Example for picking 5 of the same item:

Round 1. Pick 10 from 50. Put them back.

Round 2. Pick 10 from 50. What is the probability that FIVE of the numbers are also in the round 1 set.

posted by EduTek at 2:58 PM on January 22, 2010

Round 1. Pick 10 from 50. Put them back.

Round 2. Pick 10 from 50. What is the probability that FIVE of the numbers are also in the round 1 set.

posted by EduTek at 2:58 PM on January 22, 2010

Whoa, should have previewed. Huh.

Maybe I should explain my reasoning.

Let's determine the probability of picking 9 black ones, then 1 red one in that order.

p(9,1) =

40*39*38*37*36*35*34*33*32*10

50*49*48*47*46*45*44*43*42*41

You can see that regardless of the order, the probability is the same (different orderings just mix around the terms on the top)

So, how many orderings are there? 10P10 = 10!/9!

So,

p(1 red) =

40!/31! * 10

-------------- * 10!/9!

50!/40!

Or generalizing,

p(n red) =

40!/(30+n)! * 10

------------- * 10!/(10-n)!

50!/40!

p(n red) =

40! * 10 * 40!

----------------------------- * 10!

(30+n)! * (10-n)! * 50!

NOTE: I just noticed I have an extra "!" written in my original general answer. I still stand by my end results; it's just a typo.

posted by Salvor Hardin at 2:58 PM on January 22, 2010

Maybe I should explain my reasoning.

Let's determine the probability of picking 9 black ones, then 1 red one in that order.

p(9,1) =

40*39*38*37*36*35*34*33*32*10

50*49*48*47*46*45*44*43*42*41

You can see that regardless of the order, the probability is the same (different orderings just mix around the terms on the top)

So, how many orderings are there? 10P10 = 10!/9!

So,

p(1 red) =

40!/31! * 10

-------------- * 10!/9!

50!/40!

Or generalizing,

p(n red) =

40!/(30+n)! * 10

------------- * 10!/(10-n)!

50!/40!

p(n red) =

40! * 10 * 40!

----------------------------- * 10!

(30+n)! * (10-n)! * 50!

NOTE: I just noticed I have an extra "!" written in my original general answer. I still stand by my end results; it's just a typo.

posted by Salvor Hardin at 2:58 PM on January 22, 2010

General solution which answers your first and third question, and by repeated application can answer your second:

If you have

P(N,a,b,c) = [a!b!(N-a)!(N-b)!]/[N!c!(a-c)!(b-c)!(N-a-b+c)!]

So for your first question, the probability of picking no items in common is P(50,10,10,0) ≈ .0825, so the probability of picking at least one in common is 1 minus that, or about .9175. (matches perplexity's answer)

Exactly one item in common is P(50,10,10,1) ≈ .2662. Exactly five items in common is P(50,10,10,5) ≈ .0161.

For your question about the third draw, we'd have to first figure out the probabilties for each possible value of c (0 through 10) for the second draw, because that gives us the total number of distinct items you drew the first two times. I.e., if there were 0 common items in the first two draws, you drew 20 distinct items; if there was 1 common item, you drew 19 distinct items, and so forth.

Therefore, the probability of drawing no items drawn the third time which were drawn either of the first two times—making no assumptions about whether or not the first two draws had any items in common—is:

P(50,10,10,0) * P(50,20,10,0) +

P(50,10,10,1) * P(50,19,10,0) +

P(50,10,10,2) * P(50,18,10,0) +

P(50,10,10,3) * P(50,17,10,0) +

P(50,10,10,4) * P(50,16,10,0) +

P(50,10,10,5) * P(50,15,10,0) +

P(50,10,10,6) * P(50,14,10,0) +

P(50,10,10,7) * P(50,13,10,0) +

P(50,10,10,8) * P(50,12,10,0) +

P(50,10,10,9) * P(50,11,10,0) +

P(50,10,10,10) * P(50,10,10,0)

which Excel tells me is about .0068, or about .9932 probability that you'll pick at least one item the third time that you also picked at least one of the first two times. (matches mhum's answer)

posted by DevilsAdvocate at 3:09 PM on January 22, 2010 [2 favorites]

If you have

**N**items, draw**a**distinct items, replace them, and draw**b**distinct items, the probability that the two groups have exactly**c**items in common is:P(N,a,b,c) = [a!b!(N-a)!(N-b)!]/[N!c!(a-c)!(b-c)!(N-a-b+c)!]

So for your first question, the probability of picking no items in common is P(50,10,10,0) ≈ .0825, so the probability of picking at least one in common is 1 minus that, or about .9175. (matches perplexity's answer)

Exactly one item in common is P(50,10,10,1) ≈ .2662. Exactly five items in common is P(50,10,10,5) ≈ .0161.

For your question about the third draw, we'd have to first figure out the probabilties for each possible value of c (0 through 10) for the second draw, because that gives us the total number of distinct items you drew the first two times. I.e., if there were 0 common items in the first two draws, you drew 20 distinct items; if there was 1 common item, you drew 19 distinct items, and so forth.

Therefore, the probability of drawing no items drawn the third time which were drawn either of the first two times—making no assumptions about whether or not the first two draws had any items in common—is:

P(50,10,10,0) * P(50,20,10,0) +

P(50,10,10,1) * P(50,19,10,0) +

P(50,10,10,2) * P(50,18,10,0) +

P(50,10,10,3) * P(50,17,10,0) +

P(50,10,10,4) * P(50,16,10,0) +

P(50,10,10,5) * P(50,15,10,0) +

P(50,10,10,6) * P(50,14,10,0) +

P(50,10,10,7) * P(50,13,10,0) +

P(50,10,10,8) * P(50,12,10,0) +

P(50,10,10,9) * P(50,11,10,0) +

P(50,10,10,10) * P(50,10,10,0)

which Excel tells me is about .0068, or about .9932 probability that you'll pick at least one item the third time that you also picked at least one of the first two times. (matches mhum's answer)

posted by DevilsAdvocate at 3:09 PM on January 22, 2010 [2 favorites]

*Round 2. Pick 10 from 50. What is the probability that FIVE of the numbers are also in the round 1 set.*

Oh, I see. In that case, ignore what I said .

The answer to your question is actually in this link. So, for example, with k=5, the probability is about 1.6%. Here's a table:

0 0.082519234 (no items in common) 1 0.266191078 2 0.336898083 3 0.2177927 4 0.078469429 5 0.016142282 6 0.00186832 7 0.000115417 8 3.41696E-06 9 3.89398E-08 10 9.73494E-11 (you pick the exact same 10 items in round 2 as in round 1)

posted by mhum at 3:11 PM on January 22, 2010

Ergh. What I meant to say was "Ignore what I said here". The rest is fine.

posted by mhum at 3:22 PM on January 22, 2010

posted by mhum at 3:22 PM on January 22, 2010

« Older Does having a PowerPc processor cause programs to... | You didn't really want to post that to MeFi, did... Newer »

This thread is closed to new comments.

So there are 40 items which are "new". When you take the first item from the second set of picks, you have a 40/50 chance of picking a "new" item. Then when you take the second item from the second set of picks, you have a 39/49 chance of picking a "new" item. And so on.

So the chance of picking 10 "new" items is:

(40/50) * (39/49) * (38/48) * (37/47) * (36/46) * (35/45) * (34/44) * (33/43) * (32/42) * (31/41) = 0.082519234. So the chance of picking at least one item you already picked the first time is about 91.75%.

posted by Perplexity at 2:03 PM on January 22, 2010