What is the probability of this?

  • Thread starter EngWiPy
  • Start date
  • #1
1,367
61
Hi,

Suppose that there are M persons, and K balls number from 1 to K. Each person pulls k balls at the same time, and return them back. What is the probability that all persons pull the same k balls?

Thanks
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,836
251
Hi S_David! :wink:

tell us what you think, and then we'll comment! :smile:
 
  • #3
1,367
61
Hi S_David! :wink:

tell us what you think, and then we'll comment! :smile:

Well, if we suppose that the probability of each ball is p, then the probability of a specific k-tuple is:

[tex]{K\choose k}p^k(1-p)^{K-k}[/tex]

I guess the probability that all persons will pull the same k-tuple, will be:

[tex]\left({K\choose k}p^k(1-p)^{K-k}\right)^M[/tex]

Is that true?
 
  • #4
chiro
Science Advisor
4,797
133
Are there K balls in total for everyone or K balls per person?
 
  • #5
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,957
7,202
Well, if we suppose that the probability of each ball is p, then the probability of a specific k-tuple is:

[tex]{K\choose k}p^k(1-p)^{K-k}[/tex]

No, that's quite wrong. How many equally likely k-tuples are there? So what is the chance of picking a particular one?
 
  • #6
1,367
61
Are there K balls in total for everyone or K balls per person?

We have K balls in total, where each person pick up k balls, and then return them back to the box. So each person has the same set of balls to pick from them.
 
  • #7
1,367
61
No, that's quite wrong. How many equally likely k-tuples are there? So what is the chance of picking a particular one?

We have K choose k distinct k-tuple patterns. Right?
 
  • #8
250
0
We have K choose k distinct k-tuple patterns. Right?

The solution you gave is the probability that M persons choose the same number of balls regardless their label.

But you already know they choose the same number of balls, soooooo...
 
Last edited:
  • #9
1,367
61
The solution you gave is the probability that M persons choose the same number of balls regardless their label.

But you already know they choose the same number of balls, soooooo...

clue:

Imaging 3 balls

000
001
010
011
100
101
110
111

What is the probability M persons choose 010?

I am sot sure!! Another tip?
 
  • #10
chiro
Science Advisor
4,797
133
We have K balls in total, where each person pick up k balls, and then return them back to the box. So each person has the same set of balls to pick from them.

In this case, this is wrong.

I think what you mean and correct me if I'm wrong, is that you have K balls and everyone takes turns in picking the balls out, not that they all pick out the balls at once.

If they all pick them out at once, then there is no way they can pick out the same balls and the question is not even valid.

So lets assume that they take turns and they pick out the balls each turn for each person and the process is random in line with the assumptions of a binomial distribution. We assume that after the person has picked the balls, they put them back in and the whole thing starts again.

Then under this situation, you will need to calculate the probability of getting the specific number of balls taking into account that the balls are in fact the same. This should be done from first principles to clarify and not just using a distribution in ad-hoc manner.

So in terms of calculating the probability of getting a particular set of balls you need to calculate the following probability:

Let your balls be a,b,c and let A be the first ball, B the second, and C the third. Let X = (A = a) XOR (B = a) XOR (C = a), Y = (A = b) XOR (B = b) XOR (C = b) and Z = (A = c) XOR (B = c) XOR (C = c). Then you need to calculate the probability P(X AND Y AND Z).

Where the XOR is the symmetric difference which is given by (A XOR B) = (A\B) OR (B\A).

You can use the fact that A,B,C are not the same since picking up a ball and not replacing it until the round has finished will change what happens to later balls in the same round.

If you expand the probabilities then you should get a simpler expression which will give you the actual probability and you will have to use the Kolmogorov axioms at some point for calculating P(A OR B) and so on.

You could also do it with the binomial in the way that you calculate getting k balls out of K and then factoring in that you only get one type of permutation of your balls out of all permutations, but still account for getting them in any order. This would probably be easier to calculate in contrast with the raw probability calculation above.
 
  • #11
250
0
I am sot sure!! Another tip?

Well, I guess that one was not a very good tip anyway :tongue:

OK, imaging you have 5 balls 1,2,3,4,5 what is the probability you pick, let's say 2 and 4? Well, then you would go the pair 2,4 divided by all the possible pairs in 1,2,3,4,5.

So that is the probability you choose one particular pair, so that is the probability anyone else chooses that one particular pair... sooo... what is your guess now?
 
  • #12
1,367
61
Well, I guess that one was not a very good tip anyway :tongue:

OK, imaging you have 5 balls 1,2,3,4,5 what is the probability you pick, let's say 2 and 4? Well, then you would go the pair 2,4 divided by all the possible pairs in 1,2,3,4,5.

So that is the probability you choose one particular pair, so that is the probability anyone else chooses that one particular pair... sooo... what is your guess now?

If we suppose that probability of ball i is pi, then the probability of choosing 2 and 4 is:

[tex]p_2p_4(1-p_1)(1-p_3)(1-p_5)[/tex]

Right?

Then why we need to divide by the number of pairs?
 
  • #13
250
0
If we suppose that probability of ball i is pi, then the probability of choosing 2 and 4 is:

[tex]p_2p_4(1-p_1)(1-p_3)(1-p_5)[/tex]

Right?

Then why we need to divide by the number of pairs?

What? :yuck: that's a new problem! in the way you stated your problem every ball is equally likely to be chosen...

Suppose that there are M persons, and K balls number from 1 to K. Each person pulls k balls at the same time, and return them back. What is the probability that all persons pull the same k balls?

OK, the solution for the this problem as it is stated is [itex] {{K}\choose{k}}^{-M}[/itex]
 
  • #14
chiro
Science Advisor
4,797
133
If we suppose that probability of ball i is pi, then the probability of choosing 2 and 4 is:

[tex]p_2p_4(1-p_1)(1-p_3)(1-p_5)[/tex]

Right?

Then why we need to divide by the number of pairs?

Dude, just do a binomial distribution and adjust for the different permutations so that you only select 1 permutation since all permutations are equally likely.

How many permutations can you have when you select 3 balls from 6? (Hint: Use combinatorics)
 
  • #15
1,367
61
What? :yuck: that's a new problem! in the way you stated your problem every ball is equally likely to be chosen...



OK, the solution for the this problem as it is stated is [itex] {{K}\choose{k}}^{-M}[/itex]

What if they are not equally likely? Why p does not appear in the probability?

Actually, this is not the real question. The real question is: if each person pulls different number of balls, what is the probability that the cardinality of the intersection between all users is m?
 
  • #16
chiro
Science Advisor
4,797
133
What if they are not equally likely? Why p does not appear in the probability?

Actually, this is not the real question. The real question is: if each person pulls different number of balls, what is the probability that the cardinality of the intersection between all users is m?

Maybe you should ask the real question first off :)

Use probability axioms for situations that don't fit existing models.
 
  • #17
1,367
61
Maybe you should ask the real question first off :)

Use probability axioms for situations that don't fit existing models.

I am sorry, but it is just so confusing to me. I am not so good in probability, and I am trying to figure this out, since I am working on an engineering model, that has the same issue.

Thanks
 
  • #18
250
0
What if they are not equally likely? Why p does not appear in the probability?

Actually, this is not the real question. The real question is: if each person pulls different number of balls, what is the probability that the cardinality of the intersection between all users is m?

It does not appear because every ball is supposed to have the same probability and that simplifies the problem.

In the new stated problem, again, every ball has the same probability! so you already know your new solution won't have any p either.... :wink:
 
  • #19
1,367
61
It does not appear because every ball is supposed to have the same probability and that simplifies the problem.

In the new stated problem, again, every ball has the same probability! so you already know your new solution won't have any p either.... :wink:

OK, I will try to figure it out. Thanks
 
  • #20
250
0
OK, I will try to figure it out. Thanks

No problem, but anyway, the answer for your new problem, assuming equally likely every cardinality, is [itex]K^{-M}[/itex].
I think you've just suffered enough :tongue:
 
  • #21
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,957
7,202
OK, the solution for the this problem as it is stated is [itex] {{K}\choose{k}}^{-M}[/itex]
Not quite. That's the prob they all pick the same prespecified k. But it could be any k, so long as they all pick the same k:
[itex] {{K}\choose{k}}^{-M+1}[/itex]
 
  • #22
250
0
Not quite. That's the prob they all pick the same prespecified k. But it could be any k, so long as they all pick the same k:
[itex] {{K}\choose{k}}^{-M+1}[/itex]

The original problem was:
Suppose that there are M persons, and K balls number from 1 to K. Each person pulls k balls at the same time, and return them back. What is the probability that all persons pull the same k balls?

That pretty much specifies the number of balls that are extracted: k

But even if you transform the problem and turn "each person pulls k" into "each person pulls the same number" being those numbers equally likely, and the question "What is probability that all persons pull the same k balls?" into "What is probability that all persons pull the same balls?" Then the answer would be:

[tex]\frac{1}{K+1}\sum_{k=0}^K {{K}\choose{k}}^{-M}[/tex]
 
  • #23
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,957
7,202
Let's take a simple example. K=2, k=1, M=2.
Each takes one ball and replaces it. What's the probability they take the same ball?
You say
[itex] {{K}\choose{k}}^{-M}[/itex] = 2-2 = 1/4
I say [itex] {{K}\choose{k}}^{-M+1}[/itex] = 2-1 = 1/2
Think about it.
 
  • #24
250
0
Let's take a simple example. K=2, k=1, M=2.
Each takes one ball and replaces it. What's the probability they take the same ball?
You say
[itex] {{K}\choose{k}}^{-M}[/itex] = 2-2 = 1/4
I say [itex] {{K}\choose{k}}^{-M+1}[/itex] = 2-1 = 1/2
Think about it.

Oh, by 'prespecified' k you meant the actual set contained within the k balls, yes, then you're right. :smile:
 
  • #25
36
0
Hi,

Suppose that there are M persons, and K balls number from 1 to K. Each person pulls k balls at the same time, and return them back. What is the probability that all persons pull the same k balls?

Thanks

David, I understand that this is not the actual problem you want to solve. What is the problem you want to solve?
 
Top