What is the probability of this?


by S_David
Tags: probability
S_David
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#19
May25-12, 07:09 PM
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Quote Quote by viraltux View Post
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....
OK, I will try to figure it out. Thanks
viraltux
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#20
May25-12, 07:13 PM
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Quote Quote by S_David View Post
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
haruspex
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#21
May25-12, 07:37 PM
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Quote Quote by viraltux View Post
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]
viraltux
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#22
May26-12, 04:49 AM
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Quote Quote by haruspex View Post
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]
haruspex
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#23
May26-12, 05:50 AM
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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.
viraltux
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#24
May26-12, 07:05 AM
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Quote Quote by haruspex View Post
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.
ClifDavis
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#25
May26-12, 05:04 PM
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Quote Quote by S_David View Post
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?
S_David
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#26
May26-12, 05:23 PM
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Quote Quote by ClifDavis View Post
David, I understand that this is not the actual problem you want to solve. What is the problem you want to solve?
If each person pulls different number of balls, what is the probability that the cardinality of the intersection between all users is m?
ClifDavis
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#27
May26-12, 05:39 PM
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Quote Quote by S_David View Post
If each person pulls different number of balls, what is the probability that the cardinality of the intersection between all users is m?
Okay let me restate the problem to see if I understand it correctly.

There are M people and K distinct balls. Each of those M people will pick a number k at random between one and K inclusive and then pick k balls at random. An reliable observer will record which balls they picked. Then the selected balls are put back for the next person. After all M of these people have done this the reliable observer informs us that none of the M people picked the same number k of balls to select.

Someone gives us an integer m. We want to give them the odds that the intersection of the ball picks contains exactly m balls.

Is this a correct statement of the problem???
S_David
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#28
May26-12, 05:59 PM
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Quote Quote by ClifDavis View Post
Okay let me restate the problem to see if I understand it correctly.

There are M people and K distinct balls. Each of those M people will pick a number k at random between one and K inclusive and then pick k balls at random. An reliable observer will record which balls they picked. Then the selected balls are put back for the next person. After all M of these people have done this the reliable observer informs us that none of the M people picked the same number k of balls to select.

Someone gives us an integer m. We want to give them the odds that the intersection of the ball picks contains exactly m balls.

Is this a correct statement of the problem???
Ok, let me elaborate more: We have M person and K distinct balls numbered from 1 to K. Each person will pull a number of balls, i.e.: the number of picked balls is not known a priori. Then he will return the balls to the next person (after taking the balls' numbers by a reliable observer) who will do the same thing. At the end we will find the intersection between the balls picked by all persons. I need to find the probability that the cardinality of this intersection is m, where m is an integer between 1 and K.

I hope it is more clear now.

Thanks
ClifDavis
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#29
May26-12, 07:39 PM
P: 36
Quote Quote by S_David View Post
Ok, let me elaborate more: We have M person and K distinct balls numbered from 1 to K. Each person will pull a number of balls, i.e.: the number of picked balls is not known a priori. Then he will return the balls to the next person (after taking the balls' numbers by a reliable observer) who will do the same thing. At the end we will find the intersection between the balls picked by all persons. I need to find the probability that the cardinality of this intersection is m, where m is an integer between 1 and K.

I hope it is more clear now.

Thanks
Quote Quote by S_David View Post
Ok, let me elaborate more: We have M person and K distinct balls numbered from 1 to K. Each person will pull a number of balls, i.e.: the number of picked balls is not known a priori. Then he will return the balls to the next person (after taking the balls' numbers by a reliable observer) who will do the same thing. At the end we will find the intersection between the balls picked by all persons. I need to find the probability that the cardinality of this intersection is m, where m is an integer between 1 and K.

I hope it is more clear now.

Thanks
In order to answer the question it is not necessary to know the number of picked balls a priori. It is however necessary to know the probability distribution of the number of balls picked.

If there are M=2 people and K=5 distinct balls and there is a very high probability of 4 or 5 balls getting picked and none at all of 1 or 2 balls, then it is a very different situation than if there is no probability of 4 or 5 balls being picked and a high probability ofr 1 or 2 balls.

If I understand your question, there is not enough information to answer it.
S_David
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#30
May26-12, 07:55 PM
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Quote Quote by ClifDavis View Post
In order to answer the question it is not necessary to know the number of picked balls a priori. It is however necessary to know the probability distribution of the number of balls picked.

If there are M=2 people and K=5 distinct balls and there is a very high probability of 4 or 5 balls getting picked and none at all of 1 or 2 balls, then it is a very different situation than if there is no probability of 4 or 5 balls being picked and a high probability ofr 1 or 2 balls.

If I understand your question, there is not enough information to answer it.
All balls are equiprobable to be picked up. Is it still not complete?
haruspex
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#31
May26-12, 08:01 PM
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ClifDavis is right, of course: you need to specify the probability distribution for k.
But it sounds from an earlier post that k is supposed to have a uniform distribution from 1 to K. If so, it sounds like a very nasty problem.
haruspex
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#32
May26-12, 08:02 PM
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Quote Quote by S_David View Post
All balls are equiprobable to be picked up. Is it still not complete?
No, it's still not complete. That piece of information all sets of k balls are equally likely once k is chosen. You still haven't said how k is chosen.
S_David
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#33
May26-12, 08:11 PM
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Quote Quote by haruspex View Post
No, it's still not complete. That piece of information all sets of k balls are equally likely once k is chosen. You still haven't said how k is chosen.
You can assume a distribution to simplify it. I just need to see the idea.
haruspex
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#34
May26-12, 08:19 PM
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In that case I select a binomial distribution. This means I can rephrase it as "each ball is chosen independently with probability p".
The probability that a given ball is selected by all M is then pM.
The number of balls selected by all is binomial with parameter pM.
However, I have violated the statement that k must be 1 to K. I've made it 0 to K. Is that alright?
S_David
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#35
May26-12, 08:29 PM
P: 590
Quote Quote by haruspex View Post
In that case I select a binomial distribution. This means I can rephrase it as "each ball is chosen independently with probability p".
The probability that a given ball is selected by all M is then pM.
The number of balls selected by all is binomial with parameter pM.
However, I have violated the statement that k must be 1 to K. I've made it 0 to K. Is that alright?
Yeah.
viraltux
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#36
May27-12, 08:44 AM
P: 250
Quote Quote by haruspex View Post
In that case I select a binomial distribution. This means I can rephrase it as "each ball is chosen independently with probability p".
The probability that a given ball is selected by all M is then pM.
The number of balls selected by all is binomial with parameter pM.
However, I have violated the statement that k must be 1 to K. I've made it 0 to K. Is that alright?
The last time S_David explained the problem it seemed he is worried about "the cardinality of the intersection" which I think hardly relates to the "The number of balls selected by all"

Rephrasing the problem this way turns it into a different one. When a problem states thing like "Each person will pull a number of balls" the most common approach is to assume every number is equally likely. It would be like asking people to choose a number from 0 to K, if we do not consider things like psychological biases (like people choosing their favorite number) the most reasonable is to assume every number is equally likely. The moment you assign a probability p to a ball to be chosen then it implies things like "it is more likely people choose 4 balls than 5", which I see this nowhere in the problem. Not to mention that with this approach you need to estimate p which is not given by the problem.

Also, in any case, when a given ball is selected by all M is then pM, but the probability that all M select the same ball is pM-1... unless the problem has changed again which wouldn't be surprising at this point.


Quote Quote by S_David View Post
Ok, let me elaborate more: We have M person and K distinct balls numbered from 1 to K. Each person will pull a number of balls, i.e.: the number of picked balls is not known a priori. Then he will return the balls to the next person (after taking the balls' numbers by a reliable observer) who will do the same thing. At the end we will find the intersection between the balls picked by all persons. I need to find the probability that the cardinality of this intersection is m, where m is an integer between 1 and K.

I hope it is more clear now.

Thanks
I don't think there is information missing in this problem, the key difficulty lays in the "cardinality of this intersection" part. If we had the distribution for the intersection of cardinalities of M subsets within a set of cardinality K the problem is done.

I did calculate such distribution of m for any given cardinalities C1 and C2 considering M=2 and is already huge, and for M>2 it only gets worse

I think would I go Montecarlo on this one.


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