What is the probability of this?

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  • #36
viraltux
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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?

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. :tongue:


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 :tongue:

I think would I go Montecarlo on this one.
 
  • #37
ClifDavis
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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?

Now that we know what the actual problem is we can see where his
[tex]{K\choose k}p^k(1-p)^{K-k}[/tex] came from. You have the probability of getting k picks followed by K-k non-picks, which gives you one instance of picking k balls and since any selection of k balls is equally probable you multiply by the number of ways of getting k balls. Without having given it tons of thought, that looks reasonable to me. But then he goes off the tracks a little when he looks at all of the people doing the same thing.

David, suppose that I have M people and they all flip a coin. The probability of heads is 1/2 and the probability of tails is 1/2. Is the probability they all get the same result when they flip their coin = [tex](1/2)^M[/tex] Look at it if I have 3 people or 2.

(edited because I don't really understand how the tex stuff works)
 
  • #38
ClifDavis
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Now that we know what the actual problem is we can see where his
[tex]{K\choose k}p^k(1-p)^{K-k}[/tex] came from. You have the probability of getting k picks followed by K-k non-picks, which gives you one instance of picking k balls and since any selection of k balls is equally probable you multiply by the number of ways of getting k balls. Without having given it tons of thought, that looks reasonable to me. But then he goes off the tracks a little when he looks at all of the people doing the same thing.

David, suppose that I have M people and they all flip a coin. The probability of heads is 1/2 and the probability of tails is 1/2. Is the probability they all get the same result when they flip their coin = [tex](1/2)^M[/tex] Look at it if I have 3 people or 2.

(edited because I don't really understand how the tex stuff works)

David, going back and rereading what I wrote I can see where I could give the impression that simply correcting the last part of your expression to [tex]({(K\choose k}p^k(1-p)^{K-k})^{M-1}[/tex] would fix all you problems. It wouldn't. That expression would give the probability that all M people would pick the same k balls. Which isn't what you want at all.

Each of the M people could pick extra balls that all don't pick and you would still have k balls in in the intersection.

But let's go back to your original logic in coming up with the expression
[tex]{K\choose k}p^k(1-p)^{K-k}[/tex]
You had the probability that the first k balls were picked times the probability that the next K-k balls were not picked by the first person to get a specific instance of k balls being picked by one person which you then multiplied to find the probability that one person would pick k balls.

Now suppose instead you had taken the probability that the first k balls were picked [By all M people] followed by the probability that the final K-k balls were [not picked by all M people] to get the probability that a particular set of k balls were picked by everyone and then multiplied by the number of ways to get k balls -- would that not do it?
 
  • #39
haruspex
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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"
A ball is in the intersection if it was chosen by all. The cardinality of the intersection is the number of balls chosen by all.
 
  • #40
haruspex
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Now suppose instead you had taken the probability that the first k balls were picked [By all M people] followed by the probability that the final K-k balls were [not picked by all M people] to get the probability that a particular set of k balls were picked by everyone and then multiplied by the number of ways to get k balls -- would that not do it?
To do that calculation, you still need to know the probability distribution of k.
But it seems David_S is not that wedded to his "pick k, then pick k balls" formulation (which is what made the problem hard).
Given that he's happy with a binomial distribution for k, do you agree that allows us to rephrase it as "pick each ball independently with the same probability p"? Do you further agree that this gives a binomial distribution for the cardinality of the intersection, parameter pM?
 
  • #41
ClifDavis
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To do that calculation, you still need to know the probability distribution of k.
But it seems David_S is not that wedded to his "pick k, then pick k balls" formulation (which is what made the problem hard).
Given that he's happy with a binomial distribution for k, do you agree that allows us to rephrase it as "pick each ball independently with the same probability p"? Do you further agree that this gives a binomial distribution for the cardinality of the intersection, parameter pM?

I agree. In fact judging by his initial attempt at a solution it really seems to be what he had in mind from the beginning.

In fact, my question to him should be read, "given that each ball is picked by each person with a probability p, if you had taken the probability that the first k balls were picked [By all M people] followed by the probability that the final K-k balls were [not picked by all M people] to get the probability that a particular set of k balls were picked by everyone (and so constitute the intersection) and then multiplied by the number of ways to get k balls -- would that not provide the probability that exactly k balls were in the intersection?"

Edited for Clarification.
 
Last edited:
  • #42
ClifDavis
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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"

A ball is in the intersection if it was chosen by all. The cardinality of the intersection is the number of balls chosen by all.

Yeah, I will admit to being curious about what viraltux thinks "cardinality of the intersection" does relate to.
 
  • #43
EngWiPy
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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?

OK, when I thought about this in my system model, this solution makes sense. Thanks
 

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