
#19
May2512, 07:09 PM

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#20
May2512, 07:13 PM

P: 250

I think you've just suffered enough 



#21
May2512, 07:37 PM

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[itex] {{K}\choose{k}}^{M+1}[/itex] 



#22
May2612, 04:49 AM

P: 250

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
May2612, 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. 



#24
May2612, 07:05 AM

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#25
May2612, 05:04 PM

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#26
May2612, 05:23 PM

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#27
May2612, 05:39 PM

P: 36

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??? 



#28
May2612, 05:59 PM

P: 590

I hope it is more clear now. Thanks 



#29
May2612, 07:39 PM

P: 36

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. 



#30
May2612, 07:55 PM

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#31
May2612, 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. 



#32
May2612, 08:02 PM

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#33
May2612, 08:11 PM

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#34
May2612, 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 p^{M}. The number of balls selected by all is binomial with parameter p^{M}. However, I have violated the statement that k must be 1 to K. I've made it 0 to K. Is that alright? 



#35
May2612, 08:29 PM

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#36
May2712, 08:44 AM

P: 250

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 p^{M}, but the probability that all M select the same ball is p^{M1}... unless the problem has changed again which wouldn't be surprising at this point. I did calculate such distribution of m for any given cardinalities C_{1} and C_{2} 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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