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Say you have $2n+2b$ balls where $2n$ balls are colored white, $b$ balls are colored blue and $b$ balls are colored red.

You have two urns. You randomly choose $n+b$ balls and throw in urn $1$ while you place the remaining $n+b$ balls in urn $2$.

What is the probability that the blue balls and red balls are in separate urns?

I am most interested in case $\frac{n}b\rightarrow\infty$ such as $b=n^{\frac1c}$ with $c>1$ being fixed and in case $\frac{n}b\rightarrow c$ such as $b={\frac nc}$ with $c>1$.

joriki
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  • Why is this tagged [tag:polya-urn-model]? – joriki Sep 07 '15 at 12:09
  • is it tagged wrongly? –  Sep 07 '15 at 12:11
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    Yes, I thought that was implied in my question. If you don't know what the Pólya urn model is, please remove the tag. If you do, please explain why it's relevant. – joriki Sep 07 '15 at 12:12
  • could you please tag this properly then? –  Sep 07 '15 at 12:14
  • @Arul the tag says it's about white and black balls, you have white, blue and red. – null Sep 07 '15 at 12:14
  • I do not see much distinction. I could rewrite $blue=white$, $red=black$ and $white=colorless$ or $numbered$ –  Sep 07 '15 at 12:14
  • @Arul could you please read the Wikipedia entry that joriki has linked. – tomglabst Sep 07 '15 at 12:15
  • @Arul: It's bad style to change the question substantially after two answers have already been given. It's confusing to readers and causes unnecessary effort for the respondents. If it didn't occur to you before asking the original question that you're interested in a certain limit, you should accept one of the answers to your original question and ask a new question about that limit. – joriki Sep 07 '15 at 12:29

2 Answers2

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There are $\binom{2n}n$ ways of selecting $n$ of the $2n$ white balls to go with the red balls. Thus, for $b\gt0$ there are $2\binom{2n}n$ ways of separating the red and blue balls, where the factor $2$ occurs because the red balls can be in either of the two urns. There are $\binom{2n+2b}{n+b}$ ways to select balls for one of the urns, so this is the total number of outcomes. Since the selections are equiprobable, the probability for a separating selection is

$$ \frac{2\binom{2n}n}{\binom{2n+2b}{n+b}}\;. $$

For $b=0$, the factor of $2$ should be omitted, and the probability of (trivial) separation is $1$.

joriki
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  • What does trivial mean? –  Sep 07 '15 at 12:18
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    Trivial here means that you don't need the deduction from above to come to this conclusion. Since there are no blue and red balls, you can never fail to separate them. – tomglabst Sep 07 '15 at 12:19
  • So probabilty of separating red and blue balls as $\frac{n}b\rightarrow\infty$ approaches $1$? –  Sep 07 '15 at 12:21
  • @Arul: No, that conclusion is not warranted based on the answer I gave. The case $b=0$ is a special case, because the symmetry factor $2$ doesn't occur in this case. Having all red balls in the first urn and all blue balls in the second is the same as having all blue balls in the first urn and all red balls in the second, if there are none of either. That says nothing about large values of $\frac nb$ for $b\ne0$. In fact as $\frac nb\to\infty$ for fixed $b$, the probability of separation approaches $2^{-(2b-1)}$ (the otherwise incorrect answer that Aretino deleted). – joriki Sep 07 '15 at 12:24
  • The scenario I am thinking is posted above. –  Sep 07 '15 at 12:28
  • why is there a factor of $2b-1$ in exponent instead of $b$? –  Sep 07 '15 at 12:45
  • @Arul: That sort of question is hard to answer if one doesn't think that it should have been $b$. To make it answerable, you'll have to explain why you thought it should be $b$. – joriki Sep 07 '15 at 12:46
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Let's pretend for a moment that the balls are numbered (therefore distinct), the total number of possible choices is ${2n+2b}\choose{n+b}$. In how many of these do you get all of the blue balls and none of the red balls in the first urn? You have to pick $n$ of the $2n$ white balls to place in the first urn and then add all pf the blue ones (no choice), that is ${2n}\choose n$. Now you only have to multiply this by two (for symmetry between the urns or the colors, depending on your point of view) and end up with $2\frac{{2n}\choose n}{{2n+2b}\choose{n+b}}$.

Shahar Even-Dar Mandel
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  • What happens at $b\rightarrow 0$? –  Sep 07 '15 at 12:19
  • Oops! I see @joriki beat me with the answer and also addressed the case of $b=0$ (in which the question becomes kind of meaningless). – Shahar Even-Dar Mandel Sep 07 '15 at 12:22
  • Actually I am looking for what happens at $\frac{n}b\rightarrow\infty$ which case $b>0$ is allowed. Does probability that you separate red and blue balls approach $1$? It is counterintuitive. Am I right? –  Sep 07 '15 at 12:23
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    Not at all. Just think of the $b=1$ case with $n$ very large, the probability of separation in this case will be very close to $\frac{1}{2}$. – Shahar Even-Dar Mandel Sep 07 '15 at 12:25
  • Could you please provide calculations for this. The exact scenario I am thinking is updated. –  Sep 07 '15 at 12:26
  • Yes but your answer does not provide the resolution I need. That is why I calrified the scenario. Sorry about this. I should have posted this first. –  Sep 07 '15 at 12:29