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Suppose there are $K$ independent boxes, each of which has the same number of different colored balls, denoted as $N$. For the first case, we assume the colors of the balls in the $K$ boxes are the same. For the second case, we assume that the balls in the $K$ boxes are of different colors. For an example, the possible colors of balls are $\{\text{ red} ,\text{blue} ,\text{green} \}$. Assume that each box is of size $2$. Then, for the first case, each box is of balls $\{\text{ red} ,\text{blue} \}$. They are symmetric. For the second case, an example is $\{\text{red} ,\text{blue} \},\{\text{red} ,\text{green} \} $ and $ \{\text{blue} , \text{green} \}$. There is no necessity for the color permutation. But they are asymmetric.

Consider the experiment where each time we pick a single ball from each box randomly and get $K$ colored balls totally. Then, we would like to compare the expected number of colors for the $K$ colored balls, denoted as $\{B_1,B_2\}$, picked independently from the $K$ boxes under the above mentioned two cases. Intuitively, $B_1 \le B_2$. I have demonstrated it when $K = 2$. However, when $K$ goes larger, the case becomes more complicated. I am still confused by how to prove it theoretically.

Any references or approximations would be appreciated. Thanks a lot!

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  • $\begingroup$ In the first case, the number and the color of the balls in each basket are the same. That is, the balls in the baskets are exactly the same. To compare with it, the second case has the baskets which are of the same number of balls as that of the first case, but not necessarily the same color. What I want to know is which case has relatively larger average number of colors for the $K$ balls by drawing a ball at random from each basket? $\endgroup$ – user372097 Sep 27 '16 at 1:53
  • $\begingroup$ We only assume the color and number for the 2 cases are fixed. In general, we want to prove it in the case where there are $K$ baskets and each basket are of $N$ balls. And the number of colors is $F$. Actually, what we really matter is the symmetry and asymmetry of the color of the balls in the baskets, irrespective of the exact color distribution. $\endgroup$ – user372097 Sep 27 '16 at 2:06
  • $\begingroup$ In the first case, the expected number of balls of each colour is $K/N$. In the 2nd case, the expected number of balls of each colour is $K/B$ where $B$ is the number of equally frequently occurring balls. Any colour occuring twice or more in each box will have its frequency multiplied accordingly. $\endgroup$ – samerivertwice Sep 27 '16 at 11:24
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In the first case, the expected number of balls of each colour is $K/N$ where $N$ is both the number of balls in each box and the number of different colours. In the 2nd case, the expected number of balls of each colour is $K/N$ where $N$ is the number of different balls, provided they permutate equally frequently as shown in your example. Assuming the number of colours in case 2 must exceed the number of boxes, as implied, then your inequality must be correct.

This is because as follows:

Let $K$ be the number of boxes. Let $N$ be the number of balls in each box. Let $b_m\in B$ be the list of counts of each distinct colour in the box, such that:$$\sum\limits_{b\in B} b_m=N$$ be the number of distinct colours.

Therefore the first case is the special case of the 2nd case in which $b_m=1\forall b\in B$.

In a single draw from a single box, probability colour $m$ is chosen, is $b_m/N$.

In the first case, $b_m=1\forall b\in B$: For one box, so the probability of choosing colour $m$ from a boxis $1/N$

Probability of success or failure is the same for every box so this is a binomial distribution for every colour. In a binomial, $E(X)=np$. We are drawing one from each box so the number of trials is $K$, and probability is $1/N$ so so the expected number is $E(X)=K/N$

Moving on to the 2nd part of the question, when there may be different numbers of each colour in each box. Here we may permit any $b_m\geq 0$, and also potentially different sets $N_p$ for every box. However in the example you give, every permutation of two choices selected from {red, blue, green} is shown. This has he effect that the expected value of each must be equal. Therefore we have:$$\sum\limits_mE(X_m)=E(\sum\limits_m X_m)$$

Therefore for every colour $m$: $$E(X_m)=K/N$$

Where this time, if we use your notation $B$ to replace $N$, is the total number of colours permutated.

Provided you do not permit fewer colours than boxes, which again is implied but not stated in the question, $B\geq N \implies K/B\leq K/N$ so you are correct in your assertion which you stated using $B_n$ that $B_1\leq B_2$.

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  • $\begingroup$ Thank you very much for your explicit explanation and proof. However, I think what I have not make clear are that: first, each color in each box is fixed to only number one and the colors of balls in a specific box are different from each other; secondly, the second scenario is only supposed to have boxes of balls of different colors but the same number. So there is no necessity that it is permutation selected. $\endgroup$ – user372097 Sep 27 '16 at 13:32
  • $\begingroup$ In other words, we can explain the two different cases as the symmetric case (balls in the boxes are of the same number and exactly the same colors) and the asymmetric case (balls in the boxes are of the same number but not necessarily the same colors). $\endgroup$ – user372097 Sep 27 '16 at 13:36
  • $\begingroup$ @user372097 ok well my answer caters for the symmetric case, whether each box contains the same or whether there is an unbiased, symmetric pattern of balls equally distributed between boxes. In either of those cases, if there are more colours than boxes then your inequality is correct. But in the asymmetric case, it will depend on the number of balls of that particular colour in every box and whether there are on a weighted average basis, more than $1$ ball of that colour in each box. $\endgroup$ – samerivertwice Sep 28 '16 at 8:59
  • $\begingroup$ Yeah, I assume that the number of balls of a particular color in every box is fixed to number one. $\endgroup$ – user372097 Sep 29 '16 at 1:40

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