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Suppose we have $12$ identical balls. Think of the balls as having distinct ID numbers, $1-12$. We have $3$ identical boxes, each of which can contain exactly $4$ balls. One after the other the balls are thrown at random into one of the three boxes.

Suppose Mr. A will buy one ball from each box, so he will buy a total of three balls from the three boxes. Habit of Mr. A is to buy that ball from the box which ID is smallest of all other balls in that box.

For example, the randomly thrown balls in a box have ID $4$, $8$, $9$, and $12$. So Mr. A will buy the ball with ID $4$ from that box.

Then Mr. A goes to another box and here the balls have ID $1$, $3$, $6$, and $11$. He will buy the ball with ID $1$ from that box.

At last, Mr. A goes to the last box and here the balls have ID $2$, $5$, $7$, and $10$. He will buy the ball with ID $2$ from that box.

My question is:

What is the probability that Mr. A will buy the ball with ID $i$, where $i=1,2,\ldots,12$?

What I think is that we need to consider the following two things to calculate the probability:

$(1)$ the position of a ball in the box. Mr. A will buy the ball with ID number $2$ if and only if the box does not contain the ball with ID number $1$.

$(2)$ If Mr. A buy a ball with smallest ID from the first box in front of him, then are there $11$ balls remaining to buy another two balls or are there $8$ balls (excluding all $4$ balls from the first box)?

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    The probability that Mr. A will buy a ball? Isn't that 1? Because he will buy a ball from each of the boxes if i'm interpreting this correctly – 2017-01-10
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    Under what condition does Mr. A not buy a ball? The way you described it, Mr. A always buys exactly 3 balls. – 2017-01-10
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    @suomynonA I edited my question – 2017-01-10

3 Answers 3

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For $1\lt j\lt k$, there are $3!\binom{12-k}{3}\binom{8-j}{3}$ ways to buy $1,j,k$.

$3!$ due to the arrangements of $1,j,k$. $\binom{12-k}{3}$ due to the number of ways to choose $3$ balls greater than $k$. $\binom{8-j}{3}$ due to the number of ways to choose $3$ balls greater than $j$ from those left over from choosing $3$ greater than $k$.

The number of ways for $n$ to be the greatest is $$ \begin{align} 3!\binom{12-n}{3}\sum_{j=2}^{n-1}\binom{8-j}{3} &=3!\binom{12-n}{3}\left[\binom{7}{4}-\binom{9-n}{4}\right]\\ &=3!\binom{7}{4}\binom{12-n}{3}-3!\binom{7}{4}\binom{12-n}{7} \end{align} $$ The number of ways for $n$ to be the middle is $$ \begin{align} 3!\binom{8-n}{3}\sum_{k=n+1}^9\binom{12-k}{3} &=3!\binom{8-n}{3}\binom{12-n}{4}\\ &=3!\binom{7}{4}\binom{12-n}{7} \end{align} $$ Since the total number of arrangements is $\frac{12!}{(4!)^3}$, and $\frac{3!\binom{7}{4}}{\frac{12!}{(4!)^3}}=\frac1{165}$, the probability of buying the ball labeled $n$ would be $$ p(n)=\frac1{165}\binom{12-n}{3} $$

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    Although it doesn't appear to be, this turns out to be the same as Nafiz Ishtiaque Ornob's answer. – 2017-01-10
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    I didn't see bof's answer until after I posted. Of course, this matches their answer. – 2017-01-10
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I got the following answer:

The probability that Mr. A will buy the ball with id $i$ is, $$p(i) := \left( \begin{array}{c} 8 \\ i-1 \end{array} \right) \frac{(12-i)!(i-1)!}{11!}\,. $$

If you know this answer to be incorrect, please feel free to stop reading :)

Total number of ways the balls can be distributed among three boxes is $$n := \frac{12!}{(4!)^3}\,.$$

Let us assume that there are $12$ numbered slots, one slot is to be occupied by exactly one ball, slots $1$ to $4$ constitute box $1$, slots $5$ to $8$ constitute box $2$ and the rest constitute box $3$. At the beginning we will consider all the slots to be distinguishable but at the end we will consider all the slots in the same box to be indistinguishable from each other.

We will count the number ways the $i$'th ball can be bought from box $1$. The $i$'th ball can be placed in either of the $4$ slots in box $1$. The balls with id $1$ to $(i-1)$ can be placed in box $2$ and $3$ in $\left( \begin{array}{c} 8 \\ i-1 \end{array} \right)(i-1)!$ ways. Balls with id $(i+1)$ to $12$ can be arranged in the rest of slots in $(12-i)!$ ways. Multiplying all these numbers we get $4 \left(\begin{array}{c} 8 \\ i-1 \end{array}\right) (i-1)! (12-i)!$, in order to impose indistinguishability of the slots in the same box we must divide this number by $(4!)^3$. So, the number ways the $i$'th ball can be bought from box $1$ is: $$n_1(i) := \frac{4}{(4!)^3} \left(\begin{array}{c} 8 \\ i-1 \end{array}\right) (i-1)! (12-i)!\,.$$ Therefore, the probability of the $i$'th ball being bought from box one is: $$p_1(i) := \frac{n_1(i)}{n}\,.$$ The $i$'th ball can be bought from either of the three boxes, so the net probability of the $i$'th ball being bought is: $$p(i) := 3 p_1(i) = \left( \begin{array}{c} 8 \\ i-1 \end{array} \right) \frac{(12-i)!(i-1)!}{11!}\,.$$

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    Though the balls are identical, but giving those ID numbers doesn't make them distinct? Why shouldn't we consider the positions of the balls within a box so that the total number of ways the balls can be distributed among three boxes is $n = 12!$ ? – 2017-01-14
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    Is it because any of the 12 balls can get the ID number 1? – 2017-01-14
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    @user81411 The balls are indeed distinct due to the ID numbers, but positions inside a box aren't, which is what makes the capacity of a box 4. You care about which ID numbers are inside a box, but not about how they are oriented inside the box. If the positions inside a box were distinguishable then that box would be better described as four different boxes with capacity one each. Consider the difference between having 12 boxes where each can contain one ball and having 3 boxes where each can contain 4 balls. – 2017-01-14
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    But in a bookshelf there are 3 shelf. Each self can contain 4 books. we have 12 different books. We can arrange 4 books on a shelf in 4! ways. Why does here position matter? – 2017-01-14
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    Quite right, this is a matter of choice after all, as you said, you can arrange 12 different books in 3 different shelves in 12! ways, but then saying that you have 3 different shelves is redundant, you are simply considering all possible arrangements of 12 different objects, why bother with the shelves at all. In the given problem, the extra information of 3 different boxes has been _presumed_ to imply that we should forget about the arrangements inside a given box as an extra layer of added complexity to the problem. – 2017-01-14
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    That is, the total number of ways I can arrange 12 different books in 12! ways if the bookshelf has only one shelf. Similarly, for this particular problem, the total number of ways I can arrange 12 distinct balls in 12! ways if there were only one box. Is it? – 2017-01-14
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    Well, I guess it's better put this way, suppose you have 3 shelves, but you are messy and you just throw the books in the shelves instead of arranging them in order. Then you wouldn't care about how the books are arranged in a shelf only which books are in which shelf. On the other hand, if you are extremely tidy and you even arrange the balls neatly in a line inside the boxes then you should consider 12! ways of ordering the balls. So it's not exactly a question of "how many containers", rather "are the objects in a container organized". – 2017-01-14
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    *"The $i$'th ball can be bought from either of the three boxes..."* But my concern is: the probability of $i$th ball being bought from the first box is not same as the probability of $i$th ball being bought from the second box. Because after the 1st box, we have 8 more balls from which 2 balls will be bought. – 2017-01-16
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    @ABC you can show that the number of arrangements of balls where the $i$'th ball will be bought from box $a$ is the same as the number of arrangements of the balls where the $i$'th ball will be bought from box $b$ for any $a, b \in \{1,2,3\}$. Because for any arrangement where the $i$'th ball will be bought from box $a$, you can make a new arrangement by swapping the balls between box $a$ and $b$, and vice versa. – 2017-01-16
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Ball $i$ is thrown into a box with $3$ other balls. There are $\binom{11}3$ possible combinations for the three balls in the same box with ball $i,$ and all of them are equally likely. The man will buy ball $i$ if the three other balls have numbers greater than $i$. There are $12-i$ balls with numbers greater than $i,$ so there are $\binom{12-i}3$ favorable cases, and the probability that the man buys ball $i$ is $$\frac{\binom{12-i}3}{\binom{11}3}=\frac{(12-i)(11-i)(10-i)}{990}.$$ This is equivalent to the answer posted earlier by Nafiz Ishtiaque Ornob.

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    much better than mine :) – 2017-01-10
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    *"There are $\binom{11}3$ possible combinations for the three balls in the same box with ball $i,$"*. But if $i$ ball is in the 2nd box, then aren't there $\binom{7}3$ possible combinations for the three balls in the 2nd box with ball $i$? Because $4$ balls of the 1st box is no more under consideration and we have $8$ balls left to buy another 2 balls. – 2017-01-16
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    *"There are $12-i$ balls with numbers greater than $i,$"*. But if $i$ is in the second box, we have only $8$ balls remaining whose id is within 12. – 2017-01-16
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    @ABC Irrelevant. The question is about *absolute* probability. It's not about the *conditional* probability that ball $i$ is bought, *given* that ball $i$ is in a certain box and certain other balls are in other boxes. By the way, "One after the other the balls are thrown at random into one of the three boxes."; so the boxes are *not* filled in order, not that that matters. – 2017-01-16
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    @ABC Here's a simple illustration of the kind of mistake you're making, maybe it will help. Shuffle a deck of cards, deal a card to me, then deal one to yourself. What's the probability that I get a red card? Hmm, 52 cards, all equally likely, 26 of them red, so my chance of getting a red card is 26/52=1/2, right? OK, now **what's the probability that you get a red card**? Bear in mind that I get my card before you get yours. Is your chance of getting a red card less than 50% or is it more than 50%? – 2017-01-16
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    `Settings A`: One after the other the balls are thrown at random into one of the three boxes. `Settings B`: 12 balls, 1-12. ID 1 ball is thrown into one of the three boxes. Since no other ball appears in any boxes before ID 1 ball, Mr. A will buy that ball. Now ID 2 ball is thrown at random into one of the three boxes . Mr. A will buy the ball with ID 2 if and only if the box does not contain the ball with ID 1. Then ID 3 ball is thrown at random into one of the three boxes. He will buy the ball with ID 3 iff the box does not contain ID 1 or 2. Are the boxes in settings B filled in order? – 2017-01-17
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    @ABC My solution assumes that the balls are all treated the same, i.e., one after another you pick up a ***random*** ball and throw it into a random box; and that the labels are only used to determine which balls Mr. A will buy. If the balls are thrown in order of their labels, then you need to clarify how the box in which a ball lands is randomized. When ball #2 is thrown, what's the probability that it lands in the same box with #1? Is it $1/3$ or is it $3/11$? – 2017-01-17
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    If the probability that ball #2 lands in the same box with #1 is $1/3$, then the definition of randomization can be defined as `one after another I pick up a random ball and throw it into a random box`. Isn't it? – 2017-01-18
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    @ABC Nope. If you pick up one *random* ball after another, then all $12$ balls are on the same footing, and the probability that ball #1 and ball #2 end up in the same box is $3/11.$ In this case, the $\frac{12!}{4!4!4!}$ ways of packing the balls into $3$ boxes ($4$ to a box) are all equally likely (since one differs from another only in the labels), and the number of ways in which #1 and #2 are in the same box is $3\cdot\frac{10!}{2!4!4!}.$ – 2017-01-18
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    @ABC If balls are being picked up and thrown at random, you might as well assume that you start with unlabeled balls and apply the labels to the balls at random. And it makes no difference if the labels are applied before you start, or after all $12$ balls have been thrown. Consider the latter procedure. Start with label #1 and apply it to a random ball. Now take label #2; there are $11$ unlabeled balls to apply it to, and $3$ of them are in the same box with $1.$ – 2017-01-18
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    @ABC Let $p$ be the probability that two designated balls land in the same box. By symmetry, $p$ is the same for each of the $\binom{12}2=66$ pairs of balls. Let $X$ be the number of pairs of balls which land in the same box. By the method of indicator variables (linearity of expectation) $E(X)=66p.$ But $X$ is a constant, $X=3\binom42=18.$ So $E(X)=66p=18$ so $p=18/66=3/11.$ – 2017-01-18