# Balls in a box probabilities

A box is filled out by $1,000$ balls. The box can be thought of as containing $V$ sites and $V$ balls, with $V=1,000$. The box is repeatedly shaken, so that each ball has enough time to visit all $1,000$ sites. The ball are identical, except for being uniquely numbered from $1$ to $1,000$.

What is the probability that all of the balls labeled from $1$ to $100$ lie in the left hand side of the box?

What is the probability that exactly $P$ of the balls labeled $1$ to $100$ lie in the left hand side of the box?

Using Stirling's approximation, show that this probability is approximately Gaussian. Calculate the mean of $P$. calculate the root mean square fluctuations of $P$ about the mean. Is the Gaussian approximation good?

Any insight is greatly appreciated.

• Imagine that the balls are given random positions one at a time. So first ball number 1 is placed in a random position. Then ball number 2 is placed in one the 999 remaining positions, and so on. This point of view makes the problem easier. – Daan Michiels Oct 23 '12 at 15:33
• I understand that, but don't know hat the probability of finding balls 1-100 on the left hand side would be... 500 balls on lhs, need 1-100. I don't know how to set up the distribution. And then how this distribution would change for P. i understand the question, just can't set up the distributions properly. – Alex Trent Oct 23 '12 at 15:37
• What are these "sites" mentioned in the problem? – Thomas Andrews Oct 23 '12 at 15:47
• You can think of the box containing V balls and V sites, with V=1000 – Alex Trent Oct 23 '12 at 15:54
• In your suggested comment, you were taking "on the left" to be the extreme left. The rest of the problem suggests that it should be "on the left side of the box". I have made that more explicit. – Ross Millikan Oct 23 '12 at 16:29

Hint: for the first, think of looking at the balls in order. What is the chance that ball $1$ is on the left side of the box? Given that it is on the left, and occupying one site, what is the chance that ball $2$ is on the left side of the box? You should be able to write this in terms of combinations, which will help with the thinking on the second part.

For the second, choose $P$ of the balls $1-100$ to be on the left side of the box. How many ways are there to do that? Then choose where they go on the left. How many ways are there to do that? Then choose where the $100-P$ go on the right. How many ways to do that? The product is the number of ways to get $P$ out of $1-100$ on the left.

• Comment from Alex Trent: I'm sorry, I don't follow what you said. There are 1000 balls and 1000 sites. 500 of these sites are on the left hand side. balls 1-100 represent 1/10 of the total number of balls, and therefore occupy 1/10 of the sites. but on the left hand side, they occupy 1/5 of the sites. so the probability that balls 1-100 are on the left hand side is 1/5? – Brian M. Scott Oct 23 '12 at 17:08
• @Alex Trent: No, the first ball has chance $\frac {500}{1000}$ to be on the left. The second $\frac {499}{999}$ to be on the left as one site is taken. – Ross Millikan Oct 23 '12 at 18:28

Note:   "Not otherwise distinguishable" simply means that there is no bias in where balls land all distinct outcomes — all distinct arrangements of balls in the box are equally likely.   So we may measure probability by comparing counts of permutations of arrangements: favoured versus total.

What is the probability that all of the balls labelled from $1$ to $100$ lie in the left hand side of the box?

• Favoured space: count ways to select all $100$ of the $100$ special balls and $500-100$ of $1000-100$ non special balls. (Special meaning labelled 100 or less)
• Total space: count ways select any $500$ of all $1000$ balls.

What is the probability that exactly $P$ of the balls labelled $1$ to $100$ lie in the left hand side of the box?

• Favoured space: count selections of only $P$ of the $100$ special balls and $500-P$ of $1000-100$ non-special balls.
• Total space: count selections of any $500$ of all $1000$ balls.

Verification: when $P = 100$ the answers should be the same.