# Probability and identical balls problem

Suppose we have an urn with $2$ white and $2$ black balls then probability of drawing a black ball is $1/2$ .

In text books solutions to more difficult versions of this kind of ball problems are explained by naming black balls as $b_1 b_2 ..$ and whites as $w_1 w_2 ..$ etc..

My question:

These balls are identical, i.e blacks are identical in them and so does whites. Why do we name them as $b_1 b_2 ... w_1 w_2...$ is it a correct way to deal with problem how can an identical ball be numbered and thought that way ? How do you know which one you picked ? This confuses me.

The labeling here is unimportant. The random variables for each color possess the property of being exchangeable, roughly meaning that permuting their labels will not change the result. If you permute $b_1$ and $b_2$, $w_1$ and $w_2$ (call $b_1$ by the label $b_2$ etc..., You will still end up with th same answer.

So basically, the labels here are only for mathematical convenience.

If you wouldn't name or label identical balls, you wouldn't be able to count them: ball #1, ball #2, etc. But otherwise, what the other answer says.

In counting problems it plays a huge rôle whether the objects one is dealing with, or the actions one performs, are labeled or not. Stanley in his Enumerative combinatorics, Vol. 1, distinguishes 12 ways to count maps $f:\ A\to B$, depending on whether $f$ is supposed to be arbitrary, injective, or surjective, and whether the elements of $A$, resp. $B$, are labeled or not.

If you want to derive some combinatorial formula for unlabeled objects it may helpful to label them anyway and then forget about the labeling in a second step. Consider the following example:

In how many ways can 3 red, 5 blue, and 2 yellow balls be arranged in a row? Label the red balls from 1 to 3, the blue balls from 1 to 5, and the yellow balls 1 and 2. Then we have 10 distinguishable balls which can be arranged in $10!$ ways. Now the 3 red balls among them can be arranged in $3!=6$ ways which cannot be distinguished when we rip of the labels, and similarly the blue balls can be arranged among them in $5!=120$ ways, the yellow balls in $2$ ways. It follows that the total number $N$ of linear arrangements of the colored, but unlabeled balls is given by $$N={10!\over 3!\cdot 5!\cdot 2!}=2520\ .$$

• Isn't there an nicely illustrative subtlety here that you might have missed? RRRBBBBBYY and YYBBBBBRRR both contribute to your $N$. If you cannot make out the "direction" of the row (if the ends of the row aren't labelled start and end), you should count them as one. Commented Jan 20, 2013 at 9:07
• @Gugg: I'd say that in such problems a "row" is directed if nothing is said to the contrary. Commented Jan 20, 2013 at 9:43