Justification for the calculation of expectation on the first indicator random variable of a series of mutually dependent experiments.

Question

I am not asking for a proof of linearity of expectation, which is available in multiple posts.

Instead I would like to develop a more robust understanding as to why the calculation of the expectation of the number of $BB$ when pairing up $N$ balls ($W$ and $B,$ such that $W + B = N$) can be performed by considering simply the probability of getting the first pair $BB,$ which is obviously $\frac{B}{B+W}\frac{B-1}{B+W-1}$ times the number of pairs, as in this answer by @lulu, which I am sure it is correct:

This sort of thing is best handled by indicator variables, exploiting the fact that expectation is linear, regardless of any possible dependence between the variables.

$$E=E\left[ \sum X_i\right]=\sum E\left[X_i\right]=\frac {B+W}2\times \frac B{B+W}\times\frac {B-1}{B+W-1}=\boxed {\frac {B(B-1)}{2(B+W-1)}}$$

My intuitive check on it is that each pair of balls is equally likely to be the first one to be chosen, but this is in conflict with the fact that the probability of getting a second $BB$ pair after a first $BB$ will necessarily go down.

Answer 1

Suppose that you have a deck of cards. The probability of the second card being a king is the same as the probability of the first card to be a king, they are completely symmetric. Similarly, the probability of any card being a king is the same and equal. However, once the first card is dealt and opened, the conditional probability given its face value, changes the probability of the second card to be king.

The same here. The indicator $X_i$ looks only on the $i$th experiment. If you don't know anything that happened before, this is exactly the same as $X_1$. You are correct that $X_2\vert X_1$ is different from $X_2$, but as you calculate the unconditional $X_2$, it is the same. The conditional would go up or down based on the result of the first pair, but when computing all options you'll get back to the unconditional distribution.

A stronger intuition can be made using the following idea: for each pairing here is another pairing where all balls are paired the same but pair #1 and pair #i are switched.

A full calculation: $\Pr(X_1=1)=\tfrac{{B \choose 2}}{{n \choose 2}}=\tfrac{B(B-1)}{n(n-1)}$ since you need to choose two black balls to form the first pair. We all agree on that. Now, for $X_2$: $$\Pr(X_2=1)=\tfrac{{B \choose 2}}{{n \choose 2}}\cdot\tfrac{{B-2 \choose 2}}{{n-2 \choose 2}}+\tfrac{{W \choose 2}}{{n \choose 2}}\cdot\tfrac{{B \choose 2}}{{n-2 \choose 2}}+\tfrac{{B \choose 1}{W \choose 1}}{{n \choose 2}}\cdot\tfrac{{B-1 \choose 2}}{{n-2 \choose 2}}=\tfrac{B(B-1)(B-2)(B-3)}{n(n-1)(n-2)(n-3)}+\tfrac{W(W-1)B(B-1)}{n(n-1)(n-2)(n-3)}+2\tfrac{B(B-1)W(B-2)}{n(n-1)(n-2)(n-3)}=\tfrac{B(B-1)}{n(n-1)}\left[\tfrac{(B-2)(B-3)}{(n-2)(n-3)}+\tfrac{W(W-1)}{(n-2)(n-3)}+2\tfrac{W(B-2)}{(n-2)(n-3)} \right] $$ Now note that $(B-2)(B-3)+W(W-1)+2W(B-2)=(B-2)(W+B-3)+W(W-1+B-2)=(B-2)(n-3)+W(n-3)=(n-3)(W+B-2)=(n-2)(n-3)$ so the $[\ldots]=1$ and $\Pr(X_2=1)=\Pr(X_1=1)$

Answer 2

I still think it is best to convince yourself that random pair selection means that each ball is equally likely to be chosen into any particular pair. But if you want to do it explicitly:

(sketch)

You can do this in gory detail if you like. Just use induction. It is easy to see this if $B,W$ are small (just enumerate all the possible outcomes and check by hand). Let us then assume that if $B+W<n$ that all couples have the same probability of being Black/Black, namely $P(B,W)=\frac B{B+W}\times \frac {B-1}{B+W-1}$. And now consider an ensemble for which B+W=n$.

Now we draw the first pair. There are three possible outcomes for that and we know their probabilities. It follows that $$P_i(B,W)=\frac B{B+W}\times \frac {B-1}{B+W-1}\times P_{i-1}(B-2,W)\;+$$ $$2\frac B{B+W}\times \frac W{B+W-1}\times P_{i-1}(B-1,W-1)\;+$$+$$\frac W{B+W}\times \frac {W-1}{B+W-1}\times P_{i-1}(B,W-2)$$

Where the subscript denotes which pair we are considering.

Now, the induction hypothesis applies to all the $P_{i-1}(*,*)$ terms on the right. Thus the right hand is independent of $i$, so the left hand is as well. Moreover we know, via the induction hypothesis, what each of those $P_{i-1}(*,*)$ terms on the left are, so we can just write it all out. It's a little messy, but not difficult, and we confirm the general form as desired.

Should note: once you have shown it is independent of $i$, then the answer must have the given form, since that is the form for $P_1(B,W)$.

Answer 3

A given pair, say the pair $n$ is BB if and balls $2n-1$ and $2n$, in the order of drawing, are both black. There are $$\frac{(N-2)!}{W!(B-2)!}$$ drawing orders that give this result, regardless of the value of $n$.