This is a problem solved by bernoulli, and is given as an example in my textbook.But I'm having a little trouble understanding it.
In a box containing 2n cards, there are 2 cards marked 1, 2 cards marked 2, 2 marked 3 and so on. A set of m cards are drawn at random from the box. Find the expected number of pairs that still remain in the box.
My book says: "Define $X_i$ = 1 or 0 according as the ith pair remains in the box or not, i=1,...n." This makes sense to me, because Bernoulli solved this problem, so $X_i=1$ is the ith pair is there, $X_i=0$ is it's not. There are 2n cards, so n pairs.
"We want to calculate $E(X_1+...+X_n)$. But $E(X_i)=P(X_i=1)=$"
But then I am not sure how they got this: Probably from some definition?
$$\frac{{2n-2 \choose m}}{{2n \choose m} } $$
I get the denominator, we have 2n cards and are choosing m. But I don't udnerstand why we have 2n-2 in the numerator
So I worked out the calculation myself: $$=\frac{\frac{(2n-2)!}{m!(2n-2-m)!}}{\frac{2n!}{m!(2n-m)!}}$$ $$=\frac{(2n-2)!m!(2n-m)!}{m!(2n-2-m)!2n!}$$
Got the same answer as the book: $$=\frac{(2n-m)(2n-m-1)}{(2n(2n-1))}$$
And the book then somehow gets this: $$=\frac{(2n-m)(2n-m-1)}{(4n-2)}$$ Where did the n in the denominator go?
If you could help explain this I would be very greatful!
There are $\binom{2n}{m}$ equally likely ways to pick $m$ objects from the $2n$ available. There are $\binom{2n-2}{m}$ ways to pick these objects so that a particular pair is unpicked. This is the number of "favourables." For the probability that the pair of $i$'s remains, divide.
The $n$ "disappeared" because we are calculating $E(X_1+\cdots +X_n)$, which by the linearity of expectation is $E(X_1)+\cdots +E(X_n)$. The expectations of the $X_i$ are all the same, they are what you calculated. So we obtain the expectation of the sum by multiplying $E(X_1)$ by $n$. So the denominator changes from $2n(2n-1)$ to $2(2n-1)$.