Expected number of unbroken pairs...

[johnk]

Hello,

I need some help with this type of problem:
In a box there are two cards labeled "1", two labeled "2", etc. up to "N", so there are 2N cards in total.
When we take M cards randomly out of the box, what is the expected number of unbroken pairs left in the box?

I tried solving it for M = 1, M = 2 and looking at the pattern, but it gets too complex... can't wrap my mind around this.
Thanks for any help.

 

[johnk]

Never mind, I think it's
\(\displaystyle N\cdot \frac{2N-M}{2N}\cdot \frac{2N-M-1}{2N-1}\),
as in (number of pairs) * (probability that the first card is not one of the removed cards) * (probability that the second card is not one of the other removed cards)