Intuition vs Result on this probability question. Question is mentioned so many times on the internet and actually is an example in Sheldon Ross' textbook 'A First Course in Probability' (p 39, 8th ed.).

A football team consists of 20 offensive and 20 defensive players. The players are to be paired in groups of 2 for the purpose of determining roommates. If the pairing is done at random, what is the probability that there are no offensive-defensive roommate pairs? What is the probability that there are 2i offensive-defensive roommate pairs when i = 1,2,3...,10?

I came to an answer which is: 
$$
P_{2i} = \cfrac{ {{20}\choose{2i}}(2i)!\Big[\cfrac{(20-2i)!}{{{2}^{10-i}}(10-i)!}\Big]^2 }{ \cfrac{(40)!}{{2^{20}}{(20)!}} }
$$
where $P_n$ denote the probability that there are n pair(s) of offensive-defensive players as roommate. 
With the Stirling's approximation, the answer for several situations goes like this: 
$$P_0 \approx 1.3403 \times 10^{-6}$$
$$P_{10} \approx .345861$$
$$P_{20} \approx 7.6068 \times 10^{-6}$$
And there goes my thoughts:
The numerical results are against my intuition. If there are 20 - 20 offense defense players and they are to be grouped, I would surmise there should be 50% chance of having $P_{10}$ situation, which is the probability that only 10 pairs of offense/defense players being assigned as teammates and rest of them being assigned with players with the same position (defense-defense, offense-offense). But the likeliness of this happening is less than 50%. 
Furthermore, the asymmetry between $P_0$ and $P_{10}$ bothers me as well. Why is the probability of having none of players being matched with different position 5 times less than the probability of having all of the players being matched with different position? My intuition says those probabilities should be the same and I feel it's equally hard to assign them completely segregated or integrated.
I am not questioning the math here, but still can't wrap my head around with how to explain the seemingly-counter-intuitive numeric result. Is there anyone who could shed me light on how to explain this logically? 
Thanks.
 A: If you're pairing $n$ offensive players with $n$ defensive players, you might start with your first offensive player, and choose one of $n$ possible defensive players to pair him with.  Then for the second offensive player you have $n-1$ possible defensive players to choose from, and so on.
The total number of choices is $n (n-1) ... (1) = n!$.
On the other hand, suppose you are pairing the $n$ offensive players with offensive players and the $n$ defensive players with the defensive players (of course, $n$ had better be even).  You have only $n-1$ choices to match with the first offensive player (since you can't pair him with himself).
Then take the first defensive player and match him with one of $n-1$ choices.  That leaves you with $n-2$ offensive and $n-2$ defensive players still unmatched.  Continue in this way, alternating offensive and defensive. Your total number of choices is
$$(n-1)(n-1)(n-3)(n-3) \ldots (1)(1)$$
Note that when matching an offensive player you have one fewer choice than 
you had at the same stage when you matched offensive to defensive players
($n-1$ versus $n$, $n-3$ versus $n-2$, etc), while when matching a defensive player you have the same number of choices.  Thus the second way
gives you fewer total choices than the first way.
A: @RobertIsrael already gave a great answer why $P_{20} > P_0$.  This answer addresses why $P_{10} < 0.5$.
While your intuition might have told you that having $10$ pairs is "typical" or "average" or "most likely" etc, that does not mean it will happen with $0.5$ probability!  The "typical/average/most likely" event does not happen with $0.5$ probability at all.  (BTW I put "typical/average/most likely" in quotes because it is not clear to me that $10$ pairs is actually the expected number or the most probable case.  In fact based on Robert's reasoning I expect the average & most probable case both to skew slightly higher than $10$.)
The following two similar examples might help clarify...?
First, flip a fair coin $20$ times.  $10$ heads is indeed the average, and also the most probable.  But $P(10 \ heads) = {20 \choose 10} / 2^{20} \approx 0.18 \neq 0.5$.
Second, back to the football team.  Imagine it has $20000$ offensive players and $20000$ defensive players, and you do the pairing.  What is $P_{10000}$?  To hit $10000$ pairs exactly has got to be a very, very improbable event, right?  I don't know the value but I would be surprised if it is even $1$%.
