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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.

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.

@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 $1$%.