Let's say I'm playing a game of Mau Mau with my friend.
Each of us draws 5 cards out of the standard, 52-card pile in turns. What is the probability of me getting, for example, 4 Kings?
I know that, when drawing alone, for any chosen face it's $4/52 \times 3/51 \times 2/50 \times 1/49$
The fact that the cards are being dealt to two players in interleaving order does not matter.
You just want the probability that you draw 4 from 4 kings and 1 from 48 other cards, in any order, when selecting 5 cards from a standard deck of 52.
$$\binom{4}{4}\binom{48}{1}\Big/\binom{52}{5}$$
Or by your method, count the probability to select 4 kings then 1 other card when interleaved with the other player selecting 5 other cards. Then multiply the ways you can order 5 kings and 1 other card.
$$\dfrac{~4~\times 48\times ~3~\times 47\times ~2~\times 46\times ~1~\times 45\times 44\times 43}{52\times 51\times 50\times 49\times 48\times 47\times 46\times 45\times 44\times 43}\times\dfrac{5!}{4!\times 1!}$$
Which turns out to be the same thing.
That's for 4 kings. There are thirteen cards in a suit, so if you want the probability that you select 4 of the same value, that would be.$$\dfrac{\binom {13}1\binom 44\binom {48}1}{\binom{52}5}$$
That's for you selecting 4 of the same value. If you want the probability that either player selects 4 cards of the same value, use the rule $\mathsf P(A\cup B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)$.
$$\dfrac{2\binom {13}1\binom 44\binom {48}1}{\binom{52}5}-\dfrac{\binom {13}1\binom{12}1\binom 44\binom 44\binom{44}1\binom{43}{1}}{\binom{52}{10}}$$