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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 $4/52 \times 3/51 \times 2/50 \times 1/49$

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    What is the probability that specifically player $A$ gets a four-of-a-kind? or what is the probability that *at least one player* gets a four-of-a-kind? If its specifically asking about the first player the problem is exactly the same as the usual poker question since the fact that cards are dealt to the second player don't affect the probabilities in any way. – 2017-01-16
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    Per your edit: "*I know that when drawing alone for any chosen face its $4/52\times \dots 1/49$*" That is incorrect. That is probability of drawing all four of **a specific** denomination in **four** draws, not all four of **any denomination** in **five** draws. – 2017-01-16
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    You just answered my questions in a different way, because I was initially wondering if dealing for the other person does affect probability. Soit would be 4/52 * 3/50 * 2/48 * 1/46? – 2017-01-16
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    @JMoravitz yeah I meant **specific** ("chosen") and forgot to mention 4 cards being drawn :) – 2017-01-16
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    "so it would be $4/52\cdot 3/50\cdots 1/46$" ...No, why are you going down by two in your denominator? I just said that the fact that there is a second or third or however many players doesn't in any way affect the probabilities of the first player getting a four-of-a-kind. The number of players doesn't change the calculations at all. It is still going to be $48\cdot 13/\binom{52}{5}$ – 2017-01-16

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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 $\mathsf P(A\cup B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)$.

$$\dfrac{2\binom {13}1\binom 44\binom {48}1}{\binom{52}5}-\dfrac{\binom {13}1\binom{12}1\binom 44\binom 44\binom{44}1\binom{43}{1}}{\binom{52}{10}}$$