Two players pick cards from standard 52 card deck without replacement...

Solution 1:

In a given round of two draws, you start with $n$ cards of which $4$ are kings. The first player wins with probability $\frac 4n$. The second player wins with probability $\frac {n-4}n\cdot \frac 4{n-1}=\frac 4n\cdot\frac {n-4}{n-1}$ because they need the first player not to draw a king and there is one less card in the pack for their draw. As the last factor is less than $1$, on each round the first player has a greater chance to win than the second, so the first player has a greater chance to win overall.

I made a spreadsheet to compute the probability. I find the first player wins about $51.98\%$ of the time.

Solution 2:

Let $E_k$ be the event that the kings are all in the last $k$ cards. Then a win in turn $52-k$ is the event $E_{k+1}\setminus E_k$. Since $E_k\subset E_{k+1}$, we have $\mathsf P(E_{k+1}\setminus E_k)=\mathsf P(E_{k+1})-\mathsf P(E_k)$, so the probability for the first player to win is

$$ \sum_{j=2}^{26}\left(\mathsf P(E_{2j})-\mathsf P(E_{2j-1})\right)=\sum_{k=4}^{52}(-1)^k\mathsf P(E_k)=\sum_{k=4}^{52}(-1)^k\frac{\binom k4}{\binom{52}4}=\frac{433}{833}\approx51.98\%\;. $$

You can tell that this is greater than the second players win probability without computing the sum by noting that each summand $\mathsf P(E_{2j})-\mathsf P(E_{2j-1})$ is greater than the corresponding summand $\mathsf P(E_{2j-1})-\mathsf P(E_{2j-2})$ for the second player.