In a four hand draw without replacement, what is the probability that no subsequent ranks are the same?

Question

As an example, $J,K,3,K$, or $K,3,4,K$, or $K,J,A,2$ are allowed, but not $K,K,J,3$ or $4,10,K,K$.

I was thinking that maybe I could approach this question by counting the number of ways such that a rank does occur twice in a row, divide this by $\frac{52!}{48!}$ and subtract the result by 1 but I am unsure.

Any help would be very much appreciated.

Thank you in advance.

Answer 1

Well, the first card we draw does not depend on any previous draw, so we just assign it a $\frac{52}{52}$ probability. The second draw cannot be the same rank as the previous, and we still have $51-3=48$ cards left in the deck that do not have the previous rank, so we assign our second draw a $\frac{48}{51}$ probability.

Now the difficult part:

Suppose our first three draws were $K,Q,K$ (for example). The probability of getting the same card on our third draw as we did on our first draw is $\frac{3}{50}$. Now, for the fourth draw, there are only $49-2=47$ cards left in the deck that are not Kings, so we assign our fourth draw to be $\frac{47}{49}$.

Now suppose our first three draws were $K,Q,J$ (for example). The probability of not getting the same card on our third draw as we did on our first is $\frac{44}{50}$ (since there are still $50-3-3=44$ cards left in the deck that are not Queens or Kings). For the final draw, there are $49-3=46$ cards still left in our deck that are not Jacks, so we assign this probability to be $\frac{46}{49}$.

If we combine these two cases in general, then our ultimate probaility becomes:

$\frac{52}{52}\cdot \frac{48}{51}\cdot \frac{3}{50}\cdot\frac{47}{49}+\frac{52}{52}\cdot \frac{48}{51}\cdot \frac{44}{50}\cdot\frac{46}{49}=\frac{3464}{4165}\approx83.17\%$

Answer 2

I would do a direct calculation conditioning on whether the third card matches the first. What is the chance the second card does not match the first? What is the chance the third card matches neither? In that case, what is the chance the fourth does not match the third? What is the chance the third card matches the first (and therefore does not match the second)? In that case, what is the chance the fourth does not match the third?