3
$\begingroup$

Imagine we have a well-shuffled deck of cards and we keep drawing cards until there is at least one full house in the drawn cards. How many cards will we draw on average? I would be interested in both the exact solution (which is something around 12) but, more importantly, in a good quick way of estimating this value.

$\endgroup$
  • $\begingroup$ If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind $\endgroup$ – user132290 Dec 14 '18 at 23:48
  • $\begingroup$ And the value of 12 quoted above comes from my computer simulations. $\endgroup$ – user132290 Dec 15 '18 at 12:13
  • $\begingroup$ My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$. $\endgroup$ – Jens Dec 15 '18 at 13:13
  • $\begingroup$ @Jens Indeed. The exact expectation I got is $13.893157\dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it).. $\endgroup$ – fedja Dec 15 '18 at 17:39
  • $\begingroup$ How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow? $\endgroup$ – user132290 Dec 17 '18 at 21:51
0
$\begingroup$

A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.

To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:

Case 1 - no doubles/ triples:

$$ 4^n \cdot \binom{13}{n} $$

Case 2 - a triple/quadruple:

$$ 13 \cdot \left(4 \cdot 4^{n-3} \binom{12}{n-3} + 4^{n-4} \binom{12}{n-4}\right) $$

Case 3 - having k doubles:

$$ \binom{13}{k} \binom{13-k}{n-2k} \binom{4}{2}^k 4^{n-2k} $$

Thus, the chance that we get a straight flush drawing n cards is:

$$ \frac{\binom{52}{n}- \left( 4^n \binom{13}{n} + 13 \cdot \left(4 \cdot 4^{n-3} \binom{12}{n-3} + 4^{n-4} \binom{12}{n-4}\right) + \sum_{k=1}^n \binom{13}{k} \binom{13-k}{n-2k} \binom{4}{2}^k 4^{n-2k} \right)}{\binom{52}{n}}$$

Putting this into WolframAlpha, we get these values:

enter image description here

enter image description here

Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.

Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.

Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $\binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).

$$ g(n) = f(n)(52-n)! - g(n-1) $$

Then, simply take the sum:

$$ \sum_{n=1}^{52} \frac{n \cdot g(n)}{52!} $$

Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.