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.

  • $\begingroup$ 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

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:

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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 :)


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