Riskier Risk, Now with Wilds!

The full deck of Risk cards also contains two wildcards, which can be used as any of the three types of cards (infantry, cavalry, and artillery) upon trading them in. Thus, the full deck consists of $44$ cards.

You must have at least three cards to have any shot at trading them in. Meanwhile, having five cards guarantees that you have three you can trade in.

If you are randomly dealt cards from a complete deck of $44$ one at a time, how many cards would you need, on average, until you can trade in three? (Your answer should be somewhere between three and five. And no, it’s not four.)

Let $S$ be the random number of cards you have when you can first turn in a set of three cards. Obviously we have $3\leq S\leq 5,$ so lets go about calculating $p_s = \mathbb{P} \{ S= s\},$ for $s \in \{3, 4, 5\},$ which we can then compute $$\mathbb{E} [S] = \sum_{s=3}^5 sp_s.$$

From the Classic wildless Risk problem, we see that there remain $3836$ possible combinations of three card sets without using any wilds. To this we must at $\binom{2}{1} \binom{42}{2} = 1722$ possible combinations of three card sets with ome wild card and $\binom{2}{2}\binom{42}{1}=42$ possible combinations of three card sets with two wild cards. In total, there are $5600$ possible sets of three cards, out of a total $\binom{44}{3}=13244$ possible hands of three Risk cards, so we have $$p_3 = \frac{5600}{13244}=\frac{200}{473}\approx 42.283\dots\%$$

Rather than the messiness of figuring out the exact combinations for $S=4$, let's instead tackle $S=5.$ We see that if you do not get a set on your fourth card then you will automatically get a set by your fifth card, so we only need to figure out what situations lead you to not have a set after four cards. This can only happen whenever you have two of one type and two of a different type (with no wilds, obviously). There are $\binom{14}{2}\binom{14}{2}\binom{3}{2}=24843$ possible hands that have exactly two cards in two different types. Since there are $\binom{44}{4} = 135751$ possible combinations of four distinct Risk cards, we thus have $$p_5=\frac{24843}{135751}=\frac{3549}{19393}\approx 18.300\dots\%$$

Since we can reason that $p_4=1-p_3-p_5 = \frac{7644}{19393}\approx 39.416\dots\%,$ we have the expected number of Risk cards needed to turn in a set is $$\mathbb{E}[S]= 3 \frac{200}{473} + 4 \frac{7644}{19393} + 5 \frac{3549}{19394} = \frac{72921}{19393} \approx 3.760171\dots$$