# Probabilities of hand draws in Keyforge

Keyforge is a new game by Richard Garfield (who created Magic: The Gathering, King of Tokyo, etc.) that utilizes 36 card decks composed of three houses, 12 cards of each.

I'm trying to figure out the probability of different makeups of first hand draws, but have absolutely no idea where to begin.

How would I calculate the probability of drawing a hand that was 6-0-0, 5-0-1, 4-0-2, 4-1-1, etc.?

First, there are $$\binom{36}{6}$$ total possible hands, each equally probable. We choose $$6$$ of the $$36$$ cards to be in the hand. Then, there are $$\binom{12}{3}$$ ways to choose three cards from the first house, $$\binom{12}{2}$$ ways to choose two cards from the second house, and $$\binom{12}{1}$$ ways to choose one card from the third house. Multiply them together for $$\binom{12}{3}\cdot\binom{12}{2}\cdot\binom{12}{1}$$ total ways. Except - that's not all. To get that, I chose which houses had $$3,2$$, and $$1$$ cards from them in advance. We want all of the 3-2-1 hands, so that's $$3!=6$$ ways to choose which house gets which number of cards. We multiply by $$6$$ to get the total number of 3-2-1 hands.
Then, for the probability, we divide the number of 3-2-1 hands by the total number of hands: $$P_{321}=\frac{\binom{12}{3}\cdot\binom{12}{2}\cdot\binom{12}{1}\cdot 6}{\binom{36}{6}} = \frac{1980}{3689}\approx 0.537$$ Yes, more than half of all hands have 3-2-1 counts.
How does this change for other possible hands? Well, those $$\binom{12}{k}$$ factors will of course depend on how many cards we're putting in each house. More subtly, the order factor at the end ($$6$$ in the case I worked) depends on the symmetry of the hand. If all three houses have different counts, it's $$6$$. If two of the houses are the same, it's $$3$$ - we care which house is different. If all three ouses are the same, it's $$1$$.