If you have both Sauna and Avanto in your hand, and you know you have at least one other copy of each in your deck, which should you play first? Does it matter?

I don't know the answer to this, but it seems like something that can be solved with math.

Depends what you are trying to do. If you are just trying to maximize the chance of chaining all four cards together, then you should play Avanto first. Why? Let's assume your deck size is x >= 7.

Suppose you play Sauna, then Avanto.

The chance of finding your other Sauna is 4/x. Given you find it, the chance of having drawn the second Avanto already or of drawing it from the second Sauna is 4/(x-1). (Why? You have drawn 5 cards so far, one of which is Sauna (and thus not Avanto).) So the chance of chaining all four cards is 16/(x

^{2}-x).

Now suppose you play Avanto, then Sauna.

The chance of finding your other Avanto is 4/x. Given you find it, the chance of having drawn the second Sauna already or of drawing it from the second Avanto is 6/(x-1). (Why? You have drawn 7 cards so far, one of which is Avanto (and thus not Sauna).) So the chance of chaining all four cards is 24/(x

^{2}-x).

Hmm I just noticed you said

*at least* one other copy. That complicates things enough that the math gets hard.