One of the most fascinating things about Hearthstone is that despite the usual terminology, it is not actually a “trading card game,” in that you cannot trade. Trading is functionally replaced by a crafting system that allows you to inefficiently transform cards into any other cards whenever you like. There are advantages and disadvantages to this from a player’s perspective. The obvious disadvantage is that you can’t shape your collection without destroying value. The advantage lies primarily in not exposing players to the vagaries of a secondary market as a requirement to managing their collections. This should be a big draw for people who have never played a TCG before, as every acquisition of a new card won’t involve the feeling that as a non-expert you might be getting cheated. Related is the topic of this post: since growing your Hearthstone collection is a solo endeavor, we can compute the rate at which it happens without reference to any market conditions or other exogenous factors.

Left out of the post is a discussion of Arena rewards and the efficiency of playing Arena. I want to add this to a follow-up post as soon as I have data on Arena rewards at each tier, particularly since there are a number of good reasons to spend your gold or money on Arena rather than buying packs.

For today, however, the question we explore is how many packs you must expect to buy or otherwise acquire in order to collect any desired set of cards.

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The Monty Hall Problem

A certain probability puzzle is well-known in math circles for its unusual ability to cause people to refuse to believe the answer when it is explained to them. It’s usually known as the “Monty Hall problem” (after the host of Let’s Make a Deal):

Monty Hall has given you a choice of three identical doors. Behind one is a car and behind the other two are goats. You choose a door, but before it’s opened to reveal your prize, Monty adds a twist. He opens one of the other doors to reveal a goat (he always does this to add to the suspense). He then offers you the choice of staying with the door you chose, or switching to the remaining unopened door. Should you stay, or should you switch, and what’s your chance of winning in either case?

The answer is surprising to most: switching doubles your odds of winning the car (2/3 chance of winning, as opposed to 1/3 if you stay). The key fact is that Monty’s knowledge of which of the other doors (if any) was a car causes him to always remove a goat from the prize pool. The chance that the initial door you chose contained a car was 1/3 to start, and it’s unchanged by Monty’s ritual. But if the car was behind either of the other two (2/3 probability in total), Monty will remove the losing door and leave the winning one, and switching will win.

(If you don’t buy that the probability is anything other than 50% when everything started out equal and there are now two doors remaining, there are myriad sources on the internet trying to explain in different ways).

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