The question
You want every card in a set. Each card you open is random, and duplicates are common. How many cards must you open, on average, before you have one of each?
This is the coupon collector problem, a classic in probability. It was named for promotions in which each box contained one of several coupons and the customer wanted a complete collection. Card sets, sticker albums and gacha collections follow the same pattern, with some differences covered below.
The simplest model
Assume a set of n different cards, and each card you open is equally likely to be any of them, independently of the others. That is a strong set of assumptions, and the later sections discuss when they fail. Under them, the calculation is neat.
Break the collection into stages. In stage one you have no cards, and the first one is new for certain. In stage two you hold one distinct card, so the chance that the next card is new is (n minus 1) divided by n. In general, when you hold k distinct cards, the chance that a draw is new is (n minus k) divided by n.
How long does each stage take? The wait for a success with probability p averages 1 divided by p draws, which is the logic of the geometric distribution. So the stage with k cards in hand takes n divided by (n minus k) draws on average. Adding up all the stages gives:
expected draws = n times (1 + 1/2 + 1/3 + … + 1/n)
The bracket is the nth harmonic number. For large n it is close to the natural logarithm of n plus about 0.58, so the expected number of draws grows slightly faster than n itself.
Worked numbers
- A set of 10 cards. Stages cost 1, then 10/9, 10/8, and so on down to 10/1. The total is about 29.3 draws, a little under three times the size of the set.
- A set of 50 cards. About 225 draws, about 4.5 times the set size.
- A set of 100 cards. About 519 draws, a bit over five times the set size.
- A set of 200 cards. About 1,176 draws, a bit under six times the set size.
The ratio of expected draws to set size increases slowly as the set grows, which is the effect of the logarithm.
The last cards cost the most
The stages are not equal. At the start, almost every card is new, so progress is quick. At the end, most draws are duplicates.
In a 100-card set, when you are missing only one card, each draw has a 1 in 100 chance of being the one you need, so that final stage averages 100 draws on its own. That is roughly a fifth of the entire expected total of about 519. Missing the last five cards, the final five stages together average about 100 times (1 + 1/2 + 1/3 + 1/4 + 1/5), roughly 228 draws, which is a little under half the total.
This is why completing a collection feels like it gets harder and harder: it does. The first half of the set is cheap and the last few cards account for a large share of the cost.
The average is not a guarantee
The expected number of draws is a mean, and the spread is wide. For large sets, the standard deviation is roughly 1.3 times n. For a 100-card set, that is on the order of a hundred and some draws either side of the average of about 519, so completing the set in a few hundred fewer or more draws would not be unusual.
A rough way to think about the upper tail is that the chance of still being incomplete after n times (ln n plus c) draws is around e to the power of minus c. For n = 100 and c = 3, the draws are about 100 times (4.6 + 3), around 760, and the chance of still missing something is roughly 5 percent. This is an approximation for large sets, but it conveys the point: finishing by the average is closer to a coin flip than a certainty, and bad luck can cost a great deal more.
That also makes this a good case for the difference between mean and median outcome: the distribution has a long right tail.
Packs, not single cards
Packs contain several cards, so you might divide the draw count by the pack size. As a rough guide this helps, but it is an approximation with at least two problems.
First, a pack may not contain duplicates, or may have slots set aside for different tiers. If a pack cannot repeat a card, you do slightly better than the single-draw formula suggests. Second, and more importantly, pack structure often breaks the equal-probability assumption.
Why real sets usually take longer
Equal odds for every card is rarely true. Most sets mix common and rare cards, and a rare card may appear far less often than a common one. The total is dominated by the rarest cards, because the wait for a card with probability p averages 1 divided by p. A single card that is ten times rarer than the rest takes roughly ten times as long to find on its own.
Other factors include:
- Alternate versions or foil copies that count as separate cards, which enlarge n.
- Rates that are not published, so the model has to be guessed.
- Guarantees such as a pity timer, which shorten the tail and reduce the spread.
- Trading, which changes the problem entirely. Trading with others can reduce the cost, since duplicates become currency. The stand-alone model describes opening alone.
What to take from it
The formula is useful for sanity-checking claims and for setting expectations. If a set has 200 cards with equal odds, expecting to finish in 200 or even 400 packs of one card each is unrealistic. The mathematics says over 1,100 draws, on average.
Because the final cards are so costly, many collectors buy missing cards directly, which converts an open-ended gamble into a known price. Whether that makes sense depends on the market.
Finally, a word of caution about spending. The model shows costs that are uncertain and potentially large, and opening more packs does not make any particular pack more likely to contain what you want. Decide on a budget in advance, and treat money already spent as irrelevant to what to do next; continuing because you have already invested a lot is a form of sunk cost thinking.