I was working on creating a card game with a friend, and we wanted to understand how card draws would play out.
We have a set of 18 distinct cards. Each card participates four times in some pre-defined combination of three cards (order doesn't matter) for a total of 24 combinations. How many cards do we need to draw in our opening hand, to ensure that one of the combinations exists in our opening hand?
We thought this had to do with combinatorics, maybe related to this question, but we get stuck thinking about it, because each of our cards is distinct, so we don't actually have 24 identical cards to choose from, but rather 24 different sets of 3 cards out of 18 unique cards.
Best Answer
Presumably any possible two 3-card combinations of cards share at most $1$ card. So if you take an initial card, all the combinations involving that card would be another $4\cdot 2 = 8$ cards. So adding the initial card, that's half the pack - is there an "anti-card" to our initial choice for which all of its combinations complete the pack? In which case you could certainly hold $1+4+1+4=10$ cards without a combination & need at least $11$ cards to have a combination.
Coming from the other side, each card can only break $4$ combinations, so you'd need to omit $6$ cards to have any chance of breaking all combinations. That would imply that $13$ cards would always be enough, depending on how the combinations are organized. The requirement for $13$ cards can be firm if a lot of combinations have some common cards, for example with combinations:
$\{abm, abn, abo, abp, $ $cdq, cdr, cdm, cdn, $ $efo, efp, efq, efr, $ $ghm, ghn, gho, ghp, $ $ijq, ijr, ijm, ijn, $ $klo, klp, klq, klr \}$
you could have zero combinations unless at least one of $m,n,o,p,q,r$ are held.