Suppose you have a drawer with cards on which a number $1$ through $18$ is written. You can pick cards from the drawer with your eyes closed. What is the minimum number of cards you have to draw to guarantee that two of the drawn cards sum up to $9$ and two of the other drawn cards sum up to $27$.
I have found the number of ways to sum $9$ and $27$ but I am stuck on how to prove a minimum number of cards being drawn. Even if I find the probability of the sum of $27$ being drawn by two cards, and the probability of that of $9$, given conditional probability of $P(27|9)$, I would still not be able to find the minimum. Hence, is this using the pigeonhole principle? If so, how?
$9: (4,5) , (3,6), (2,7) , (1,8)$
$27: (13,14), (12,15), (11,16),(10,17), (9,18)$
What is the minimum number of cards you have to draw to guarantee that two of the drawn cards sum up to 9 and two of the other drawn cards sum up to 27 ?
The answer is $15$ cards.
Indeed, if you have drawn only $14$ cards, then the $4$ remaining ones might be : $1,2,3,4$ (or $8,2,6,5$ : there are 16 possibilities). Then no pair of cards sums to $9$.
If you have drawn at least $15$ cards, then for at least one of the four pairs summing to $9$, both cards are included. Similarly, at least two of the five possible pairs summing to $27$ is complete.
Yes, this can be presented with the pigeonhole principle: if the holes are the cards remaining in the drawer, and the pigeons are the pairs summing to $9$, then four pairs could be present in three holes only with two pairs in the same hole: since the pairs are mutually disjoint, this is impossible.