It is dark in my bedroom, and I want to get at least two socks of each color…

Concept: This is a worst-case (adversarial) counting problem. To guarantee at least k items of every category drawn from a mixed pool, first find the maximum number of draws possible while still missing the target count in some category — this is the worst case. Drawing one sock beyond the worst case forces every category to reach the target, since no further avoidance is possible. The worst case itself is found by fully exhausting every category except one, drawing only k-1 items of that one category, and choosing which category to under-draw so as to maximize the total.

Application: Here k = 2 (at least two socks of each color) and the drawer holds 26 red, 24 blue, and 34 brown socks. Check each color as the one left under-drawn:

• Leave blue under-drawn: take all 26 red + all 34 brown + 1 blue = 61.

• Leave red under-drawn: take all 24 blue + all 34 brown + 1 red = 59.

• Leave brown under-drawn: take all 26 red + all 24 blue + 1 brown = 51.

The largest of these three totals is the true worst case: 61 socks (leaving blue under-drawn), since red + brown fully drawn already total 60.

Cross-check: If 62 socks are drawn, only 84 minus 62 = 22 socks remain in the drawer. Having fewer than 2 drawn of some color would require leaving at least (that color's count minus 1) socks behind: at least 23 for blue, 25 for red, or 33 for brown — all greater than the 22 socks left. So it is impossible to draw 62 socks without getting at least 2 of every color, confirming the worst case of 61 is tight.

Minimum required = 61 + 1 = 62.