smallest number of socks to guarantee that the selection contains at least $10$ pairs A drawer in a darkened room contains $100$ red socks, $80$ green socks, $60$ blue socks and $40$ black socks. 
A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn.
What is the smallest number of socks that must be selected to guarantee that the selection contains at lest $10$ pairs?
(A pairs of socks is two socks of the same color. No sock may be counted in more than one pair.)
What is the smallest number of socks that must be selected to guarantee that the selection contains at least $10$ pairs?
My attempt:
$100 = a $
$ 60 = b$
$ 40 = c$
$ 80 = d$
$a + b +c +d =280 $ socks
I know the probability of choosing each color on the first try are :
$p(a) = 0,3571;\,\,  p(b) = 0,2142;\,\, p(c) = 0,1428;\,\, p(d) = 0,2857$
How can I find  the smallest number of socks that must be selected to guarantee that the selection contains at lest $10$ pairs?  
 A: Let the different types of socks be represented by $\text{A},\text{B},\text{C}$ and $\text{D}$
Suppose that you wish to draw one pair of socks from the drawer. Then you would pick $5$ socks (one of each kind, plus one (Let $\text{A}$) to guarantee atleast one pair). 
Notice that in the worst possible situation, you will continue to draw the same sock $A$
$\big($Because as you draw any other sock, let $\text{B}$, it will combine from previously selected $\text{B}$ sock and will result in a pair just by adding one sock, while in the case if selecting sock $\text{A}$ you'll have to select a total $2$ sock after those $5$ to make a pair$\big)$ , until you get $10$ pairs. This is because drawing the same sock results in a pair every $2$ of that sock, whereas drawing another sock creates another pair. Thus the answer is 

$$\underbrace{5}_{\text{Previously Selected}}+\underbrace{2}_{\text{2 socks make a pair}}\times \Big(\underbrace{10}_{\text{Total number of pairs required}}-\underbrace{1}_{\text{Already selected 1}}\Big) = \boxed{23}$$

A: My first guess would be $23$. 
In a rather unfortunate scenario, you could pick $19$ of one color socks. Then, you would get $1$ more of each sock, leading to a total of $22$. Right now, we have $9$ pairs, and one more of each sock. Regardless of the next sock, we will have $10$ pairs. Therefore $22+1 = 23$.
A: It takes $20$ socks to make the $10$ pairs.  The last sock you drew must have made a pair because you stopped drawing, so you don't have any of that color left.  You can have at most one sock of each of the three other colors left, so the maximum number you need to draw is $23$.  John Lou has presented a draw that requires $23$, so we are done.
A: Suppose you put $n$ socks into $4$ color boxes such that there are a total of exactly $k$ pairs of socks in the $4$ color boxes.  Then, $n$ must be at least $2k$ because that many socks are required for the $k$ pairs.  The maximum possible value of $n$ is $2k+4$ because each of the $4$ color boxes can have an odd number of socks and so there can be one unused socks in each color box.  Thus, $2k \le n \le 2k+4$.  If $k=9$, then $18 \le n \le 22$.  Thus, the maximum number of socks we can have in the $4$ color boxes and still not have guaranteed $10$ pairs is $22$.   So the answer to your question is $23$.
