Cards of Probabilities

Now suppose that the condition where the probability of the sum of the two cards is divisible by $3$ is not less than $\cfrac{1}{3}$ is ignored. The least possible sum of the 12 numbers $=10+11+12+...+21=\cfrac{31\cdot12}{2}=186$

From $10$ to $21$ ,

$12,15,18,21\equiv0(mod3)$ ,

$10,13,16,19\equiv1(mod3)$ ,

$11,14,17,20\equiv2(mod3)$

The possible combinations of sum divisible by $3$ come from choosing from the groups $n\equiv0(mod3)$ twice or $n\equiv1(mod3)$ and $n\equiv2(mod3)$ once respectively.

$\therefore$ The probability of the sum of the two cards is divisible by $3$ $=\cfrac{{4 \choose 1}\cdot{4 \choose 1}+{4 \choose 2}}{{12 \choose 2}}=\cfrac{4\cdot4+\cfrac{4\cdot3}{2}}{\cfrac{12\cdot11}{2}}=\cfrac{1}{3}$

Hence, the least possible sum of the 12 numbers where the probability of the sum of the two cards is divisible by 3 is not less than $\cfrac{1}{3}$ is $186$ .

One interesting fact worth pointing out is that when there are $3n$ positive numbers for $n \in N$ and there are $n$ numbers respectively such that the remainders when they are divided by $3$ are $0,1,2$ respectively,

the probability of the sum of the two cards is divisible by $3$ $=\cfrac{{n \choose 1}\cdot{n \choose 1}+{n \choose 2}}{{3n \choose 2}}=\cfrac{n^{2}+\cfrac{n(n-1)}{2}}{\cfrac{3n(3n-1)}{2}}=\cfrac{2n^{2}+n(n-1)}{3n(3n-1)}=\cfrac{3n^2-n}{9n^2-3n}=\cfrac{1}{3}$