B. A fair cube die has the numbers $1,2,3,4,5,6$ on its six faces. The die is rolled four times, and the numbers obtained are $a, b, c, d$ respectively. The probability that one of the numbers $a$, $a+b$, $a+b+c$, $a+b+c+d$ equals 4 is ( ).
(A) $\frac{33}{1296}$
(B) $\frac{334}{1296}$
(C) $\frac{343}{1296}$
(D) $\frac{433}{1296}$
B. C.

If $a=4$, the probability is $\frac{1}{6}$;
From $4=1+3=2+2=3+1$, we know that the probability of $a+b=4$ is $\frac{3}{6^{2}}$;

From $4=1+1+2=1+2+1=2+1+1$, we know that the probability of $a+b+c=4$ is $\frac{3}{6^{3}}$;

If $a+b+c+d=4$, then it can only be $a=b=c=d=1$, with a probability of $\frac{1}{6^{4}}$.

Notice that, in the above four scenarios, any two cannot be true simultaneously.

Therefore, the probability of $a=4$ or $a+b=4$ or $a+b+c=4$ or $a+b+c+d=4$ is
$$
\frac{1}{6}+\frac{3}{6^{2}}+\frac{3}{6^{3}}+\frac{1}{6^{4}}=\frac{1}{6}\left(1+\frac{1}{6}\right)^{3}=\frac{343}{1296} .
$$