HMMT-Feb 2023 guts, Problem 6

Let $A, E, H, L, T$, and $V$ be chosen independently and at random from the set $\left\{0, \frac{1}{2}, 1\right\}$. Compute the probability that $\lfloor T \cdot H \cdot E\rfloor=L \cdot A \cdot V \cdot A$.

Solution: Solution: There are $3^{3}-2^{3}=19$ ways to choose $L, A$, and $V$ such that $L \cdot A \cdot V \cdot A=0$, since at least one of $\{L, A, V\}$ must be 0 , and $3^{3}-1=26$ ways to choose $T, H$, and $E$ such that $\lfloor T \cdot H \cdot E\rfloor=0$, since at least one of $\{T, H, E\}$ must not be 1 , for a total of $19 \cdot 26=494$ ways. There is only one way to make $\lfloor T \cdot H \cdot E\rfloor=L \cdot A \cdot V \cdot A=1$, namely setting every variable equal to 1 , so there are 495 total ways that work out of a possible $3^{6}=729$, for a probability of $\frac{55}{81}$. $\fbox{\frac{55}{81}}$.

Answer: \frac{55}{81}