Transform The Minimum Score

All possible outcomes of the two first rolls can be arranged in a $4 \times 4$ table

$\begin{array}{ccccccc} & 1 & 2 & 3 & 4 & X_1 \\ 1 & (1,1) & (1,2) & (1,3) & (1,4) \\ 2 & (2,1) & {\color{#D61F06}(2,2)} & {\color{#D61F06}(2,3)} & {\color{#D61F06}(2,4)} \\ 3 & (3,1) & {\color{#D61F06}(3,2)} & {\color{#3D99F6}(3,3)} & {\color{#3D99F6}(3,4)} \\ 4 & (4,1) & {\color{#D61F06}(4,2)} & {\color{#3D99F6}(4,3)} & {\color{#20A900}(4,4)} \\ X_2 \end{array}$

For all black results, $U_1 = 1$ , for red results $U_1 = 2$ and so on for $U_1 = 3,4$ .

Hence, we derive the values

$P(U_1 = 1) = \frac {7}{16}$

$P(U_1 = 2) = \frac {5}{16}$

$P(U_1 = 3) = \frac {3}{16}$

$P(U_1 = 4) = \frac {1}{16}$

With this, we can now work out the product

$\begin{aligned} & \displaystyle \left\lceil \left( \prod_k P(U_1 = k) \right) ^{-1} \right\rceil \\ =& \left\lceil \left( \frac {7}{16} \cdot \frac {5}{16} \cdot \frac {3}{16} \cdot \frac {1}{16} \right) ^{-1} \right\rceil \\ =& \left\lceil \frac {65536}{105} \right\rceil \\ =& \boxed{625} \end{aligned}$