Two throws: ordered outcomes and conditioning

The first throw labels columns and the second labels rows. For independent throws all 36 ordered pairs are equally likely. The full sum table is:

Method 1 counts the single cell $(6,6)$: $1/36$. Method 2 multiplies independent probabilities: $(1/6)(1/6)=1/36$.

Method 1 excludes the fifth row and column, leaving 25 cells. Method 2 gives $(5/6)(5/6)=25/36$.

Method 1 counts $(5,6)$ and $(6,5)$: $2/36=1/18$. Method 2 counts one specified order, then multiplies by $2!$: $(1/6)(1/6)\times2=1/18$.

The five pairs are $(1,5),(2,4),(3,3),(4,2),(5,1)$, so the probability is $5/36$.

The pairs with sum six and first throw smaller are $(1,5),(2,4)$. Thus

$\begin{aligned}&P(\text{first}<\text{second}\mid\text{sum}=6)\\ & =\frac{2/36}{5/36}\\ & =\frac25.\end{aligned}$

Restrict to the five equally likely sum-six outcomes. Exactly two satisfy first<second, giving $2/5$. The six reconstructed grid panels retain both conditional methods and the source orientation.