We can use complementary probability. If we consider a single die, a throw of 5 or higher is required for the product of the two dice to be at least 20, and a throw of 8 is required for the product to exceed 40. Thus, to make the numbers slightly easier, we consider the probability that the product of the two numbers is less than 40. We tabulate the conditions for product less than 40 with different styles denoting simultaneous conditions when necessary:
\[\]
\[
\begin{array}{c||c|c|c|c}
\text{First Die} & \multicolumn{4}{c}{\text{Second Die}} \\
\hline
1,2 & \text{All} & - & - & - \\
\hline
3 & 1\text{-}7,\textbf{\textit{8}} & 1\text{-}3,\textbf{\textit{4}} & 1\text{-}2,\textit{\text{5}} & - \\
\hline
4 & 1\text{-}6,\textbf{\textit{7}} & 1\text{-}4,\textbf{\textit{5}} & 1\text{-}2,\textit{\text{6}} & - \\
\hline
5 & 1\text{-}5,\textbf{\textit{6}} & 1\text{-}3,\textbf{\textit{7}} & \text{All on }\textit{8} & - \\
\hline
6,\textbf{\textit{8}} & \text{All} & \text{All} & \text{All} & - \\
\end{array}
\]
\[\]
There are $\text{7, 4, 5, 6, 5, 7}$ multiplicities for $\text{First Die throws 1-6}$, so the probability is $\frac{7+4+5+6+5+7}{48} = \frac{34}{48}$ meaning an odds of $\boxed{\frac{7}{17}}$.
An octahedral die has eight faces numbered 1 to 8, each of which comes up with equal likelihood (you can see a picture of one in Diagram 3 of the Test 1 Supplement).
Two such dice are rolled, and the product of the numbers is noted. In lowest terms, what are the odds against the product being greater than or equal to 40? Justify your answer.
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