# Rolling a pair of six-sided dice, what is the probability that the larger number is at least 5?

You have two six-sided dice. What is the probability that the larger number rolled will be at LEAST 5?

I have listed out all the possible rolls. You can set dice 2 equal to 5, and roll dice one. The possible rolls are (1,5), (2,5), (3,5), (4,5). Do the same thing, but this time dice 2 is set to 6. We have (1,6), (2,6), (3,6), (4,6), and (5,6). These 9 possible rolls can happen with setting the 1st dice to both 5 and 6 as well. Giving us a total of 18 possible rolls $$\therefore P=18/36$$

However, the answer I'm given states that $$P=20/36$$. I'm having trouble understanding why.

You have covered the $$18$$ cases in which the two numbers are different and the larger is at least $$5$$. The $$2$$ other cases counted by the answer you're given are $$(5,5)$$ and $$(6,6)$$.
Another way to solve the problem is by complementary counting. The negation of "the larger number is at least $$5$$" is "both numbers are at most $$4$$". This leaves $$4$$ possibilities for each number, for a probability of $$\frac{4^2}{6^2} = \frac{16}{36}$$. Now subtract this from $$1$$ to get the answer of $$\frac{20}{36}$$.
• I agree that talking about "the larger number" is a tiny bit ambiguous in this case, so I don't blame you about being unsure. However, the more likely interpretation of "The larger of $x$ and $y$ is at least $5$" is $\max\{x,y\} \ge 5$, which includes $(5,5)$ and $(6,6)$. Jan 28 at 21:41