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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.

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1 Answer 1

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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 $\frac{20}{36}$.

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  • $\begingroup$ Interesting. I like that complementary counting POV. A had the idea of including (5,5) and (6,6), however I wasn't sure if they should be included due to the fact that neither of them are larger than the other. Thank you! $\endgroup$
    – brandon_ducks
    Jan 28 at 21:39
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    $\begingroup$ 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 $(6,6)$. $\endgroup$
    – Misha Lavrov
    Jan 28 at 21:41

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