A die is rolled 3 times, what is the probability that the third launch has an outcome that is strictly included between the other two?
An explanation would also be very appreciated as I’m not a mathematician.
Thank you!
2026-03-29 02:36:16.1774751776
One die is rolled three times - probability outcome of third roll is between first two?
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Here is a way to look at this without exhaustively making a list of all the outcomes. The only way the above is possible is when the three numbers are distinct (otherwise strict inequality won't be satisfied)
$$P(\text{distinct numbers}) = \frac{6\cdot 5\cdot 4}{6\cdot 6\cdot 6}$$
Now, in an arrangement of distinct numbers, there are two ways out of 6 total permutations that the condition can be satisfied
$$P = \frac{2}{6}\cdot P(\text{distinct numbers}) = \frac{5}{27}$$