A die is rolled 3 times, what is the probability that the third launch has an outcome that is between or equal to the other two?
Update: If the third launch had to be strictly between the first two launches, the probability would be the probability of having three distinct numbers multiplied by 1/3. Therefore this probability is equal to (1 * 5/6 * 4/6) * 1/3 = 5/27.
Now if I add the probability of the first and the third launch being equal and the probability of the second and the third being equal, and I subtract the probability of the event in which the three numbers are the same, this would be the probability I'm looking for (5/27 + 1/6 + 1/6 -1/36)?
Thank you!
Best Answer
Your answer is correct. We can verify it using a combinatorial approach. While the combinatorial approach involves more work than your approach using the Inclusion-Exclusion Principle, it does allow us to confirm your answer.
Since there are six possible outcomes for each of the three rolls, there are $6^3 = 216$ possible outcomes.
There are three favorable cases to consider:
Case 1. Three distinct numbers, with the third roll producing the middle value.
There are $\binom{6}{3}$ ways to select three different numbers. Since the third roll must be the middle number, there are $2!$ ways to arrange the smallest and largest numbers among the first and second rolls. Hence, there are $$\binom{6}{3}2!$$ favorable outcomes in this case.
Case 2: The first two rolls are distinct numbers, and the third roll matches one of them.
There are six possible outcomes for the first roll. Since the outcome of the second roll must be different from that of the first roll, there are five possible outcomes for the second roll. The third roll must match one of the first two rolls. Hence, there are $$6 \cdot 5 \cdot 2$$ favorable outcomes in this case.
Case 3: All three rolls produce the same number.
There are six ways to select the number that appears on all three dice.
Total: Since the three cases are mutually exclusive and exhaustive, the number of favorable cases is $$\binom{6}{3}2! + 6 \cdot 5 \cdot 2 + 6$$
Hence, the probability that the third roll has an outcome that is between or equal to the other two is $$\frac{\binom{6}{3}2! + 6 \cdot 5 \cdot 2 + 6}{6^3} = \frac{40 + 60 + 6}{216} = \frac{106}{216} = \frac{53}{108}$$ As you can verify, this matches your answer.