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My proposed solution follows. Is this correct? If yes, I'd like to know if there is a better way to solve this problem.

We can see that the value of the first throw could only be $1, 2, 3$ or $4$.
Throws are labeled as $T_1, T_2, T_3$.

If $T_1$ is $1$ then $T_2$, $T_3 \in \{2,3,4,5,6\}$
If $T_1$ is $2$ then $T_2$, $T_3 \in \{3,4,5,6\}$
If $T_1$ is $3$ then $T_2$, $T_3 \in \{4,5,6\}$
If $T_1$ is $4$ then $T_2$, $T_3 \in \{5,6\}$

Hence the number of favorable outcomes is $\dbinom{2}{2}+\dbinom{3}{2}+\dbinom{4}{2}+\dbinom{5}{2} = 20$

And $P(A) = \frac{20}{6\times6\times6}=\frac{5}{54}$

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  • $\begingroup$ As an aside, $\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2}=\binom{6}{3}$ can be explained by the hockey stick identity $\endgroup$
    – JMoravitz
    Apr 29, 2021 at 12:55

3 Answers 3

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Your analysis is spot on.

An alternative approach is that the answer has to be $\displaystyle \frac{\binom{6}{3}}{6^3}$ because there is a clear bijection between a group of $3$ satisfying rolls, and choosing 3 distinct numbers, without replacement, from $\{1,2,3,4,5,6\}.$

The reason that the bijection exists is:

  • Only rolls where all 3 numbers are distinct are eligible for consideration.

  • For any group of 3 distinct numbers, there is only 1 way of ordering the numbers satisfactorily.

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That method would be hard to scale up to larger numbers.

As an alternative, for an $N$ sided die: First note that the probability that the rolls are distinct is $$1\times \frac {N-1}N\times \frac {N-2}N$$

Given that the rolls are distinct, there are $6$ possible, equally probable, orders for them, only one of which is increasing. Thus the answer to your question is $$\frac 16\times 1\times \frac {N-1}N\times \frac {N-2}N=\boxed {\frac {(N-1)(N-2)}{6N^2}}$$

Note that if $N=6$ this resolves to $$\frac {5\times 4}{6\times 36}=\frac {20}{6^3}$$ as desired.

Should say: it's a numerical coincidence that, in your case, the number of ways to order $3$ distinct items happens to be the same as the number of faces on a standard die.

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You can note that whichever 3 different numbers you choose as the result of your dice throws, there is only one way to order those numbers: smallest to largest. Therefore, the number of options for 3 increasing dice throws is equivalent to ${6 \choose 3} = 20$. From there the same calculations apply as yours.

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