# A die was thrown three times. Find probability that first throw is less than the second and second is less than the third throw.

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

• 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 Apr 29, 2021 at 12:55

## 3 Answers

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.

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.

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.