Probability of Increasing Sequence of Die Rolls Remaining Increasing

Question

Context

I was talking with a friend about this simple problem and we couldn't figure it out easily. I spent a little while with paper and pen and googled a bit and checked previous questions but have not been successful finding a solution. I think I am missing some language/terms.

Problem

Let $D_i$ be the $i$th outcome of a 20-sided die event. That is, $D_i\in[1,20]$. What is the probability that $D_1<D_2<D_3$ for three sequential die events?

Scratch Work

My pen & paper work thus far (I feel like I am abusing notation):

$P(D_1)=1$. $P(D_2>D_1)=\frac{20-D_1}{20}$. $P(D_3>D_2)=\frac{20-D_2}{20}$.

$P(D_1<D_2<D_3)=P(D_1)P(D_2>D_1)P(D_3>D_2)=(1)(\frac{20-D_1}{20})(\frac{20-D_2}{20})=1-\frac{D_1+D_2}{20}+\frac{D_1D_2}{400}$.

$D_1+D_2\in[2,40]$; $D_1D_2\stackrel{?}{\in}[1,400]$.

Closing

I know that I am doing something gravely wrong and I want to know what it is. All help appreciated, especially generalizations.

Answer 1

Hint: Partition over the middle value: $$\begin{align}\mathsf P(D_1<D_2<D_3)&=\sum_{t=2}^{19}\mathsf P(D_1<t)\mathsf P(D_2=t)\mathsf P(D_3>t)\\[1ex]&=\sum_{t=2}^{19}\dfrac{(t-1)(20-t)}{20^3}\end{align}$$

Answer 2

Note that, if all three rolls are distinct, there is a $\frac{1}{3!}$ chance they come in increasing order (if they are not distinct, there is no chance). The probability of at least two of them matching is $3 \times \dfrac{1}{20} \times \dfrac{19}{20} + 20 \times \dfrac{1}{20^3}$ (splitting into cases where exactly 2 match and all 3 match). The desired probability is then $\dfrac{1}{6} \times \left( 1 - \left( 3 \times \dfrac{1}{20} \times \dfrac{19}{20} + 20 \times \dfrac{1}{20^3} \right) \right) = \boxed{\dfrac{57}{400}}$.

Answer 3

$$ P(\mbox{Increasing}) = P(\mbox{Increasing}|\mbox{All distinct})P(\mbox{All distinct})$$

$$P(\mbox{All distinct})=\frac{20*19*18}{20*20*20}= \frac{19*9}{200}.$$

As argued above, $$ P(\mbox{Increasing}|\mbox{All distinct}) = \frac{1}{3!}.$$

Therefore, the answer is $$ \frac{19*9}{200} \times \frac{1}{6} = \frac{57}{400}.$$