TITLE: Combinatorial identity on decreasing dice throws
QUESTION [5 upvotes]: Suppose I repeatedly throw fair $n$-sided dice until I throw a $1$, at which point I stop.
I want to know the probability $p(n)$ that my sequence of throws will be decreasing, such as $5-4-2-1$ or just $1$ immediately, but not $2-3-1$ which would count as a failure.
The answer clearly depends on whether I am considering strictly decreasing sequences in which case $4-2-2-1$ would count as a failure, or weakly decreasing sequences in which case it would be a success. Allowing weakly decreasing sequences would increase the probability and would mean there is no bound on the potential length of successful sequences.

For strictly decreasing sequences I have $$p_s(n)=\sum\limits_{k=1}^n {n-1 \choose k-1}\frac{1}{n^k} = \frac{(n+1)^{n-1}}{n^n}$$

For weakly decreasing sequences and $n>1$ I have $$p_w(n)=\sum\limits_{k=1}^\infty {n+k-3 \choose k-1}\frac{1}{n^k} = \frac{n^{n-2}}{(n-1)^{n-1}}$$


For example with $n=6$, I get $p_s(6)=\frac{7^5}{6^6}\approx 0.36023$ and $p_w(6)=\frac{6^4}{5^5}\approx 0.41472$. In general $$p_s(n)=p_w(n+1).$$ Is there a combinatorial argument that leads to this equality between the finite sum and the infinite sum?

REPLY [2 votes]: By making the substitution $m=n-1$, the RHS of $p_w$ becomes the RHS of $p_s$.
Use the binomial identity to change the binomial coefficient of $p_w$ to:
$$\binom{n+k-3}{n-2}$$
The binomial coefficient of $p_w$ after the substitution represents one more die:
$$\binom{n+k-2}{n-1}$$
There is a $1-1$ mapping from the collection of throws of specific numbers in  $p_w$ and the single same specific throw in $p_s$.
$\frac{1}{m^k}=\frac{1}{(n+1)^k}$ represents the adjusted probability, and $p_w$ uses the probability of an infinite run length, given by:
$$\frac{1}{n+1}+\frac{1}{(n+1)^2}+\dots$$
$$=\frac{1}{1-\frac{1}{(n+1)}}-1$$
$$=\frac{n+1}{n}-1$$
$$=\frac{1}{n}$$
Equality of $p_s$ and $p_w$ then comes from equating the two LHS's by $\frac{1}{n^k}$.