How to calculate probability distribution for combination of functions

Question

please excuse (or change, if possible) the title if it doesn't make sense.

I have a problem which is something like this:

You have a variable $i$ which starts at $i=1$. You also have a target number of $j$, and you have an n-sided die (in my case $j=6$ and $n=6$ but ideally i'd like to be able to solve this generally.)

Now you repeatedly roll the n-sided die, each time the die the lands on a number greater than $i$ you increment $i$ by 1. This stops once $i=j$.

I've already calculated the mean number of rolls until $i=j$ as:

$$\sum_{x=n-(j-1)}^{n-1} \frac{n}{x}$$

Now i'd like to be able to produce a probability distribution of how many rolls until $i=j$. I'm currently only looking at $j=6$ and $n=6$ and have come up with a solution, however it is very manual.

I can model the probability of raising $i$ from 1 to 2 (assuming 6-sided die) in $x$ rolls as:

$$p_1(x)=\frac 16^{x-1}\times \frac 56 $$

Similarly the probability of raising $i$ from 2 to 3 in $x$ rolls

$$p_2(x)=\frac{2}{6}^{x-1}\times \frac 46 $$

(I'll skip the definition of $p_3(x)$ to $p_5(x)$)

Now for the probability of raising $i$ from 1 to 6 in just 5 rolls would be

$$p_1(1) \times p_2(1) \times ... \times p_5(1)$$

But what I need to do next is the probability of raising $i$ from 1 to 6 in 6 rolls, I realise I need to do something like:

$$p_1(2) \times p_2(1) \times ... \times p_5(1) + p_1(1) \times p_2(2) \times ... \times p_5(1) + ... \times p_4(1) \times p_5(2)$$

(Not sure if I removed too much from the above equation but essentially the sum of all the different products of $p_{1-5}$ where the arguments to the functions add up to 6, does that make sense?)

And then the above but for all arguments summing to 7 and on to Infinity.

So I'm wondering how to generalise this last step? It seems to be that it could/should be able to be represented as a function. Maybe there is something similar that already exists?

Thank you in advance. Let me know if i need to edit the question to clarify anything :)

Answer 1

Clearly you need $j \le n$

As Tony Hellmuth says in the comments, this is closely related to the Coupon Collector's problem. You can make the relationship even closer if you start with $i=0$, which has the effect of adding $1$ to the number of throws. Doing that, I believe you can say that the probability that the first time $i=j$ is the $r$th throw is $$ S_2(r-1,j-1) \frac{ n!}{(n-j)! n^r}$$ where $S_2(a,b)$ is a Stirling number of the second kind

For $n=j=6$ it gives probabilities like this table for varying $r$ starting at $i=0$ (if starting from $i=1$ then subtract $1$ from each $r$); in the tail $\mathbb P(R=r+1) \approx \frac56 \mathbb P(R=r)$. The expected value is $14.7$, median is $13$ and mode is $11$.

r       Prob(R=r)

1       0
2       0
3       0
4       0
5       0
6       0.0154321
7       0.0385802
8       0.0600137
9       0.0750171
10      0.0827689
11      0.0843943
12      0.0816093
13      0.0760425
14      0.0689872
15      0.0613674
16      0.0537917
17      0.0466281
18      0.0400747
19      0.0342160
20      0.0290645
21      0.0245899
22      0.0207390
23      0.0174480
24      0.0146506
25      0.0122827
26      0.0102849
27      0.0086036
28      0.0071917
29      0.0060077
30      0.0050162
31      0.0041867
32      0.0034932
33      0.0029139
34      0.0024302
35      0.0020264
36      0.0016896
37      0.0014085
38      0.0011742
39      0.0009787
40      0.0008158
41      0.0006799
42      0.0005667
43      0.0004723
44      0.0003936
45      0.0003280
46      0.0002734
47      0.0002278
48      0.0001899
49      0.0001582
50      0.0001319