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Round trip stones

$N$ vessels initially contain $3$ stones each. What is the probability of having at least $3$ stones in the first vessel after moving a uniformly random selection from the first vessel to the second, then from the second vessel to the third, and so on back to the first vessel?

## Input
- A positive integer $N$.
- Your code must work for inputs up to and including $10$, but may crash, error, or give incorrect output for larger inputs.

## Output
- A probability $p$, so $0 \le p \le 1$.
- Your output is valid if rounding it to $6$ decimal places results in the output shown in the relevant test case.
- This is the probability of the first vessel containing greater than or equal to $3$ stones after the following process:
  - Start with $N$ vessels containing $3$ stones each.
  - For each vessel from $1$ to $N$ in order:
    - Choose uniformly randomly between $0$ and all of the stones from the vessel, and move them to the next vessel.
    - Note that when moving from vessel $N$, they are moved back to vessel $1$.

## Examples

### Input 1
When there is only $1$ vessel, all of the stones moved from it are moved back into it, resulting in exactly $3$ stones in vessel $1$, so the probability of there being at least $3$ stones in vessel $1$ is $1$.

### Input 2
When there are $2$ vessels, the process goes like this:
- $0$, $1$, $2$, or $3$ stones are taken from vessel $1$ and added to vessel $2$. For example, $2$ stones may be taken from vessel $1$ and added to vessel $2$.
- There are now $5$ stones in vessel $2$, so $0$, $1$, $2$, $3$, $4$, or $5$ stones are taken from vessel $2$ and moved back to vessel $1$.

There are $22$ ways that this process can happen, and $16$ of those ways result in $3$ or more stones in vessel $1$. This means the probability of there being at least $3$ stones in vessel $1$ is $16/22=0.727273$ (rounded to $6$ decimal places).

### Input 3
When there are $3$ vessels, the process goes like this:
- $0$, $1$, $2$, or $3$ stones are taken from vessel $1$ and added to vessel $2$. For example, $2$ stones may be taken from vessel $1$ and added to vessel $2$.
- There are now $5$ stones in vessel $2$, so $0$, $1$, $2$, $3$, $4$, or $5$ stones are taken from vessel $2$ and added to vessel $3$. For example, $5$ stones may be taken from vessel $2$ and added to vessel $3$.
- There are now $8$ stones in vessel $3$, so $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, or $8$ stones are taken from vessel $3$ and moved back to vessel $1$.

There are $140$ ways that this process can happen, and $102$ of those ways result in $3$ or more stones in vessel $1$. The probability of there being at least $3$ stones in vessel $1$ is $102/140=0.728571$ (rounded to $6$ decimal places).

## Test cases
Test cases are in the format `input : output`, and are rounded to 6 decimal places. For example, an output of `0.727272727` is equivalent to an output of `0.727273` when rounded.

```text
1 : 1.000000
2 : 0.727273
3 : 0.728571
4 : 0.736842
5 : 0.743789
6 : 0.749164
7 : 0.753344
8 : 0.756657
9 : 0.759336
10 : 0.761542
```

## Scoring
This is a [code golf challenge]. Your score is the number of bytes in your code. Lowest score for each language wins.

> Explanations are optional, but I'm more likely to upvote answers that have one.


[code golf challenge]: https://codegolf.codidact.com/categories/49/tags/4274 "The code-golf tag"