Comments on Probability of rolling all 6 dice faces
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Probability of rolling all 6 dice faces
The probability of rolling every number from 1 to 6 with $N$ six-sided dice.
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, $0 \le p \le 1$
- This is the probability of seeing every number from 1 to 6 at least once after rolling $N$ six-sided dice simultaneously.
- For inputs up to and including 10, your output is valid if rounding it to 6 decimal places results in the output shown in the test cases.
Note that this means that if you find an incorrect algorithm that happens to give the correct result when rounded to 6 decimal places for inputs from 1 to 10, that is still a valid entry.
Test cases
- Test cases are in the format
input : output.
1 : 0
2 : 0
3 : 0
4 : 0
5 : 0
6 : 0.015432
7 : 0.054012
8 : 0.114026
9 : 0.189043
10 : 0.271812
Scoring
This is a code golf challenge. Your score is the number of bytes in your code.
Explanations are optional, but I'm more likely to upvote answers that have one.
cQuents, 10 bytes ```text …
10mo ago
Dyalog APL, 19 bytes (with …
10mo ago
Thunno 2, 14 bytes ``` 6Rẉ …
10mo ago
Dyalog APL, 21 bytes ```apl …
10mo ago
Post
Dyalog APL, 19 bytes
(with index origin zero)
{-/(!∘6×⍵*⍨6÷⍨⊢)⍳7}
This isn't bruteforce! The number of cases where all six faces appear are the OEIS sequence A000920.
\[ P_n = \frac{\mathsf{OEIS}_{\text{A000920}}(n)}{6^n} \]which is
\[ \frac {6!} {6^n} \left\lbrace {n \atop 6}\right\rbrace \]where $\left\lbrace{n\atop k}\right\rbrace$ are the Stirling numbers of the second kind, which expand like this:
\[ \frac 1 {6^n} \sum_{i=0}^6 \left(-1\right)^{6-i} i^n \binom 6 i \](where $\binom n k$ are binomial coefficients) which simplifies to
\[ \sum_{i=0}^6 \left(-1\right)^i \binom 6 i\left(\frac i 6\right)^n \]Note that the $i=0$ term is zero, but it is useful for me because it means the first term of the alternate sum has a positive coefficient.
Code explanation:
-
⍳7To the list of numbers 0 through 6, apply the following function:-
!∘6$\binom 6 x$ -
×times -
6÷⍨⊢$\frac x 6$ -
⍵*⍨to the power of $n$
-
-
-/and then take the alternate sum
Kinda sad to see a bruteforce solution is smaller than the closed form, I'll try and find a language where this is even shorter :)
Side note: this seems to have some floating point error which means the results for $n \in \left\{ 3, 4, 5 \right\}$ aren't exactly zero but in the range of $10^{-16}$ with ⎕fr←647 and $10^{-34}$ with ⎕fr←1287. either of these are both within the 6 decimal places of precision required, but just in case you cared about the pure computation, I think this is just error stacking up.
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