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If I have fair 6-sided dice, how many should I roll it so that I have more than 50% chance of getting every unique side at least once?

Or, more generally,

For a given die with $d$ sides, how many times $n$ should I roll it so that I have more than $p$ chance of getting every unique side once?

What I've done so far:

If you only roll 6 dice, you odds of getting all unique is $6! \big / 6^6$, which is 1.5%. For more dice, all potential rolls are $6^n$, but how many of those have all unique faces?

HallaSurvivor
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Leo Barlach
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  • Work recursively, keeping track of how many unique faces you have seen. Or just simulate, which should be considerably easier. – lulu Aug 14 '22 at 17:32
  • I'm working on another combinatory things, and there are 230k combinations. How many time should you do it until you get all combinations, but for that number it's too big to simulate. – Leo Barlach Aug 14 '22 at 17:37
  • A better match is: https://math.stackexchange.com/questions/947599/how-many-dice-do-you-have-to-roll-to-get-your-odds-of-seeing-each-face-at-least?rq=1 Regardless, versions of this question have been asked on this site many times; it's worth doing a search. – Misha Lavrov Aug 14 '22 at 17:37

1 Answers1

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I don't know a method to directly compute n,

but since we want all the numbers $1-6$, ie all cells filled, Stirling numbers of the second kind come to mind immediately, and it is easy to find heuristically.

we want $P_{n-1} <0.5 \leq P_n$

where $P_n = \large\dfrac{6!{n\brace 6}}{6^n} $

Starting with a hunch of $12$, which yields $P_n \approx 0.4378$,

$n\;$ turns out to be $13,\;\; P_{13} \approx0.5139$

true blue anil
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