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Let's say we have a 20-sided die and a 6-sided die. In order to roll the 6-sided die one time, we must first roll a 20 on the 20-sided die. Our objective is to get to the 6-sided die and roll each number once

How many rolls should it take us to roll each number on the 6-sided die? To me, it seems to be an extension of the https://en.wikipedia.org/wiki/Coupon_collector%27s_problem

My initial hypothesis is something like (20)(6/(6-0)) + (20)(6/(6-1)) + ... (20)*(6/(6-5))

AKA SUM i=0 to n-1 with (20)*(n/(n-i)), where n is subbed in for 6

Looking good? I don't really have anywhere to validate my results, as this is for a personal project

user3210680
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  • Not sure the rules are clear. Once I throw a $20$, do I then throw the $6$ sided die only? Or do I only get to throw the $6$ sided die once after each throw of a $20$? – lulu Aug 11 '20 at 20:21
  • Nor do I understand your computation. What's $n$? – lulu Aug 11 '20 at 20:22
  • @lulu "I only get to throw the 6 sided die once after each throw of a 20?" <- that's the one. For the computation, n is 6 – user3210680 Aug 11 '20 at 20:45
  • Please edit your post to include that explanation...don't leave vital information in the comments only. – lulu Aug 11 '20 at 20:46
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    In any case, the problem is essentially equivalent to the question of how many rolls of the six sided die do we expect it to take in order to see all $6$ values. We expect it to take $20$ rolls to see a $20$ on the twenty sided die, so just add $20$ at each stage of the process. – lulu Aug 11 '20 at 20:48
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    Gotcha, updated the problem description. So it looks like 206/6 + 206/5 + 206/4 + 206/3 + 206/2 + 206/1 would aptly describe the number of expected rolls to 'finish' rolling 1-6 on the 6-sided die – user3210680 Aug 11 '20 at 20:54
  • The concept is good, but you have to watch the arithmetic. Say the six sided die were a one sided die. Then the answer is $20+1=21$ not $20\times 1=20$. I suggest working it out carefully for a two sided die before settling on a pattern. – lulu Aug 11 '20 at 20:56
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1 Answers1

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From the basic coupon collector's problem, you need an expected $\sum\limits_{i=1}^n \frac{n}{i}$ attempts to get the full set.

So here, with $n=6$, you need an expected $14.7$ rolls of the six-sided die.

For each roll of the six-sided die, you need an expected $20$ rolls of the twenty-sided die to see a $20$, using the geometric distribution. So in this case you need an expected $20\times 14.7=294$ rolls of the twenty sided die.

So overall you need an expected $14.7+294=308.7$ dice rolls.

Henry
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