Wikipedia:Reference desk/Archives/Mathematics/2017 December 16

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December 16

4 sided die procedure.

I want to roll a 4-sided die, but only have a six sided one. Is the following procedure "fair": Roll die, if 1,2,3,4 use that value. If 5,6 remember the value and re-roll. If 1,2,3,4 use that value. If 5,6 remember the value. If the remembered values are 55 use 1, 56 use 2, 65 use 3, 66 use 4. ? -- SGBailey (talk) 18:29, 16 December 2017 (UTC)

For any a in 1,2,3,4, the probability of getting a using your procedure is 1/6 + 2/6*1/6 + 1/6*1/6 = 9/36 = 1/4 (assuming the d6 is fair). So yes, fair.--108.52.27.203 (talk) 18:38, 16 December 2017 (UTC)

You can also see this by the fact that each of 1,2,3,4 is treated equivalently, and that after two throws one of the mentioned criteria will be met, so they must have equal probability, i.e., 1/4. Gap9551 (talk) 02:46, 17 December 2017 (UTC)

Why remember {5,6}, just ignore them. Simple enough! Ockham's Razor: "Entities are not to be multiplied without necessity" or do not make things (more) complicated for no reason. 110.22.20.252 (talk) 02:15, 17 December 2017 (UTC)

Maybe they only want to roll twice. Gap9551 (talk) 02:44, 17 December 2017 (UTC)

Actually, there are loads of procedures, each of which needs two throws only. Here is a procedure, significantly different from the OP's one:

Roll the die twice, and remember the sum of both results. Now:

• If the sum is any of the numbers: three or four or five, use their difference - being also their GCD - i.e. 1.
• If the sum is any of the numbers: eight or ten or twelve, use their difference - being also their GCD - i.e. 2.
• If the sum is any of the numbers: six or nine, use their difference - being also their GCD - i.e. 3.
• Else, use 4.

HOTmag (talk) 09:40, 17 December 2017 (UTC)

Something isn't right; if you roll a 2 twice then the procedure gives you 0, and I count 17 ways the procedure generates a 4 giving probability of 17/36. This seems simpler: if the sum is 4 or 7 then use 1, if the sum is 5 or 6 use 2, if the sum is 8 or 9 use 3, else use 4. --RDBury (talk) 06:39, 19 December 2017 (UTC)

What's not right? I agree that your procedure is simpler than mine; However, it seems like you misunderstood me. When I wrote "if the sum is any of the numbers...use their difference ", then by "their difference " I meant the difference between the numbers I'd just mentioned. For example, when I wrote "If the sum is any of the numbers: six or nine, use their difference ", then by "their difference " I meant the difference between the numbers six and nine. That said, what's not right?

Additionally, I didn't understand how you "count 17 ways the procedure generates a 4 ". I can only count nine ways (One 4 is generated when the sum of both results is two, two 4s are generated when the sum is eleven, and six 4s are generated when the sum is seven). HOTmag (talk) 08:34, 19 December 2017 (UTC)

HOTmag, the "their" in your procedure is ambiguous. It can easily be interpreted as referring to the difference between the two numbers that were rolled (which is how I read it the first time, and RDBury probably too), rather than the way you intended it, i.e., the difference between the numbers 3,4,5, etc. To be nitpicky, the difference between a set of three numbers is not clearly defined, but I assume you mean the difference between successive numbers when they are ordered (5-4=1, or 4-3=1, excluding 5-3=2). Just using GCD is more elegant in my opinion. Edit: and yes, your procedure is correct, giving 9/36 for each of 1,2,3,4. Gap9551 (talk) 16:53, 19 December 2017 (UTC)

In my opinion, the "their" in my procedure is unambiguous, because the sentence in which the "their" is mentioned - does not mention the numbers that were rolled - but rather only mentions the numbers I was really referring to when I wrote "their" after I'd just mentioned those numbers, so I can't notice any ambiguity there.

As for what I meant by "difference": Please notice that every arithmetic sequence has a clearly defined "difference", so again I can't notice any ambiguity there. HOTmag (talk) 17:36, 19 December 2017 (UTC)

It's called "common difference", not "difference". The article you linked to even starts by calling it "difference between the consecutive terms". This is another reason why the term "difference" in your procedure can easily be read as the difference between the two rolls, since the concept of difference is well defined for two numbers but not for three. Gap9551 (talk) 18:05, 19 December 2017 (UTC)

I agree the difference is subtle. I like your procedure, and I was just trying to be helpful in pointing out a possible way that people could misunderstand it, which would be a pity. Gap9551 (talk) 18:10, 19 December 2017 (UTC)

Nice procedure too, RDBury. Thanks everyone. Gap9551 (talk) 16:56, 19 December 2017 (UTC)

---

The OP's question has been answered by 108.52.27.203 and Gap9551; all subsequent discussion seems to only be me-too contributions that can't compete with the simple elegance of the OP's procedure. 110.22.20.252's procedure fails to show its alleged simplicity; as far as one can judge from its incomplete explanation, it rather seems to require an infinite iteration. HOTmag's procedure requires always two rolls and always a computation consisting of multiple operations; clearly inferior to the OP's solution.

To illustrate Gap9551's equivalence argument, here is a matrix with the possible outcomes:

First die ↓	second die
	⚀	⚁	⚂	⚃	⚄	⚅
⚀	1	1	1	1	1	1
⚁	2	2	2	2	2	2
⚂	3	3	3	3	3	3
⚃	4	4	4	4	4	4
⚄	1	2	3	4	1	2
⚅	1	2	3	4	3	4

— Sebastian 14:44, 19 December 2017 (UTC)