[gwnn]

I just saw this problem on youtube. The solution is actually subjective but I'm fairly sure my approach is not the best one.

You have three indistinguishable dice and you can only throw all three of them at once. This works great if you need to toss three dice, obviously. But what if you need less?

a) how can you simulate a single dice from the result of the three dice-throws? (with the exact probability)

if you're lazy, this is the solution:

Spoiler

add the three results and take mod 6.
edited to be fully rigorous: if the mod is 0 take 6.

b) how can you simulate two dice from the result of the three dice-throws? (again, with the correct probabilities)
It is NOT allowed to repeat problem a) twice.
if you're super lazy this is the solution I came up with, although it's hard to believe it's the simplest method out there:

Spoiler

I define s as 1, 2, 3, or 4 and B as 5 or 6.
Then, I have the following 5 cases:

• three identical numbers: 66
  
• BBB: 55 (nb: just 556 or 566 as 555 and 666 are already taken)
  
• ssB: ss (i.e., forget about the 5 or 6)
  
• sBB: s5 (i.e., take the small one and include a 5)
  
• sss: m6, where m is the sum of the three mod 5.

I am not sure if I'm proud or ashamed of this solution.

The video is here:
https://www.youtube....h?v=xHh0ui5mi_E
It spoils the first solution but gives no second solution, only saying that it might be ugly.

[barmar]

For a, don't forget to add 1.

[mgoetze]

Ah yes, makes sense that that's the simplest way to solve (a). I went a more complicated manual route:

Spoiler

3  1  --> 2
4  3  --> 1
5  6  --> 1
6  10 --> 2
7  15 --> 3
8  21 --> 3
9  25 --> 2
10 27 --> 1
11 27 --> 4
12 25 --> 5
13 21 --> 6
14 15 --> 6
15 10 --> 5
16 6  --> 4
17 3  --> 4
18 1  --> 5

[WesleyC]

Mgoetze: Your solution is exactly the same as the one given in the video, just expressed slightly differently. And unlike the video you actually *did* show proof that your result is correct.

One thing that wasn't mentioned is that just taking mod of the numbers only works if you're trying to simulate a die with the same number of sides due to the symmetry. For example, taking say MOD 4 of the 3* 6-sided dice will not give you a fair 4-sided die.

[StevenG]

You shut your eyes before throwing the dice and select the appropriate number of dice at random.

[WellSpyder]

StevenG, on 2016-April-14, 01:18, said:

Yuo shut your eyes before throwing the dice and select the appropriate number of dice at random.

Or mark a point on the table and take the one that lands nearest for (a) and exclude the one that lands nearest for (b)....

[Cyberyeti]

WesleyC, on 2016-April-13, 21:58, said:

Mgoetze: Your solution is exactly the same as the one given in the video, just expressed slightly differently. And unlike the video you actually *did* show proof that your result is correct.

One thing that wasn't mentioned is that just taking mod of the numbers only works if you're trying to simulate a die with the same number of sides due to the symmetry. For example, taking say MOD 4 of the 3* 6-sided dice will not give you a fair 4-sided die.

MGoetze's is functionally the same but not identical and is the way I did it.

I worked out the probabilities and took the lowest 3 and highest 3 that added up, so 3 (1) + 6 (10) + 9 (25) = 36/216

Since 3 and 18, 6 and 15, 9 and 12 have the same probabilities, there are alternatives, the modulo solution groups 3/9/15 and 6/12/18.

[Zelandakh]

Other similar problems with simple and related solutions are how to simulate a die of a lower number of sides, such as a 5-sided die from an icosahedron, how to simulate a die of a higher number of sides from 2 dice and simulating a die roll without having any die at all. Unlike the OP problems, all of these have practical uses.

[gwnn]

I learned from dburn not to use "die" as a noun and my life got instantly much better after that.

[barmar]

WesleyC, on 2016-April-13, 21:58, said:

Mgoetze: Your solution is exactly the same as the one given in the video, just expressed slightly differently. And unlike the video you actually *did* show proof that your result is correct.

Someone in the comment section said that found the solution unbelievable, but they made a spreadsheet and convinced themselves.

[kenberg]

Zelandakh, on 2016-April-14, 05:28, said:

Other similar problems with simple and related solutions are how to simulate a die of a lower number of sides, such as a 5-sided die from an icosahedron, how to simulate a die of a higher number of sides from 2 dice and simulating a die roll without having any die at all. Unlike the OP problems, all of these have practical uses.

With no die at all?? You do have something, I assume?

[gwnn]

barmar, on 2016-April-14, 09:16, said:

Someone in the comment section said that found the solution unbelievable, but they made a spreadsheet and convinced themselves.

It's actually easy to see that the solution for only 1 dice works. Of course I also made a spreadsheet first and then learned the easy way in the comment section.

Spoiler

Just imagine two of the dice have landed already and only the third one is still rolling. Whatever the sum of the first two is and what the probability of that sum is, the third dice will divide that probability in 6 equal parts with equal probabilities in the six remainders.

[kenberg]

barmar, on 2016-April-14, 09:16, said:

Someone in the comment section said that found the solution unbelievable, but they made a spreadsheet and convinced themselves.

To each his own, but the advantage of thinking it through is that you can see in some generality when an analogous coding can be done and identify cases where it cannot be done. It is probably quicker to think it through in this case, and in a case where it cannot be done it is definitely useful to think through why it cannot be done rather than try a lot of spreadsheet programs. Plus it is more fun.

But by spreadsheet or by thinking, I don't see how to simulate a die roll without having something. It reminds me of Guys and Dolls, where Big Louie wants to shoot craps with his own dice. He has had the spots removed, but not to worry, he remembers where they are.

[Zelandakh]

gwnn, on 2016-April-14, 05:58, said:

I learned from dburn not to use "die" as a noun and my life got instantly much better after that.

Using dice for the singular is simply wrong. I could easily add it to the pet peeves thread.

kenberg, on 2016-April-14, 09:24, said:

With no die at all?? You do have something, I assume?

You have 2 people and nothing else of any special note. I once watched a group of children playing diceless D&D on the train using a form of this method but it is also useful for picking a random number while avoiding the usual biases that the basic method involves.

Spoiler

The 2 most obvious methods are:
The Asker selects a number and asks the Roller to select a result. You add the 2 together and subtract the number of sides if the result is not valid; and
The Asker holds out the appropriate number of fingers and silently selects one to represent 1 and a direction. 2 ..3..4, etc are then assigned to each finger in turn. The Roller simply selects a finger. This latter method is what the children on the train were doing.
If alone, you could achieve the same effect by looking at the second hand of your watch after selecting a number and using that as the Asker "seed" value (converting the value you get to a d6 roll in some fashion (various possibilities for that)).

[gwnn]

Zelandakh, on 2016-April-14, 12:24, said:

Using dice for the singular is simply wrong. I could easily add it to the pet peeves thread.

Or you could easily consult Merriam-Webster:
http://www.merriam-w...dictionary/dice
Or Oxford Dictionaries:
http://www.oxforddic...on/english/dice
Before you say what is simply wrong. You could also find a bunch of articles written on the topic without any clear conclusion. In other sources die is indeed given as the singular, I know.

Anyway you recently told me not to engage trolling so I will refrain from talking about posters who just go to a math thread and give the wonderful reply that there are other similar but more useful problems out there.

[kenberg]

Zelandakh, on 2016-April-14, 12:24, said:

You have 2 people and nothing else of any special note. I once watched a group of children playing diceless D&D on the train using a form of this method but it is also useful for picking a random number while avoiding the usual biases that the basic method involves.

Spoiler

The 2 most obvious methods are:
The Asker selects a number and asks the Roller to select a result. You add the 2 together and subtract the number of sides if the result is not valid; and
The Asker holds out the appropriate number of fingers and silently selects one to represent 1 and a direction. 2 ..3..4, etc are then assigned to each finger in turn. The Roller simply selects a finger. This latter method is what the children on the train were doing.
If alone, you could achieve the same effect by looking at the second hand of your watch after selecting a number and using that as the Asker "seed" value (converting the value you get to a d6 roll in some fashion (various possibilities for that)).

Ok, I agree. I didn't give much thought to just what was desired here.
You do need two people. Or one person and a clock. Or something!
But yes, two people, or even a group of people, could shoot craps this way without any dice.

As to singular/plural I was surprised to see that dice is now regarded as both but I guess the dice is cast.

[Zelandakh]

kenberg, on 2016-April-14, 16:41, said:

As to singular/plural I was surprised to see that dice is now regarded as both but I guess the dice is cast.

In the same way as "to run quick" is now regarded as correct along with similar "adverbs". Note that there is nothing new in dictionaries posting "bad" English as perfect(ly) correct. My 1918 Concise Oxford definitively provides ax and stanch as alternative spellings along with many others. Even being aware of the alternative rules, the correct forms are for me still axe, staunch, quickly and, yes, die.

[gwnn]

Instead of saying "oh, interesting, I didn't realize that. I guess I was wrong about 'simply incorrect' - I should have checked around before making my sweeping claims." you can always be counted on to change your standard from "simply wrong" to "for me, incorrect" without blinking an eye. I don't get this kind of communication. Will the world collapse if you admit you were even partially wrong?

[Fluffy]

So I have a button that rolls 3 dices into a transparent box, and I want to pick a single result. This has so many solutions, I could pick the result closer to me, but that would end up being subjective, so I would just roll them enough times untill there is a double/triple result, then I would pick that one.

The mod solution is better? If you think so you haven't played board games with people who consistently failed maths. And on top of that, it works for every kind of dice, not only 6-sided ones with numbers (as long as each sides are not duplciated).

EDIT: Sorry the mod solution works on any kind of numbered dice (*dumb*)

EDIT2: I saw the video and the guy says that here are 216 posibilities when rolling 3 dice... this is technically wrong as you can't distiguish 2,2,5 from 2,5,2 not 5,3,1 from 3,1,5. The right total is 56 I think, but each will have different probability.

[mgoetze]

Fluffy, on 2016-April-15, 02:22, said:

3 dices

The next step in the evolution of the language.