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Topic: 3 Indistinguishable Dice (Read 1621 times) |
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lorentztrans
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3 Indistinguishable Dice
« on: Apr 28th, 2016, 11:16pm » |
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I want to generate an integer 2-12 with the same probability distribution as the sum of two dice. (2 with probability 1/36, 7 with probability 1/6 etc.). I only have access to three dice which must be rolled simultaneously and are indistinguishable except for their values when rolled. For example 4,4,6 cannot be distinguished from 4,6,4 as an outcome.
Is there an elegant way to use the 3 die result to generate a 2 die sum with the correct distribution of outcomes?
(This is not a lateral-thinking puzzle. It may be more appropriate for ps, but the setup is simple enough to go here. From: standupmaths "The Three Indistinguishable Dice Puzzle" who explains it much more nicely. An "ok" solution seems to exist, but it's not entirely elegant)
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towr
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Re: 3 Indistinguishable Dice
« Reply #1 on: Apr 29th, 2016, 1:29pm » |
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Can I randomly give the dice some order after I've thrown them?
And then just leave one out based on the sum of all three modulo 3
Never mind, that doesn't really make much sense. If I can give them some random order, I could just pick any two dice at random without further ado.
It ought to work with sorted ordering (or worst-case ordering).
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| « Last Edit: Apr 29th, 2016, 1:41pm by towr » |
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towr
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Re: 3 Indistinguishable Dice
« Reply #2 on: Apr 29th, 2016, 2:24pm » |
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An unelegant solution: just assign all (sorted) possible outcomes to the desired distribution. (e.g. all dice equal (6 outcomes) => 2, first two 0, last 2-5 (12 outcomes) => 3, etc)
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Hippo
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Re: 3 Indistinguishable Dice
« Reply #3 on: May 10th, 2016, 2:57pm » |
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My first tought was:
Order results and "bin" dice with index (sum mod 6).
... So decrement the sum by dice indexed by (sum mod 6).
... but this generates 2 7 times, 3 12 times, 4 13 times, 5 33 times ...
and correct solution is 2 6 times, 3 12 times, 4 18 times, 5 24 times ...
Correct solution would be any mapping of ordered triples to 2..12 giving correct weighted sums per results. AAA has weight 1, ABB has weight 3 and ABC has weight 6 (number of rolls corresponding to the ordered result).
As all the weighted sums should be multiples of 6, all AAA rolls should generate the same result.
Each result is generated by even number of ABB triples. It is almost trivial to generate a solution.
Interesting task is to generate solution with easy enough description.
And now I have read the previous post which says the same;)
... So one rather simple assignment rule set:
In the case AAA (all 3 same) use 12 as the result
In the case AAB (not all 3 same, but not all 3 different) use A+B as the result.
In the case ABC (all different) use A+B+C-4 as the result. ... that would be almost OK except one 5 should be changed to 10 ... for example 126 would be exception giving result 10.
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| « Last Edit: May 10th, 2016, 3:40pm by Hippo » |
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rmsgrey
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Re: 3 Indistinguishable Dice
« Reply #4 on: May 11th, 2016, 7:23am » |
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I saw someone's solution based on the 1-die solution - so die 1 is the 1-die solution, and die 2 is pulled from a look-up table...
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teekyman
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Re: 3 Indistinguishable Dice
« Reply #5 on: Jun 27th, 2016, 6:03am » |
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These solutions assume you can roll the 3 die more than once. Otherwise, I think mapping the distributions is the only option, and the question becomes "is there an elegant probability map from the sorted distribution of 3 die into that of the sum of 2 die)
Solution 1. Towr's first (deleted?) solution is right, you can randomly choose a die to remove using another roll of the die.
Intuitive idea: You want to be able to pick two of the three die at random to get the desired distribution. If the randomness with which you pick two die out of three is independent of the rolls of the die, then you have the desired result.
If all the randomness you are allowed is a roll of 3 dies, roll them and take the result mod 3 to get a uniform (0,1,2) random variable. Then roll the 3 die again, and remove that number die (after placing the die in sorted order).
This might worry your intuition, because conditioned on the result of the die you will throw out, (0,1,2), the resulting distribution will not be the distribution of two uniformly chosen independent die because we placed them in sorted order first. But this is ok, because we want the unconditional result.
However, from the perspective of a die, no matter what number I am rolled and how I was placed in sorted order with the other die, there is a 1/3 chance that I will be one thrown out.
Hence the key fact we need, that the randomness with which a die is thrown out is independent of the rolls of the die, holds.
The downside is that unlike the solution of mapping the probability distributions into each other, this idea requires two throws of the die instead of just 1.
Easier Solution 2:
Rolling 3(or n) die and taking the sum mod 6 has the distribution of a die roll (map 0 -> 6). Do this twice and add the results!
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| « Last Edit: Jun 27th, 2016, 6:41am by teekyman » |
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