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Modifying two standard dice

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Assume you have a standard dice with labels 1, 2, 3, 4, 5, 6 on each respectively. Pretend you are playing a simple game where you roll the two dice and add their values together. Is it possible to relabel the dice using positive integers in such a way that if you play the same dice game you would have an equal probability of rolling the same sums?

Note: What I mean by relabeling is not simply rotating all of the numbers on the dice where by each dice still contains the same six numbers but rather actually making new dice that has a different set of six integers than the original. (e.g. one dice could be 1, 2, 3, 4, 5, 9 and the other could be 1,1,1,1,1,1 ... would this pair produce the same sum probabilities as the original?--in short, no ^_^)

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[1, 2, 2, 3, 3, 4]


[1, 3, 4, 5, 6, 8]
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You can achieve equal likelihood for the sums of two dice.

Put a 1 (or the number of your choice) on each of the twelve faces.

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Is it possible to relabel the dice using positive integers in such a way that if you play the same dice game you would have an equal probability of rolling the same sums?

Does "same sums" mean sums that are the same as the ones you

would get using standard dice: 2 3 4 5 6 7 8 9 10 11 and 12?

If so, then the answer is yes.

It again is trivial to achieve all those outcomes with equal probability.

Label one of the dice with 12 13 14 15 16 17.

This insures that the sums that are the same as those produced by

standard dice will appear with equal probability: zero.

Or does "same sums" ask for re-numbered dice to produce precisely

the sums given by standard dice, and nothing else, but do so with

equal probability? i.e. p(2) = ... = p(7) = ... = p(12)?

Then the answer is no.

There are 36 distinguishable configurations of two dice.

There are 11 numerical sums to achieve.

For them to appear with equal probability,

that probability would have to be

36/11 = 3.2727272727272727272727272727273.

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Posted (edited) · Report post

Let me rephrase the op. the probability of rolling a 2 with two normal dice is 1/36. The probability of rolling a 3 is 2/36. Can you relabel the dice so that you still get these probabilities and all others using different dice configurations

Edited by BMAD
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Let me rephrase the op. the probability of rolling a 2 with two normal dice is 1/36. The probability of rolling a 3 is 2/36. Can you relabel the dice so that you still get these probabilities and all others using different dice configurations

How about (1,2,3,4,5,6) and (a,a,a,a,a,a) where a is any +ve integer?

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Assume you have a standard dice with labels 1, 2, 3, 4, 5, 6 on each respectively. Pretend you are playing a simple game where you roll the two dice and add their values together. Is it possible to relabel the dice using positive integers in such a way that if you play the same dice game you would have an equal probability of rolling the same sums?

Note: What I mean by relabeling is not simply rotating all of the numbers on the dice where by each dice still contains the same six numbers but rather actually making new dice that has a different set of six integers than the original. (e.g. one dice could be 1, 2, 3, 4, 5, 9 and the other could be 1,1,1,1,1,1 ... would this pair produce the same sum probabilities as the original?--in short, no ^_^)

Here's an answer

Let i be any integer.

Label one die with (i + 1, i+2, …, i + 6 ). Label the other die with (-i + 1, -i + 2, …, -i + 6 ). Rolling these two dice together will result in a sum probability distribution that is identical to that of two standard dice.

Note that the case of two standard die is equivalent to setting i = 0.

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Assume you have a standard dice with labels 1, 2, 3, 4, 5, 6 on each respectively. Pretend you are playing a simple game where you roll the two dice and add their values together. Is it possible to relabel the dice using positive integers in such a way that if you play the same dice game you would have an equal probability of rolling the same sums? Note: What I mean by relabeling is not simply rotating all of the numbers on the dice where by each dice still contains the same six numbers but rather actually making new dice that has a different set of six integers than the original. (e.g. one dice could be 1, 2, 3, 4, 5, 9 and the other could be 1,1,1,1,1,1 ... would this pair produce the same sum probabilities as the original?--in short, no ^_^)

Here's an answer

Let i be any integer.

Label one die with (i + 1, i+2, …, i + 6 ). Label the other die with (-i + 1, -i + 2, …, -i + 6 ). Rolling these two dice together will result in a sum probability distribution that is identical to that of two standard dice.

Note that the case of two standard die is equivalent to setting i = 0.

Nice.

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Posted · Report post

?

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?

Maybe OP meant non-negative?

I think I have a proof there is no other solution. Construct the table.

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?

Maybe OP meant non-negative?

I think I have a proof there is no other solution. Construct the table.

Missed the point about positive integers. I agree with bonanova, in that case

The sum distribution is (2,3,4,5,6,7,8,9,10,11,12)

Since the smallest sum is 2, and the die number are positive integers, then 1 face from each of the 2 die must be a '1'.

The next sum is 3 with probability 2/36, so you can either have

1) two '2' on 1 die

2) one '2' on each of the 2 dice

If you allow 1), then it quickly becomes clear that it is not possible to get a sum of 4 with probability 3/36. Therefore 1 face from each of the 2 die must have a '2' on it.

Following the same logic further, we see that the only solution to the constraints in the OP is the case of the standard dice. In other words, there are no other dice arrangement of positive integers such that their sum probability distribution is precisely the same as that of two standard dice.

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Posted · Report post

There is a solution using positive integers. I discovered something rather ...amazing while working this solution.

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Let the digits be

(die 1) = a b c d e f and

(die 2) = m n o p q r in ascending order.

All faces show positive integers.

  1. You need one 2 outcome so a = m = 1: 1-1; and no other face is a 1.
  2. You need two 3 outcomes so b = n = 2: 1-2, 2-1; and no other face is a 2.
  3. You need three 4 outcomes so c = o = 3: 1-3, 2-2, 3-1; and no other face is a 3.
  4. You need four 5 outcomes so d = p = 4: 1-4, 2-3, 3-2, 4-1; and no other face is a 4.
  5. You can't have a 13+ outcome so f < 7 > r: else d-r or f-p > 12.
  6. You need one 12 outcome so f = r = 6: 6-6; and no other face is a 6.
  7. You need two 11 outcomes so e = q = 5: 5-6, 6-5; and no other face is a 5.

Which statement is suspect?.

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Let the digits be

(die 1) = a b c d e f and

(die 2) = m n o p q r in ascending order.

All faces show positive integers.

  • You need one 2 outcome so a = m = 1: 1-1; and no other face is a 1.
  • You need two 3 outcomes so b = n = 2: 1-2, 2-1; and no other face is a 2.
  • You need three 4 outcomes so c = o = 3: 1-3, 2-2, 3-1; and no other face is a 3.
  • You need four 5 outcomes so d = p = 4: 1-4, 2-3, 3-2, 4-1; and no other face is a 4.
  • You can't have a 13+ outcome so f < 7 > r: else d-r or f-p > 12.
  • You need one 12 outcome so f = r = 6: 6-6; and no other face is a 6.
  • You need two 11 outcomes so e = q = 5: 5-6, 6-5; and no other face is a 5.
Which statement is suspect?.

Your second assumption is suspect (regarding the statement that they must be ascending).

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The probability distribution for standard dice can be generated by powers of d(x)=(x+x^2+x^3+x^4+x^5+x^6)^n.

The coeff of x^k gives the frequency out of 6^n of rolling k.

Two dice: (x+x^2+x^3+x^4+x^5+x^6)^2 = x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^10+2x^11+x^12

An alternate factorization of the polynomial is (x+x^3+x^4+x^5+x^6+x^8)*(x+2x^2+2x^3+x^4)

The first factor represents a die w faces (1, 3, 4, 5, 6, 8). The other factor, faces (1, 2, 2, 3, 3, 4).

well done witzar

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The probability distribution for standard dice can be generated by powers of d(x)=(x+x^2+x^3+x^4+x^5+x^6)^n.

The coeff of x^k gives the frequency out of 6^n of rolling k.

Two dice: (x+x^2+x^3+x^4+x^5+x^6)^2 = x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^10+2x^11+x^12

An alternate factorization of the polynomial is (x+x^3+x^4+x^5+x^6+x^8)*(x+2x^2+2x^3+x^4)

The first factor represents a die w faces (1, 3, 4, 5, 6, 8). The other factor, faces (1, 2, 2, 3, 3, 4).

well done witzar

Fascinating analysis. Thanks for posting this wonderful problem.

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