Miwin's dice

design of numbers of Miwins dice III, IV, V

Each die is named for the sum of its two lowest numbers. The dots on each die are colored blue, red or black. Each die has the following numbers:

Numbers 1 and 9, 2 and 7, and 3 and 8 are on opposite sides on all three dice. Additional numbers are 5 and 6 on die III, 4 and 5 on die IV, and 4 and 6 on die V. The dice are designed in such a way that, for every die, another will usually win against it. The probability that a given die in the sequence (III, IV, V, III) will roll a higher number than the next in the sequence is 17/36; a lower number, 16/36. Thus, die III tends to win against IV, IV against V, and V against III. Such dice are known as nontransitive.

design of numbers of Miwins dice IX, X, XI

Each die is named for the sum of its lowest and highest numbers. The dots on each die are colored yellow, white or green. Each die has the following numbers:

The probability that a given die in the sequence (XI, X, IX, XI) will roll a higher number than the next in the sequence is 17/36; a lower number, 16/36. Thus, die XI tends to win against X, X against IX, and IX against XI.

Third set:

fourth set:

The probability that a given die in the first sequence (5, 3, 1, 5) or the second sequence (6, 4, 2, 6) will roll a higher number than the next in the sequence is 5/9; a lower number, 4/9.

• 1/3 of the die-face values can be divided by three without carry over.
• 1/3 of the die-face values can be divided by three having a carry over of one.
• 1/3 of the die-face values can be divided by three having a carry over of two.

The probability for a given number with all three dice is 11/36, for a given rolled double is 1/36, for any rolled double 1/4. The probability to obtain a rolled double is only 50% compared to normal dice.

• 
  Cumulative frequency type III and IV
• 
  Cumulative frequency type III and V
• 
  Cumulative frequency type IV and V
• 
  Cumulative frequency type III and IV and V = "Miwin-Distribution"

Removing the common dots of Miwin's Dice reverses intransitivity.

III:IV = 17:16, III':IV' = 4:5   →   IV' > III'
analog: III/V, IV/V

Miwin's dice facilitate generating numbers at random, within a given range, such that each included number is equally-likely to occur. In order to obtain a range that does not begin with 1 or 0, simply add a constant value to bring it into that range (to obtain random numbers between 8 and 16, inclusive, follow the 1 – 9 instructions below, and add seven to the result of each roll).

• 1 – 9: 1 die is rolled (chosen at random): P(1) = P(2) = ... = P(9) = 1/9
• 0 – 80: 2 dice are rolled (chosen at random), always subtract 1: P(0) = P(1) = ... = P(80) = 1/9² = 1/81

Governing probability: P(0) = P(1) = ... = P(90) = 8/9³ = 8/729

To obtain an equal distribution with numbers from 0 – 90, all three dice are rolled, one at a time, in a random order. The result is calculated based on the following rules:

• 1st throw is 9, 3rd throw is not 9: gives 10 times 2nd throw (possible scores: 10, 20, 30, 40, 50, 60, 70, 80, 90)
• 1st throw is not 9: gives 10 times 1st throw, plus 2nd throw
• 1st throw is equal to the 3rd throw: gives 2nd throw (possible scores: 1, 2, 3, 4, 5, 6, 7, 8, 9)
• All dice equal: gives 0
• All dice 9: no score

Sample:

This gives 91 numbers, from 0 – 90 with the probability of 8 / 9³, 8 × 91 = 728 = 9³ − 1

Governing probability: P(0) = P(1) = ... = P(103) = 7/9³ = 7/729

This gives 104 numbers from 0 – 103 with the probability of 7 / 9³, 7 × 104 = 728 = 9³ − 1

Governing probability: P(0) = P(1) = ... = P(728) = 1 / 9³ = 1 / 729

This gives 729 numbers, from 0 – 728, with the probability of 1 / 9³.

One die is rolled at a time, taken at random.

Create a number system of base 9:

• 1 must be subtracted from the face value of every roll because there are only 9 digits in this number system ( 0,1,2,3,4,5,6,7,8 )
• (1st throw) × 81 + (2nd throw) × 9 + (3rd throw) × 1

This system yields this maximum: 8 × 9² + 8 × 9 + 8 × 9° = 648 + 72 + 8 = 728 = 9³ − 1

Examples:

3 throws, random selection of one of the dice for each throw, type is used as attribute:

5832 = 2 × 2 × 2 × 9 × 9 × 9 = 18³ numbers are possible.

Two dice are rolled, chosen at random, one at a time. Each pair is scored by multiplying the first by nine and subtracting the second from the result: 1st throw × 9 − 2nd throw.

Examples:

This variant provides numbers from 0 – 80 with a probability of 1 / 9² = 1 / 81.

Two dice are rolled, chosen at random, one at a time. The pair is scored according to the following rules:

• 1st throw is 9: gives 10 × 2nd throw − 10
• 1st throw is not 9: gives 10 × 1st throw + 2nd throw − 10

Examples

This variant provides numbers from 0 – 80 with a probability of 1 / 9² = 1 / 81.

Two dice are rolled, chosen at random, one at a time. The score is obtained according to the following rules:

• Both throws are 9: gives 0
• 1st throw is 9 and 2nd throw is not 9: gives 10 × 2nd throw
• 1st throw is 8: gives 2nd throw
• All others: gives 10 × 1st throw − 2nd throw

Examples: