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Nontransitive Dice and maths

poshnialloposhniallo Registered User regular
edited February 2011 in Help / Advice Forum
Hi

I'm not great at maths, but I understand the basics.

I'm making a game and I have this question:

I understand the maths of nontransitive dice with sides showing numbers such as:

A 3/5/7

B 2/4/9

C 1/6/8

However, the probability of each beating the next one (A->B, B->C, C->A) is only 5/9.

Is there any set of numbers, sticking with three dice, to increase that probability?

Thanks.

poshniallo on
I figure I could take a bear.

Posts

  • SmasherSmasher Starting to get dizzy Registered User regular
    edited February 2011
    I was going to attempt to write a program to investigate this problem, but it turns out somebody beat me to it.

    According to that site, assuming three 6-sided dice the highest probability for each dice pair is 7/12.

    Edit: After reading the comments on the site it looks like you can do better, depending on what exactly you're looking for. If you want each pair of dice to have the same probability I think 7/12 is the max, but if you're simply looking for the maximum cumulative inequality then for

    D1: (1,10,11,12,13,14)
    D2: (2,3,4,15,16,17)
    D3: (5,6,7,8,9,18)

    P(D3>D2)=21/36=7/12
    P(D2>D1)=21/36
    P(D1>D3)=25/36

  • poshnialloposhniallo Registered User regular
    edited February 2011
    Smasher wrote: »
    I was going to attempt to write a program to investigate this problem, but it turns out somebody beat me to it.

    According to that site, assuming three 6-sided dice the highest probability for each dice pair is 7/12.

    Edit: After reading the comments on the site it looks like you can do better, depending on what exactly you're looking for. If you want each pair of dice to have the same probability I think 7/12 is the max, but if you're simply looking for the maximum cumulative inequality then for

    D1: (1,10,11,12,13,14)
    D2: (2,3,4,15,16,17)
    D3: (5,6,7,8,9,18)

    P(D3>D2)=21/36=7/12
    P(D2>D1)=21/36
    P(D1>D3)=25/36

    Thanks!

    However, I'd like to keep the numbers fairly low for gameplay reasons. If I have a 6-sided dice with numbers repeated, it seems I can keep them fairly low but am stuck to 5/9.

    If I have completely different numbers on each side I can go to 7/12, but then I get much larger numbers (up to 18).

    Is that right? If so I'll have to swallow that lower probability and use other gameplay tweaks to magnify the disparity between the three elements.

    I figure I could take a bear.
  • SmasherSmasher Starting to get dizzy Registered User regular
    edited February 2011
    Here's an equivalent set of dice with lower face values:

    D1: (1,4,4,4,4,4)
    D2: (2,2,2,5,5,5)
    D3: (3,3,3,3,3,6)

  • poshnialloposhniallo Registered User regular
    edited February 2011
    Smasher wrote: »
    Here's an equivalent set of dice with lower face values:

    D1: (1,4,4,4,4,4)
    D2: (2,2,2,5,5,5)
    D3: (3,3,3,3,3,6)

    Those are 7/12? Thanks!

    I figure I could take a bear.
  • SmasherSmasher Starting to get dizzy Registered User regular
    edited February 2011
    poshniallo wrote: »
    Smasher wrote: »
    Here's an equivalent set of dice with lower face values:

    D1: (1,4,4,4,4,4)
    D2: (2,2,2,5,5,5)
    D3: (3,3,3,3,3,6)

    Those are 7/12? Thanks!

    P(D1>D3) is 25/36 while the other two are 7/12 (21/36). Did you want them all to be the same? If so you can go to the site I listed and convert one of the other dice combinations by labeling the runs of sequential numbers in ascending order.

  • poshnialloposhniallo Registered User regular
    edited February 2011
    Smasher wrote: »
    poshniallo wrote: »
    Smasher wrote: »
    Here's an equivalent set of dice with lower face values:

    D1: (1,4,4,4,4,4)
    D2: (2,2,2,5,5,5)
    D3: (3,3,3,3,3,6)

    Those are 7/12? Thanks!

    P(D1>D3) is 25/36 while the other two are 7/12 (21/36). Did you want them all to be the same? If so you can go to the site I listed and convert one of the other dice combinations by labeling the runs of sequential numbers in ascending order.

    I did need them to all be the same, but what I've got from you so far has been a tremendous help.

    Delving into it, I understand the complexities and limits much more now, which is also very useful.

    I think I've got enough to be going on with now, so if a passing mod would like to they are welcome to lock this.

    Thanks again Smasher.

    I figure I could take a bear.
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