| (Phil Dutre) | 28 Oct 2012 8:18 a.m. PST |
Relooking through one of my old courses on probability, I came alomg the following six-sided dice: A: 3 3 3 3 3 3
B: 0 0 4 4 4 4
C: 1 1 1 5 5 5
D: 2 2 2 2 6 6 Now, if you do the math, it turns out that B wins from A in 2/3 of the cases. C wins from B; D wins from C; and finally A wins from D.
So, we have a probabilistic equivalent of rock-papers-scissors. Of course, there''s nothing mysterious about this, but it's a funny little result that perhaps could find its use in a rule mechanic here or there. |
| Deserter | 28 Oct 2012 9:22 a.m. PST |
I can be wrong but this the first new dice mechanics I have seen in ages… thank you… |
| MajorB | 28 Oct 2012 9:35 a.m. PST |
Of course the "A" dice aboove is not really a dice at all, since the result is always "3". In other words, a mathematical constant. |
| RavenscraftCybernetics | 28 Oct 2012 9:37 a.m. PST |
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| (Phil Dutre) | 28 Oct 2012 11:00 a.m. PST |
You could always change the numbers on the dice, of course, as long as relative order is manintained. E.g. The two zeros could be replaced by 1 and 2, the three ones by 3, 4 and 5 etc. |
| MajorB | 28 Oct 2012 12:04 p.m. PST |
The two zeros could be replaced by 1 and 2, the three ones by 3, 4 and 5 etc No, that wouldn't work, the probabilities of success would change. |
| Dale Hurtt | 28 Oct 2012 4:21 p.m. PST |
I am not sure what the point is, however. B beats A 4 out of 6 times.
C beats B 5 out of 6 times.
D beats C 5 out of 6 times.
A beats D 4 out of 6 times. The probabilities are not even. |
| kreoseus2 | 29 Oct 2012 3:03 a.m. PST |
Thats Rock/paper/scissors. What about lizard/spock ? |
| MajorB | 29 Oct 2012 3:40 a.m. PST |
C beats B 5 out of 6 times.
D beats C 5 out of 6 times. Sorry, that is incorrect. With the following values:
B: 0 0 4 4 4 4
C: 1 1 1 5 5 5
D: 2 2 2 2 6 6 C beats B 4 out of 6 times.
D beats C 4 out of 6 times. |
| (Phil Dutre) | 29 Oct 2012 4:03 a.m. PST |
The two zeros could be replaced by 1 and 2, the three ones by 3, 4 and 5 etcNo, that wouldn't work, the probabilities of success would change.
It would still work, but perhaps I was unclear about what I meant. 2 zeros ==> 1 and 2
3 ones ==> 3, 4 and 5
4 twos ==> 6, 7, 8 and 9
6 threes ==> 10, 11, 12, 13, 14, 15
etc … The odds of beating another die are still the same, since the only thing that counts is whether a number is smaller or larger than the other one, not the actual values. |
| Marshal Mark | 29 Oct 2012 3:00 p.m. PST |
I have a similar set of 3 dice in a probability worksheet I use at school. |
| pellen | 29 Oct 2012 11:14 p.m. PST |
I first learned about nontransitive dice a few years ago from Martin Gardner's Collosal Book of Mathematics. Also see
link Not that I can think of any practical use for them in a game. |
| Patrice | 11 Nov 2012 8:34 a.m. PST |
Interesting and amazing. I did not think that I could still be surprised by dice statistics, but I had to draw and fill the 6x6 tables to believe this. |
