Dice probabilities

[Albert Chan]

(04-30-2024 10:07 PM)dm319 Wrote:  This is sorcery.

Dice = (x^2 + x + 1)/3
With uniform distribution, center has probability of 1/3

Dice^2 = (x^4 + 2x^3 + 3x^2 + 2x + 1)/9
We have triangular distribution. Again, center probability = 3/9 = 1/3

Dice^n, n→∞, will approach normal distribution.
Center still has highest probability, but less than 1/3.

P(Dice^5, sum=5) < 1/3 = 81/3^5

With x=100, we can numerically evaluate all coefficients of Dice^5 without overlap.

(04-29-2024 09:22 PM)Albert Chan Wrote:  >10101^5                     // (x^2+x+1)^5, for x=100
 1.05153045515E20

P(Dice^5, sum=5) = 51/3^5 ≈ 0.2099

Interestingly, this equals P(Dice^5, sum≥7) too.