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I play many Blackganmmon games, and there are too many double rollings.
Last game stat:
5 doubles/19 rolls and 6 doubles/18 rolls.
Theoretically every 6th roll could be double, isn't it?

Thats statistics...

The probability for 6 doubles in 18 rolls is about 4 percent:
18!/(6!*12!) * (1/6)**6 * (5/6)**12 = 0.04462611752 = 4.46 percent

This means, if you repeat about 100 times the experiment 18 rolls of 2 dices you get about 4 times exactly 6 doubles (and about 1 time exactly 7 doubles out of 18 rolls, with a very small probabilty there could be even a higher number like 8 or 9 doubles).

5 doubles/19 rolls you get about every 10th game (19!/(5!*14!)*(1/6)**5*(5/6)**14 = 0.11646926278 = 11.6 percent).

formula for n rolls, k doubles: n!/(k!*(n-k)!)*(1/6)**n*(5/6)**(n-k)

Guszty wrote:I play many Blackganmmon games, and there are too many double rollings.
Last game stat:
5 doubles/19 rolls and 6 doubles/18 rolls.
Theoretically every 6th roll could be double, isn't it?

Theoretically every roll could be double.
Statistically every roll could be double.

Red Nebula wrote:

I love dilbert thanks for that

Red Nebula wrote:

And so true!