@gnasape:

Thanks bigchris, what if we assume that the dice rolls of the attacker and defenders are two separate events?

16 rolls for attacker, 7 rolls for defender.

sure, that’s fine. as long as you are separating the rolls based on some arbitrary labeling and not based on their values, you can test any group/sub-group you like.

**Ho**: the dice are uniformly distributed

**Ha**: the dice are not uniformly distributed

**Attacker**: 6,5,6,5,4,6,5,6,6,4,3,5,4,6,3,6

histogram for entire sequence (16 rolls total)

6s: 6666666

5s: 5555

4s: 444

3s: 33

2s:

1s:

pearson chi-square test:

with 16 rolls, we expect 16/6=2.67 rolls in each bin according to **Ho**.

(7 - 2.67)^2 / 2.67 +

(4 - 2.67)^2 / 2.67 +

(3 - 2.67)^2 / 2.67 +

(2 - 2.67)^2 / 2.67 +

(0 - 2.67)^2 / 2.67 +

(0 - 2.67)^2 / 2.67 = 13.25

chi-square value = 13.25

degrees-of-freedom = 5

Probability (One-Tailed): 0.0211

that is, for the attacker, if **Ho** is in fact true, the probability of getting results as or more extreme than these is 0.0211. These data are sufficient evidence (at the 5% level) to reject **Ho** and adopt **Ha**.

**Defender**: 2,2,1,3,4,2,1

histogram for entire sequence (7 rolls total)

6s:

5s:

4s: 4

3s: 3

2s: 222

1s: 11

pearson chi-square test:

with 7 rolls, we expect 7/6=1.17 rolls in each bin according to **Ho**.

(0 - 1.17)^2 / 1.17 +

(0 - 1.17)^2 / 1.17 +

(1 - 1.17)^2 / 1.17 +

(1 - 1.17)^2 / 1.17 +

(3 - 1.17)^2 / 1.17 +

(2 - 1.17)^2 / 1.17 = 5.857

chi-square value = 5.857

degrees-of-freedom = 5

Probability (One-Tailed): 0.3204

that is, for the defender, if **Ho** is in fact true, the probability of getting results as or more extreme than these is 0.3204. this is insufficient evidence to reject **Ho**. therefore, we do not reject **Ho**.