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 toHo.

(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 = 13.25chi-square value = 13.25

degrees-of-freedom = 5

Probability (One-Tailed): 0.0211that is, for the attacker, if

Hois 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 rejectHoand adoptHa.

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

histogram for entire sequence (7 rolls total)

6s:

5s:

4s: 4

3s: 3

2s: 222

1s: 11pearson chi-square test:

with 7 rolls, we expect 7/6=1.17 rolls in each bin according toHo.

(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 = 5.857chi-square value = 5.857

degrees-of-freedom = 5

Probability (One-Tailed): 0.3204that is, for the defender, if

Hois in fact true, the probability of getting results as or more extreme than these is 0.3204. this is insufficient evidence to rejectHo. therefore, we do not rejectHo.

Thanks, I am basing it as usually the attacker’s dice roll is submitted first than defender.