Rolls: 6@1; Total Hits: 06@1: (2, 4, 2, 6, 3, 6)
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 distributedAttacker: 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 = 13.25chi-square value = 13.25
degrees-of-freedom = 5
Probability (One-Tailed): 0.0211that 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: 11pearson 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 = 5.857chi-square value = 5.857
degrees-of-freedom = 5
Probability (One-Tailed): 0.3204that 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.
Thanks, I am basing it as usually the attacker’s dice roll is submitted first than defender.
