My math sucks. I can grok the odds (probability?) of rolling a certain number on one die, greater (less) than or equal to a certain number on two dice an even these conditions for consecutive rolls. (You multiply them, right?)
Now I am faced with a decision of whether to take two consecutive independent 4-1 shots or a single 8-1 shot. How do I calculate this one?
Thanks in advance
johnl
Let’s do a quick count, never do this during a game, too much time wasted. I like to call a quick “36 count”, we’ll double the 4-1 count because you get two.
Count each k/ as 1/2 and each kia as 1, then give them weight according to probability. VS one squad:
The 8-1 yields a score of 4 1/2 (call the 2kia a 1kia).
Roll a 2=1kia
Roll a 3= (1 Kia) x 2 (chances for a three)=2kia
Roll a 4= (k/) x3 (chances for a four even if they cower)= 1 1/2 Kia
Total= Weighted 4 1/2 KIAs score
Let’s do 2 4-1s
Roll 2 = 1 KIA (x2 for two rolls)= 2 KIAs
Roll 3 = k/ (x2 rolls)(x2 out of 36)=2 KIAs
Total= weighted 4 KIA score
Looks like 8-1 has the edge so far. I’ll run through the Moral check stuff later, kinda busy.
Situational, the 8–1 might be better against 2 or more units with high moral (eg. Japanese banzai, human wave or multiple squad berserk stack) getting attacked because the “weighted KIA” score raises to 5 1/2.