A little math lesson (sorry Xi ).
Starting with owning no techs, on average, it takes 21 rolls of dice to get the particular tech you want (eg heavy bomber). Here is how:
First the easy part. On average it takes 6 rolls to get a six (let’s call this a tech chance). Now on each tech chance you roll to see what tech you’re getting, so the probability of getting heavy bomber on the first tech chance (let’s call it P(1)) is 1/6. But the probability of not getting it is 5/6 but you get another tech (super sub?). So on the second tech chance you will get heavy bomber 1/5 of the time, given that you miss on the first tech chance. So the probability of getting heavy bomber on the second tech chance (and missing on the first tech chance) is P(2)=(5/6)(1/5)=1/6. Similarly the probability of getting heavy bomber on the third, forth, fifth and sixth tech chance (and missing on the previous tech chances) are:
So it turns out that you are as likely to get heavy bomber on your first tech chance as the sixth (and last) tech chance, similar to rolling a dice where you are equally likely to get any of the number 1 to 6. Therefore the average number of tech chances you need to get heavy bomber (or any one particular tech you want) is 3.5, much like the average number of IPC damage done on a SBR is 3.5. Now it takes an average of 6 rolls to get one tech chance so that 3.5 tech chance equals 21 rolls. Remember we are talking about averages here.
Phew! Now raise your hand if you know the averge number of dice roll to get heavy bomber starting with axis ad, or if you know what the heck I’m talking about. :roll:
This is first year university probability math. Don’t feel bad if you don’t get it. Just take my word for it. If there’re enough demand, I’ll post the solution to axis ad.