I did a spreadsheet where the probability of zero sucesses is 1 after zero trials, and each line is another trial, with the chance of staying the same is multiplied by 0.4 and the probability of adding one success is multiplied by 0.6. I get 88.1% for 8 or more successes after 11 trials.
Beating 4 monsters that each require two successes before 1 failure would be more like 8 successes and no failures, for an 8.96% chance. Unless there is some way to soak up failures?
![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
timeasmymeasure
no subject
Date: 2008-09-27 04:57 am (UTC)Beating 4 monsters that each require two successes before 1 failure would be more like 8 successes and no failures, for an 8.96% chance. Unless there is some way to soak up failures?