2 dice roll probability formula for at least
What is the probability that one will have at least two “heartless” draws For example, in the above scenario of ten trials of some sort, calculating “at least 4” P(two rolls, no 6's) = P(“not 6” on dice #1 AND “not 6” on dice # 2).
It's obvious that the chances of a normal two -sided coin coming down heads, rather than tails, are exactly for each throw. Plainly the probability of rolling a six with a single six-sided dice (I never say 'die') is one event.
If you rolled x dice the probability of getting at least one 6 is 2 . In the case .. I'm interested in calculating the wait time to roll consecutive 2s using one die.