Originally posted by: alkemyst
Originally posted by: TuxDave
What is the probability then? We can work from there to figure out what you're really trying to say. You can assume "a series of rolls" = 99 rolls.
AGAIN FOR THE RETARDED KID: I will also assume we are talking the same 6 sided die,
The odds of getting a 6 in 99 rolls is: 1-(5/6)^99 or 99.999998551039183310360007916863%
Once again though, your math is not answering the question of:
What is the probability of rolling a 6 on the 99th roll, given that the 98 previous rolls were not sixes?
Are you familiar with conditional probabilities?
Once again, here is the step by step answer to the above question:
Let A = probability of getting a 6 on roll 100
Let B = probability of not getting a 6 on rolls 1-99
Now P(A|B) = P(AB)/P(B)
P(AB) = (5/6)^98*(1/6)^1
P(B) = (5/6)^98
Therefore, P(A|B) = 1/6
The math you just posted is not answering the same question as what you posted here:
"The odds of getting a 6 when one has not occured in a series of rolls is not 1/6. "
The odds of getting a 6 when one has not occured in a series of rolls IS 1/6, and I just proved it above for the 3rd time in this thread.
asking
"What is the probability of rolling at least 1 six in 99 rolls"
is not the same question as:
"What is the probability of rolling a 6 on the 99th roll, given that the previous 98 rolls were not sixes"
The first question can be solved using simple probability. The second is asking a different question, and is solved using conditional probabilities.
