Hi, this is my first post so bear with me if my formatting is a little unorthodox. I recently was presented with this question by a friend (in relation to a board game, I believe), and I'm not entirely sure I have the correct answer.
Problem: Imagine that you are rolling an 8 sided die to determine an outcome. If you roll 7 or 8, the outcome succeeds. If you roll 1-6 the outcome fails, but you are allowed to re-roll once. That is to say, if you roll a 2 (for instance) you will then roll the 8 sided die again to give yourself another chance at success (whereas you would not roll again if you got a 7 or 8). The question is, taking the re-roll into account, what is the overall probability of success?
My understanding is that it works like this: In the initial roll, each possible outcome has a 1/8th probability, giving a total of 2/8ths probability of success and 6/8ths probability of failure. However the re-roll turns each of those 6 failures into 8 possibilities which each have an overall 1/64th chance of happening (1/8th of 1/8th), giving a 2/64 probability of success to each of the 6 initial failures. Overall that would give a 12/64ths probability of success and 36/64ths probability of failure if a 1-6 was rolled initially. Adding in the 2/8ths probability of success from the initial roll, the total probability is 28/64ths chance of success and 36/64ths chance of failure. Is this correct?
Hi Elictricblue,
You have it right, but there is a shorter route to the answer.
Consider the over-all probability of failure; in order to fail, you have to roll 1 through 6 twice. This happens with probability (6/8)^2. So the probability of success is one minus this, 1 - (6/8)^2.