Points A, B and C lie on a circle and are arranged in clockwise order. We roll a fair 6-sided die and perform any of the 3 following operations based on the roll of the die.
- Move Clockwise - if 1 or 2 shows up.
- Move Anti-Clockwise - if 3 or 4 shows up.
- Do not move - if 5 or 6 show up.
We take A as our starting point. We roll the dice 8 times and perform the operations based on the dice roll. What is the probability that we end up getting back to A. This is how A, B & C lie on the circle
I was able to solve this question with some help.
I wanted to add some twists to the problem statement and came up with assigning different probabilities to the operations -
We pick and choose the operations at random instaad of the dice roll -
- There is a 2/6 chance of selecting the Clockwise Operation
- There is a 3/6 chance of selecting the Anti-Clockwise Operation
- There is a 1/6 chance of not moving at all (No Movement Operation)
Now, what is the probability that we end up getting back to A.
I was able to pinpoint all the valid states in the previously asked question.
As all the operations were equally likely before, we could use combinatorics to calculate all valid states and divide it by the total number of states to get the answer.
Here, I assigned probabilities to each of the valid states. For example -
P({2, 2, 4}) = $(2/6)^2 * (3/6)^2 * (1/6)^4$ - (as the events are independent) and then added all of them.
However, I strongly believe my approach is incorrect. If I use the same approach on the unmodified version of the problem, I would reach the wrong answer i.e $15 * (1/3)^8$.
How should one approach this problem ? I believe this is related to Markov Chains but still unsure on the procedure.