This is a follow-up question for the Sliding Bolt Puzzle. If you have not solved it yet, you might want to head there first, as the extended discussion of its solution in this question will contain spoilers.
In the original version of the Sliding Bolt Puzzle, I asked for the fastest sequence in terms of the maximum number of button presses it takes to definitely open the door, which was given in the accepted answer. However, I was wondering whether there might be a sequence which, while having more button presses, is still faster time-wise when you take into account the probability of the unknown initial position.
To explain a bit more what I mean: As Kendall Frey explained, there are four possible states of the bolts, numbered 1 to 4. (We will neglect state 1 as an initial position, because it is specified that the door is locked at first.) However, these four states are not equally probable as an initial position: The probabilities are
- 8/14 for state 2,
- 4/14 for state 3,
- 2/14 for state 4.
(obtained by simply listing all 16-2=14 possible initial configurations and categorizing them accordingly).
So we see that state 2 is by far the most probable one; the sequence CBCACBC will open it after four touches of a button at the earliest, while the most improbable state 4 will be opened directly.
Hence, my question is: Which is the sequence with the smallest expected value E[t], where t is the number of button pushes until the door opens, taking into account the probability distribution of the initial configuration?
Note that there is also some randomness regarding the next state when you press button A while in state 2 or button B while in state 3 (see the transition state matrix in the accepted answer), which should also be considered.