It isn't a probability; the only use of it was to note the method leading to a 1/2 solution and where I consider it to fail, specifically because the number of times you are woken is not bound in [0,1] and thus "P(W)" as used in the 1/2 conditioning is malformed, as it doesn't keep track of when you're actually woken up. In as much as it is anything, using the 1/2 argumentation, "P(W)" is the expected number of wakings.

No. You will wake on Monday with probability one. But, on a randomly selected awakening, it is more likely that it's Monday&Heads than Monday&Tails, because you are on the Heads path on 50% of experiments

Sorry, but if we're randomly selecting a waking then it is not true that you're on the heads path 50% of the time. In a pair of runs, one head, one tail, you are woken 3 times, twice on the tails path.

On a randomly selected run of the experiment, there is a 1/2 chance of being in either branch, but: Choose a uniformly random waking in a uniformly chosen random run is not the same as Choose a uniformly random waking.