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.
Why are you using the notation P(W) when you mean E(W)? And if you can get an expectation for it, you must know the probability of it.
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.
Randomly selecting a waking does not imply a uniform distribution. On the contrary, we know the distribution is not uniform.
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.
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.
Why are you using the notation P(W) when you mean E(W)? And if you can get an expectation for it, you must know the probability of it.
Randomly selecting a waking does not imply a uniform distribution. On the contrary, we know the distribution is not uniform.