So, why is it ok for a simulation of an outcome with 1⁄2 probability to have 1⁄3 frequency?
There are only two outcomes and both of them have 1⁄2 probability and 1⁄2 frequency. The code saves awakenings in the list, not outcomes
People mistakenly assume that three awakenings mean three elementary outcomes. But as the simulation shows, there is order between awakenings and so they can’t be treated as individual outcomes. Tails&Monday and Tails&Tuesday awakenings are parts of the same outcome.
If this still doesn’t feel obvious, consider this. You have a list of Heads and Tails. And you need to distinguish between two hypothesis. Either the coin is unfair and P(Tails)=2/3, or the coin is fair but whenever it came Tails, the outcome was written twice in the list, while for Heads—only once. You check whether outcomes are randomly spread or pairs of Tails follow together. In the second case, even though the frequency of Tails in the list is twice as high as Heads, P(Tails)=P(Heads)=1/2.
There are only two outcomes and both of them have 1⁄2 probability and 1⁄2 frequency. The code saves awakenings in the list, not outcomes
People mistakenly assume that three awakenings mean three elementary outcomes. But as the simulation shows, there is order between awakenings and so they can’t be treated as individual outcomes. Tails&Monday and Tails&Tuesday awakenings are parts of the same outcome.
If this still doesn’t feel obvious, consider this. You have a list of Heads and Tails. And you need to distinguish between two hypothesis. Either the coin is unfair and P(Tails)=2/3, or the coin is fair but whenever it came Tails, the outcome was written twice in the list, while for Heads—only once. You check whether outcomes are randomly spread or pairs of Tails follow together. In the second case, even though the frequency of Tails in the list is twice as high as Heads, P(Tails)=P(Heads)=1/2.