If the first sister’s experience is equivalent to the original Sleeping Beauty problem, then wouldn’t the second sister’s experience also have to be equivalent by the same logic? And, of course, the second sister will give 100% odds to it being Monday.
Suppose we run the sister experiment, but somehow suppress their memories of which sister they are. If they each reason that there’s a two-thirds chance that they’re the first sister, since their current experience is certain for her but only 50% likely for the second sister, then their odds of it being Monday are the same as in the thirder position- a one-third chance of the odds being 100%, plus a two-thirds chance of the odds being 50%.
If instead they reason that there’s a one-half chance that they’re the first sister, since they have no information to update on, then their odds of it being Monday should be one half of 100% plus one half of 50%, for 75%. Which is a really odd result.
Maybe I was a bit vague. I was trying to say that waking up SB’s twin sister on monday was a way of saying that SB’s would be equally aware of that as if her self would be awakened on monday under the conditions stipulated in the original experiment, i.e. zero recollection of the event. Or the other way around SB is awakened on monday but her twin siter on Tuesday. SB will not be aware of that here twin sister will be awakened on Tuesday. For that reason she is only awakened ONE time no matter if it is heads or tails. She will only experience ONE awakening per path. The is no cumulative effect of her being awakened 2 or a million times, every time is the “first” time and the “last” time”. If she is awake its equal chance that it is day 1 on the heads path as it would be day 56670395873966 (or any other day) on the tails path as far as she knows.
Or like this. Imagine that I flip a coin that I can see but you can not. I give you the rule that if it is heads I show you a picture of a dog. If it is tails, I show you the same picture of a dog but I might have shown this picture to thousands of people before you and maybe thousands of people after you, which you have no information about. You might be the first one to see it but you might also be the last one to see it or somewhere in the middle, i.e. you are not aware of the other observers. When I show you the picture of the dog, what chance do you give that the coin flip was heads?
But I am curious to know how a person with a thirder position argues in the case that she is awakened 999 or 8490584095805 times on the tails path, what probability should SB give heads in that case?
If the first sister’s experience is equivalent to the original Sleeping Beauty problem, then wouldn’t the second sister’s experience also have to be equivalent by the same logic? And, of course, the second sister will give 100% odds to it being Monday.
Suppose we run the sister experiment, but somehow suppress their memories of which sister they are. If they each reason that there’s a two-thirds chance that they’re the first sister, since their current experience is certain for her but only 50% likely for the second sister, then their odds of it being Monday are the same as in the thirder position- a one-third chance of the odds being 100%, plus a two-thirds chance of the odds being 50%.
If instead they reason that there’s a one-half chance that they’re the first sister, since they have no information to update on, then their odds of it being Monday should be one half of 100% plus one half of 50%, for 75%. Which is a really odd result.
Maybe I was a bit vague. I was trying to say that waking up SB’s twin sister on monday was a way of saying that SB’s would be equally aware of that as if her self would be awakened on monday under the conditions stipulated in the original experiment, i.e. zero recollection of the event. Or the other way around SB is awakened on monday but her twin siter on Tuesday. SB will not be aware of that here twin sister will be awakened on Tuesday. For that reason she is only awakened ONE time no matter if it is heads or tails. She will only experience ONE awakening per path. The is no cumulative effect of her being awakened 2 or a million times, every time is the “first” time and the “last” time”. If she is awake its equal chance that it is day 1 on the heads path as it would be day 56670395873966 (or any other day) on the tails path as far as she knows.
Or like this. Imagine that I flip a coin that I can see but you can not. I give you the rule that if it is heads I show you a picture of a dog. If it is tails, I show you the same picture of a dog but I might have shown this picture to thousands of people before you and maybe thousands of people after you, which you have no information about. You might be the first one to see it but you might also be the last one to see it or somewhere in the middle, i.e. you are not aware of the other observers. When I show you the picture of the dog, what chance do you give that the coin flip was heads?
But I am curious to know how a person with a thirder position argues in the case that she is awakened 999 or 8490584095805 times on the tails path, what probability should SB give heads in that case?