You’ll need to describe that better. If you replace (implied by “instead”) step 1, you are never wakened. If you add “2.1 Put a ball into the box” and “2.2 Remove balls from the box. one by one, until there are no more” then there are never two balls in the box.
I mean that there are no sleeping or awakenings, instead there are balls in a box that follow the same logic:
Two coins are tossed, if both are Heads, nothing happens, otherwise a ball is put into a box. Then the second coin is placed the other side and once again, the ball is placed into the box unless both coins are Heads. Then you are randomly given a ball from the box.
Should you reason that there is another ball in a box with probability 2/3? After all, there are four equiprobable combinations: HH, TT, HT, TH. Since the ball, you were given, was put into the box, it couldn’t happen when the outcome was HH, so we are left with HT, TH and TT.
This variation of my two-coin is just converting my version of the problem Elga posed back into the one Elga solved. And if you leave out the amnesia step (you didn’t say), it is doing so incorrectly.
The entire point of the two-coin version was that it eliminated the obfuscating details that Elga added. So why put them back?
So please, before I address this attempt at diversion in more detail, address mine.
Do you think my version accurately implements the problem as posed?
Do you think my solution, yielding the unambiguous answer 1⁄3, is correct? If not, why not?
Your Two Coin Toss version is isomorphic to classical Sleeping Beauty problem with everything this entails.
The problem Elga solved in his paper isn’t actually Sleeping Beauty problem—more on it in my next post.
Likewise, the solution you propose to your Two Coin Toss problem is actually solving a different problem:
Two coins are tossed if the outcome is HH you are not awakened, on every other outcome you are awakened. You are awakened. What is the probability that the first coin came Heads?
Here your reasoning is correct. There are four equiprobable possible outcomes and awakening illiminates one of them. Person who participates in the experiment couldn’t be certain to experience an awakening and that’s why it is evidence in favor of Tails. 1⁄3 is unambiguously correct answer.
But in Two Coin Toss version of Sleeping Beauty this logic doesn’t apply. It would proove too much. And to see why it’s the case, you may investigate my example with balls being put in the box, instead of awakenings and memory erasure.
My problem setup is an exact implementation of the problem Elga asked. Elga’s adds some detail that does not affect the answer, but has created more than two decades of controversy.
You’ll need to describe that better. If you replace (implied by “instead”) step 1, you are never wakened. If you add “2.1 Put a ball into the box” and “2.2 Remove balls from the box. one by one, until there are no more” then there are never two balls in the box.
I mean that there are no sleeping or awakenings, instead there are balls in a box that follow the same logic:
Two coins are tossed, if both are Heads, nothing happens, otherwise a ball is put into a box. Then the second coin is placed the other side and once again, the ball is placed into the box unless both coins are Heads. Then you are randomly given a ball from the box.
Should you reason that there is another ball in a box with probability 2/3? After all, there are four equiprobable combinations: HH, TT, HT, TH. Since the ball, you were given, was put into the box, it couldn’t happen when the outcome was HH, so we are left with HT, TH and TT.
This variation of my two-coin is just converting my version of the problem Elga posed back into the one Elga solved. And if you leave out the amnesia step (you didn’t say), it is doing so incorrectly.
The entire point of the two-coin version was that it eliminated the obfuscating details that Elga added. So why put them back?
So please, before I address this attempt at diversion in more detail, address mine.
Do you think my version accurately implements the problem as posed?
Do you think my solution, yielding the unambiguous answer 1⁄3, is correct? If not, why not?
Your Two Coin Toss version is isomorphic to classical Sleeping Beauty problem with everything this entails.
The problem Elga solved in his paper isn’t actually Sleeping Beauty problem—more on it in my next post.
Likewise, the solution you propose to your Two Coin Toss problem is actually solving a different problem:
Here your reasoning is correct. There are four equiprobable possible outcomes and awakening illiminates one of them. Person who participates in the experiment couldn’t be certain to experience an awakening and that’s why it is evidence in favor of Tails. 1⁄3 is unambiguously correct answer.
But in Two Coin Toss version of Sleeping Beauty this logic doesn’t apply. It would proove too much. And to see why it’s the case, you may investigate my example with balls being put in the box, instead of awakenings and memory erasure.
My problem setup is an exact implementation of the problem Elga asked. Elga’s adds some detail that does not affect the answer, but has created more than two decades of controversy.
The answer of 1⁄3.