## Friday, May 17, 2024

### Yet another argument for thirding in Sleeping Beauty?

Suppose that a fair coin has been flipped in my absence. If it’s heads, there is an independent 50% chance that I will be irresistably brainwashed tonight after I go to bed in a way that permanently forces my credence in heads to zero. If it’s tails, there will be no brainwashing. When I wake up tomorrow, there will be a foul taste in my mouth of the brainwashing drugs if and only if I’ve been brainwashed.

So, I wake up tomorrow, find no taste of drugs in my mouth, and I wonder what I should to my credence of heads. The obvious Bayesian approach would be to conditionalize on not being brainwashed, and lower my credence in heads to 1/3.

Next let’s evaluate epistemic policies in terms of a strictly proper scoring accuracy rule (T,F) (i.e., T(p) and F(q) are the epistemic utilities of having credence p when the hypothesis is in fact true or false respectively). Let’s say that the policy is to assign credence p upon observing that I wasn’t brainwashed. My expected epistemic utility is then (1/4)T(p) + (1/4)T(0) + (1/2)F(p). Given any strictly proper scoring rule, this is optimized at p = 1/3. So we get the same advice as before.

So far so good. Now consider a variant where instead of a 50% chance of being brainwashed, I am put in a coma for the rest of my life. I think it shouldn’t matter whether I am brainwashed or put in a coma. Either way, I am no longer an active Bayesian agent with respect to the relevant proposition (namely, whether the coin was heads). So if I find myself awake, I should assign 1/3 to heads.

Next consider a variant where instead of a coma, I’m just kept asleep for all of tomorrow. Thus, on heads, I have a 50% chance of waking up tomorrow, and on tails I am certain to wake up tomorrow. It shouldn’t make a difference whether we’re dealing with a life-long coma or a day of sleep. Again, if I find myself away, I should assign 1/3 to heads.

Now suppose that for the next 1000 days, each day on heads I have a 50% chance of waking up, and on tails I am certain to wake up, and after each day my memory of that day is wiped. Each day is the same as the one day in the previous experiment, so each day I am awake I should assign 1/3 to heads.

But by the Law of Large Numbers, this is basically an extended version of Sleeping Beauty: on heads I will wake up on approximately 500 days and on tails on 1000 days. So I should assign 1/3 to heads in Sleeping Beauty.

Maximillian said...

Hello Dr . Pruss . What do you think of this argument .
The new argument is this. If thirding is correct in Sleeping Beauty, then ‘zeroing’ is
correct in Immortal Beauty. After being informed that she is in an Immortal Beauty
experiment, Beauty should assign credence 0 to Heads (singular incarnation) and cre-
dence 1 to Tails (infinite reincarnation).
I’ll make this argument by going through four intermediate cases, successively
closing the gap between the two setups. Consider, first, this case:
Modified Sleeping Beauty. The protocol is as in Sleeping Beauty, except that if the coin comes
up Heads then Beauty is woken either on Monday or on Tuesday (never on both days). The
scientists have received prior instructions specifying on what day Beauty should be woken
on Heads (Beauty doesn’t know these instructions).
Next consider this:
Created Sleeping Beauty. After several medical breakthroughs, the scientists are able to create
adult humans in vitro. Beauty does not yet exist on Sunday. The scientists flip a coin on Sunday
night. If Heads, they create Beauty in vitro on either Monday or Tuesday. Which day she is
created is specified by previously received instructions. Upon creation, Beauty is initially
asleep; she is then woken and, some time after, put back to sleep until Wednesday. If Tails,
the scientists create Beauty on Monday, again initially asleep. She is woken and, some time
after, put back to sleep, and any memory that she has made on Monday is erased. The scientists
wake her again on Tuesday, before putting her back to sleep until Wednesday.
Next up: Reincarnated Beauty. Pleased with the team’s scientific progress, God decides to grant the
scientists some of Her divine powers: the scientists gain the ability to reincarnate the deceased.
They rush to test out the newfound powers by setting up a case like Created Sleeping Beauty,
with a few (morbid) changes. The Heads protocol remains unchanged. On Tails, however,
rather than putting Beauty back to sleep on Monday, they give her a deadly potion. The scien-
tists then create a perfect duplicate of Beauty’s body, as it was first created, and, with their new-
found powers, reincarnate Beauty into this body on Tuesday.
Next:
Multiply Reincarnated Beauty. The scientists are no longer satisfied with mere two-day exper-
iments. They decide to set up a Reincarnated-Beauty-like case, except that it is run over n days.
More specifically, if the coin comes up Heads, Beauty is incarnated and woken precisely once
during the next n days (according to previously received instructions). If the coin comes up
Tails, Beauty is incarnated, before being given the deadly potion, on each of m of the next n
days (again, according to previously received instructions).
The argument from thirding in Sleeping Beauty to zeroing in Immortal Beauty goes as
follows.
Immortal Thirders
(1) If thirding is correct in Sleeping Beauty, thirding is correct in Modified Sleeping Beauty.
(2) If thirding is correct in Modified Sleeping Beauty, thirding is correct in Created Sleep-
ing Beauty.
(3) If thirding is correct in Created Sleeping Beauty, thirding is correct in Reincarnated
Beauty.
(4) If thirding is correct in Reincarnated Beauty, 1/(m+1)-ing is correct in Multiply
Reincarnated Beauty.
(5) If 1/(m+1)-ing is correct in Multiply Reincarnated Beauty, zeroing is correct in
Immortal Beauty.
(C) If thirding is correct in Sleeping Beauty, zeroing is correct in Immortal Beauty.

Alexander R Pruss said...

This is very creative, but I don't see what we get out of all this complication. Why not just do a version of Sleeping Beauty where on heads there is one wake-up, on Monday, and on tails there are infinitely many wake-ups? Then zeroing (or infinitesimaling) follows?

There is a technical issue in the infinite cases (both yours and mine), namely that there is a countable infinite fair lottery on heads, and we know that countably infinite fair lotteries lead to paradoxes.