Clearly what we need to do first is to turn this into a Sleeping Beauty variant.
Let C = 6/pi^2, and suppose that we choose to wake Sleeping Beauty up k times with probability C/k^2. Then Sleeping Beauty is put in the awkward position that the expected number of times she wakes up is infinite. When asked “what is the probability that you were only woken up once?” the SSA, of course, suggests that Sleeping Beauty should answer C = 6/pi^2, while the SIA sort of gives up and maybe suggests 0 as a possible answer (If you object to 0 as a probability, recall that we’re dealing with an infinite sample space, which you should also refuse to believe in).
I argue that this is a legitimate answer to give. Why? Recall that in the standard Sleeping Beauty problem, there is a way to specifically elicit SIA-based probabilities for Sleeping Beauty. We ask “Would you rather receive $1 if the coin came up heads (and $0 otherwise), or $1 if the coin came up tails?” By answering “heads”, Sleeping Beauty earns $1 twice over the course of the experiment, for a total of $2; by answering tails, she earns $1. We can vary the payoffs to confirm that her best strategy is to act as though the probability the coin came up heads is 1⁄3.
Now let’s consider the infinite case. We ask “Would you rather receive a googolplex dollars if you are to be woken up only once, or $1 if you are to be woken up more than once?” This is actually a bit awkward because money has nonlinear utility; but it’s trivial to see that Sleeping Beauty maximizes her expected winnings by choosing the second option. So she is acting as though the probability she’s only woken up once is less than 1/googolplex, and similarly for any other large number. The only probability consistent with this is 0.
I’m to lazy to work out how, but it seems very easy to work out an infinite series of bets to give her such that she’ll get 0 with certainty no matter what number of times she is woken up if this is the case. And if the experiential really does never end there still never comes a time when she gets to enjoy the money.
Actually, you’re right about the infinite series of bets. Let N be the number of times Sleeping Beauty is to be woken up. Suppose (edit: on each day she wakes up) Sleeping Beauty is offered the following bets:
$10 if N=1, or $1 otherwise.
$10 if N=2, or $0.50 otherwise.
$10 if N=3, or $0.25 otherwise.
$10 if N=4, or $0.125 otherwise.
And so on.
In each individual bet, the second option has an infinite expectation, while the first has a finite expectation. However, if Sleeping Beauty accepts all the first options, she gets $10 every day she wakes up, for a total of $10N; if Sleeping Beauty accepts all the second options, she gets less than $2 every day she wakes up, for a total of $2N. Even though both options yield infinite expected money, this is still clearly inferior.
I suspect though, that this is a problem with the infinite nature of the experiment, not with Sleeping Beauty’s betting preferences.
That’s not what I meant. I meant… ugh I’m really tired right now and can’t think straight.
maybe:
Pot starts at 1$, each iteration she bets the pot against adding one dollar to it if N is greater than the number of iterations so far, with if needed the extra rule that if she gets woken up an infinite number of time she really gets infinite $.
To sleep deprived to check if the math actually works out like I think it does.
I don’t believe the first point, but I’m not entirely certain you’re wrong, so if you think you have such a construction, I’d like to see it.
As for your second point, the number of times that Sleeping Beauty wakes up is always finite, so no matter what, the experiment does end. It’s just that, due to the heavy tail of the distribution, the expected value is infinite (see also: St. Petersburg paradox). Of course, we would have to adjust rewards for inflation; also, the optimal strategy changes if the universe (or Sleeping Beauty) has a finite lifespan. So there’s a few implementation problems here, yes.
Clearly what we need to do first is to turn this into a Sleeping Beauty variant.
Let C = 6/pi^2, and suppose that we choose to wake Sleeping Beauty up k times with probability C/k^2. Then Sleeping Beauty is put in the awkward position that the expected number of times she wakes up is infinite. When asked “what is the probability that you were only woken up once?” the SSA, of course, suggests that Sleeping Beauty should answer C = 6/pi^2, while the SIA sort of gives up and maybe suggests 0 as a possible answer (If you object to 0 as a probability, recall that we’re dealing with an infinite sample space, which you should also refuse to believe in).
I argue that this is a legitimate answer to give. Why? Recall that in the standard Sleeping Beauty problem, there is a way to specifically elicit SIA-based probabilities for Sleeping Beauty. We ask “Would you rather receive $1 if the coin came up heads (and $0 otherwise), or $1 if the coin came up tails?” By answering “heads”, Sleeping Beauty earns $1 twice over the course of the experiment, for a total of $2; by answering tails, she earns $1. We can vary the payoffs to confirm that her best strategy is to act as though the probability the coin came up heads is 1⁄3.
Now let’s consider the infinite case. We ask “Would you rather receive a googolplex dollars if you are to be woken up only once, or $1 if you are to be woken up more than once?” This is actually a bit awkward because money has nonlinear utility; but it’s trivial to see that Sleeping Beauty maximizes her expected winnings by choosing the second option. So she is acting as though the probability she’s only woken up once is less than 1/googolplex, and similarly for any other large number. The only probability consistent with this is 0.
I’m to lazy to work out how, but it seems very easy to work out an infinite series of bets to give her such that she’ll get 0 with certainty no matter what number of times she is woken up if this is the case. And if the experiential really does never end there still never comes a time when she gets to enjoy the money.
Actually, you’re right about the infinite series of bets. Let N be the number of times Sleeping Beauty is to be woken up. Suppose (edit: on each day she wakes up) Sleeping Beauty is offered the following bets:
$10 if N=1, or $1 otherwise.
$10 if N=2, or $0.50 otherwise.
$10 if N=3, or $0.25 otherwise.
$10 if N=4, or $0.125 otherwise.
And so on.
In each individual bet, the second option has an infinite expectation, while the first has a finite expectation. However, if Sleeping Beauty accepts all the first options, she gets $10 every day she wakes up, for a total of $10N; if Sleeping Beauty accepts all the second options, she gets less than $2 every day she wakes up, for a total of $2N. Even though both options yield infinite expected money, this is still clearly inferior.
I suspect though, that this is a problem with the infinite nature of the experiment, not with Sleeping Beauty’s betting preferences.
That’s not what I meant. I meant… ugh I’m really tired right now and can’t think straight.
maybe:
Pot starts at 1$, each iteration she bets the pot against adding one dollar to it if N is greater than the number of iterations so far, with if needed the extra rule that if she gets woken up an infinite number of time she really gets infinite $.
To sleep deprived to check if the math actually works out like I think it does.
I don’t believe the first point, but I’m not entirely certain you’re wrong, so if you think you have such a construction, I’d like to see it.
As for your second point, the number of times that Sleeping Beauty wakes up is always finite, so no matter what, the experiment does end. It’s just that, due to the heavy tail of the distribution, the expected value is infinite (see also: St. Petersburg paradox). Of course, we would have to adjust rewards for inflation; also, the optimal strategy changes if the universe (or Sleeping Beauty) has a finite lifespan. So there’s a few implementation problems here, yes.