It may be silly to continue this here, since I’m not sure anybody’s still reading, but at least I’m writing it down at all this way, so… here’s “Nick’s Sleeping Beauty can be Dutch Booked” (by Nick’s own rules)
In his Sleeping Beauty paper, Nick considers the ordinary version of the problem: Beauty is awakened on Monday. An hour later, she is told that it is Monday. Then she is given an amnesia drug and put to sleep. A coin is flipped. If the coin comes up tails, she is awakened again on Tuesday (and can’t tell the difference to Monday). Otherwise, she sleeps through to Wednesday.
Nick distinguishes five possible observer-moments: Beauty wakes up on Monday (h1 and t1, depending on heads/tails); Beauty is told that it’s Monday (h1m and t1m); Beauty wakes up on Tuesday (t2). Let P-(x) := P(x | h1 \/ t1 \/ t2), and P+(x) := P(x | h1m \/ t1m).
There are two possible worlds, heads-world (h1,h1m) and tails-world (t1,t1m,t2). Within each of the groups (h1,t1,t2) and (h1m,t1m), Nick assigns equal probabilities to each observer-moment in a given possible world. This gives:
In his paper, Nick considers the following Dutch book, suggested by a referee (I’m quoting from the paper):
Upon awakening, on both Monday and Tuesday, before either knows what day it is, the bookie offers Beauty the following bet: Beauty gets $10 if HEADS and MONDAY. Beauty pays $20 if TAILS and MONDAY. (If TUESDAY, then no money changes hands.)
On Monday, after both the bookie and Beauty have been informed that it is Monday, the bookie offers Beauty a further bet: Beauty gets $15 if TAILS. Beauty pays $15 if HEADS.
If Beauty accepts these bets, she will emerge $5 poorer.
Nick dismisses this argument because if the coin falls tails, Beauty will accept the first bet twice, once on Monday and once on Tuesday. Now, on Tuesday no money changes hands, so what’s the difference? Well, Nick thinks it’s very interesting that it could make a difference, but clearly it does, you see, because otherwise Sleeping Beauty could be Dutch booked if she accepts his probability assignments!
Instead of trying to argue that it makes no difference, let me just exhibit a variation where Beauty only accepts every bet at most once in every possible world.
Before Beauty is put to sleep, we throw a second fair coin, labelled A and B. If it comes up A, then on Monday, we tell Beauty, “It’s day A!” And if we wake her up on Tuesday, we tell her, “It’s day B!” If the coin comes up B, Monday is B, and Tuesday is A.
We now have doubled the number of worlds and observer-moments. The worlds are HA, HB, TA, and TB, each with probability 1⁄4; the observer-moments are ha1, ha1m; hb1, hb1m; ta1, ta1m, ta2; tb1, tb1m, tb2. P- and P+ are defined analogously to before, and again, we assign equal probability to each of the awakenings in every possible world (and make them sum to the probability of that world). This gives:
The sets of observer-moments that Beauty cannot distinguish are: {ha1,ta1,tb2}; {hb1,tb1,ta2}; {ha1m,ta1m}; {hb1m,tb1m}. (E.g., on {ha1,ta1,tb2}, Beauty just knows that she’s been awakened and that it’s “Day A.” In world B, Tuesday is Day A, thus tb2 is in this set.)
Note well that in none of these sets, there is more than one observer-moment from the same possible world. I exhibit the following variation of the above Dutch Book.
On {ha1,ta1,tb2}, the Bookie offers Beauty the first bet above: Beauty gets $10 if HEADS and MONDAY. Beauty pays $20 if TAILS and MONDAY. (If TUESDAY, then no money changes hands.)
On {ha1m,ta1m}, the Bookie offers the second bet: Beauty gets $15 if TAILS. Beauty pays $15 if HEADS.
Beauty now loses $5 if the day-label-coin comes up A, and breaks even if it comes up B. Every bet is accepted exactly once in every possible world in which it is offered at all. We could add symmetrical additional bets to make sure that Beauty also loses money in B worlds, but I think I’ve made my point. Nick can create his priors over observer-moments without violating the axioms of probability, but if it worries him if Beauty can be Dutch-booked in the way he discusses in his paper, I do believe he needs to be worried...
It may be silly to continue this here, since I’m not sure anybody’s still reading, but at least I’m writing it down at all this way, so… here’s “Nick’s Sleeping Beauty can be Dutch Booked” (by Nick’s own rules)
In his Sleeping Beauty paper, Nick considers the ordinary version of the problem: Beauty is awakened on Monday. An hour later, she is told that it is Monday. Then she is given an amnesia drug and put to sleep. A coin is flipped. If the coin comes up tails, she is awakened again on Tuesday (and can’t tell the difference to Monday). Otherwise, she sleeps through to Wednesday.
Nick distinguishes five possible observer-moments: Beauty wakes up on Monday (h1 and t1, depending on heads/tails); Beauty is told that it’s Monday (h1m and t1m); Beauty wakes up on Tuesday (t2). Let P-(x) := P(x | h1 \/ t1 \/ t2), and P+(x) := P(x | h1m \/ t1m).
There are two possible worlds, heads-world (h1,h1m) and tails-world (t1,t1m,t2). Within each of the groups (h1,t1,t2) and (h1m,t1m), Nick assigns equal probabilities to each observer-moment in a given possible world. This gives:
P-(h1) = 1⁄2; P-(t1) = 1⁄4; P-(t2) = 1⁄4
P+(h1m) = 1⁄2; P+(t1m) = 1⁄2
In his paper, Nick considers the following Dutch book, suggested by a referee (I’m quoting from the paper):
Nick dismisses this argument because if the coin falls tails, Beauty will accept the first bet twice, once on Monday and once on Tuesday. Now, on Tuesday no money changes hands, so what’s the difference? Well, Nick thinks it’s very interesting that it could make a difference, but clearly it does, you see, because otherwise Sleeping Beauty could be Dutch booked if she accepts his probability assignments!
Instead of trying to argue that it makes no difference, let me just exhibit a variation where Beauty only accepts every bet at most once in every possible world.
Before Beauty is put to sleep, we throw a second fair coin, labelled A and B. If it comes up A, then on Monday, we tell Beauty, “It’s day A!” And if we wake her up on Tuesday, we tell her, “It’s day B!” If the coin comes up B, Monday is B, and Tuesday is A.
We now have doubled the number of worlds and observer-moments. The worlds are HA, HB, TA, and TB, each with probability 1⁄4; the observer-moments are ha1, ha1m; hb1, hb1m; ta1, ta1m, ta2; tb1, tb1m, tb2. P- and P+ are defined analogously to before, and again, we assign equal probability to each of the awakenings in every possible world (and make them sum to the probability of that world). This gives:
P-(ha1) = P-(hb1) = 1⁄4
P-(ta1) = P-(ta2) = P-(tb1) = P-(tb2) = 1⁄8
P+(ha1m) = P+(hb1m) = P+(ta1m) = P+(tb1m) = 1⁄4
The sets of observer-moments that Beauty cannot distinguish are: {ha1,ta1,tb2}; {hb1,tb1,ta2}; {ha1m,ta1m}; {hb1m,tb1m}. (E.g., on {ha1,ta1,tb2}, Beauty just knows that she’s been awakened and that it’s “Day A.” In world B, Tuesday is Day A, thus tb2 is in this set.)
Note well that in none of these sets, there is more than one observer-moment from the same possible world. I exhibit the following variation of the above Dutch Book.
Beauty now loses $5 if the day-label-coin comes up A, and breaks even if it comes up B. Every bet is accepted exactly once in every possible world in which it is offered at all. We could add symmetrical additional bets to make sure that Beauty also loses money in B worlds, but I think I’ve made my point. Nick can create his priors over observer-moments without violating the axioms of probability, but if it worries him if Beauty can be Dutch-booked in the way he discusses in his paper, I do believe he needs to be worried...