It seems to me that you are working very hard to justify your solution. It’s a solution by argument/intuition. Why don’t you just do the math?
The experimenters fix 2 unique constants, k1,k2, each in {1,2,..,20}, sedate you, roll a D20 and flip a coin. If the coin comes up tails, they will wake you on days k1 and k2. If the coin comes up heads and the D20 that comes up is in {k1,k2}, they will wake you on day 1.
I just used Bayes rule. W is an awakening. We want to know P(H|W), because the question is about her subjective probability when (if) she is woken up.
To get P(H|W), we need the following:
P(W|H)=2/20 (if heads, wake up if D20 landed on k1 or k2)
P(H)=1/2 (fair coin)
P(W|T)=1 (if tails, woken up regardless of result of coin flip)
With your approach, you avoid directly applying Bayes’ theorem, and you argue that it’s ok for credence to be outside of [0,1]. This suggests to me that you are trying to derive a solution that matches your intuition. My suggestion is to let the math speak, and then to figure out why your intuition is wrong.
You and I both agree on Bayes implying 1⁄21 in the single constant case. Considering the 2 constant game as 2 single constant games in series, with uncertainty over which one (k1 and k2 the mutually exclusive “this is the k1/k2 game”)
This is the logic that to me drives PSB to SB and the 1⁄3 solution. I worked it through in SB by conditioning on the day (slightly different but not substantially).
I have had a realisation. You work directly with W, I work with subsets of W that can only occur at most once in each branch and apply total probability.
Formally, I think what is going on is this: (Working with simple SB) We have a sample space S = {H,T}
“You have been woken” is not an event, in the sense of being a set of experimental outcomes. “You will be woken at least once” is, but these are not the same thing.
“You will be woken at least once” is a nice straightforward event, in the sense of being a set of experimental outcomes {H,T}. “You have been woken” should be considered formally as the multiset {H,T,T}. Formally just working thorough with multisets wherever sets are used as events in probability theory, we recover all of the standard theorems (including Bayes) without issue.
What changes is that since P(S) = 1, and there are multisets X such that X contains S, P(X) > 1.
Hence P({H,T,T}) = 3⁄2; P({H}|{H,T,T}) = 1⁄3.
In the 2 constant PSB setup you suggest, we have S = {H,T} x {1,..,20}
W = {(H,k1),(H,k2), (T,1),(T,1),(T,2),(T,2),....,(T,20),(T,20)}
And P(H|W) = 1⁄21 without issue.
My statement is that this more accurately represents the experimental setup; when you wake, conditioned on all background information, you don’t know how many times you’ve been woken before, but this changes the conditional probabilities of H and T. If you merely use background knowledge of “You have been woken at least once”, and squash all of the events “You are woken for the nth time” into a single event by using union on the events, then you discard information.
This is closely related to my earlier (intuition) that the problem was something to do with linearity.
In sets, union and intersection are only linear when the working on some collection of atomic sets, but are generally linear in multisets. [eg. (A υ B) \ B ≠ A in general in sets]
Observe that the approach I take of splitting “events” down to disjoint things that occur at most once is precisely taking a multiset event apart into well behaved events and then applying probability theory.
What was concerning me is that the true claim that P({H,T}|T) = 1 seemed to discard pertinent information (ie the potential for waking on the second day). With W as the multiset {H,T,T}, P(W|T) = 2. You can regard this as expectation number of times you see Tails, or the extension of probability to multisets.
The difference in approach is that you have to put the double counting of waking given tails in as a boost to payoffs given Tails, which seems odd as from the point of view of you having just been woken you are being offered immediate take-it-or-leave-it odds. This is made clearer by looking at the twins scenario; each person is offered at most one bet.
Credence isn’t constrained to be in [0,1]???
It seems to me that you are working very hard to justify your solution. It’s a solution by argument/intuition. Why don’t you just do the math?
I just used Bayes rule. W is an awakening. We want to know P(H|W), because the question is about her subjective probability when (if) she is woken up.
To get P(H|W), we need the following:
P(W|H)=2/20 (if heads, wake up if D20 landed on k1 or k2)
P(H)=1/2 (fair coin)
P(W|T)=1 (if tails, woken up regardless of result of coin flip)
P(T)=1/2 (fair coin)
Using Bayes rule, we get:
P(H|W)=(2/20)(1/2) / [(2/20)(1/2)+(1)*(1/2)] = 1⁄11
With your approach, you avoid directly applying Bayes’ theorem, and you argue that it’s ok for credence to be outside of [0,1]. This suggests to me that you are trying to derive a solution that matches your intuition. My suggestion is to let the math speak, and then to figure out why your intuition is wrong.
You and I both agree on Bayes implying 1⁄21 in the single constant case. Considering the 2 constant game as 2 single constant games in series, with uncertainty over which one (k1 and k2 the mutually exclusive “this is the k1/k2 game”)
P(H | W) = P(H ∩ k1|W) + P(H ∩ k2|W) = P(H | k1 ∩ W)P(k1|W) + P(H|k2 ∩ W)P(k2|W) = 1⁄21 . 1⁄2 + 1⁄21 . 1⁄2 = 1⁄21
This is the logic that to me drives PSB to SB and the 1⁄3 solution. I worked it through in SB by conditioning on the day (slightly different but not substantially).
I have had a realisation. You work directly with W, I work with subsets of W that can only occur at most once in each branch and apply total probability.
Formally, I think what is going on is this: (Working with simple SB) We have a sample space S = {H,T}
“You have been woken” is not an event, in the sense of being a set of experimental outcomes. “You will be woken at least once” is, but these are not the same thing.
“You will be woken at least once” is a nice straightforward event, in the sense of being a set of experimental outcomes {H,T}. “You have been woken” should be considered formally as the multiset {H,T,T}. Formally just working thorough with multisets wherever sets are used as events in probability theory, we recover all of the standard theorems (including Bayes) without issue.
What changes is that since P(S) = 1, and there are multisets X such that X contains S, P(X) > 1.
Hence P({H,T,T}) = 3⁄2; P({H}|{H,T,T}) = 1⁄3.
In the 2 constant PSB setup you suggest, we have S = {H,T} x {1,..,20} W = {(H,k1),(H,k2), (T,1),(T,1),(T,2),(T,2),....,(T,20),(T,20)}
And P(H|W) = 1⁄21 without issue.
My statement is that this more accurately represents the experimental setup; when you wake, conditioned on all background information, you don’t know how many times you’ve been woken before, but this changes the conditional probabilities of H and T. If you merely use background knowledge of “You have been woken at least once”, and squash all of the events “You are woken for the nth time” into a single event by using union on the events, then you discard information.
This is closely related to my earlier (intuition) that the problem was something to do with linearity.
In sets, union and intersection are only linear when the working on some collection of atomic sets, but are generally linear in multisets. [eg. (A υ B) \ B ≠ A in general in sets]
Observe that the approach I take of splitting “events” down to disjoint things that occur at most once is precisely taking a multiset event apart into well behaved events and then applying probability theory.
What was concerning me is that the true claim that P({H,T}|T) = 1 seemed to discard pertinent information (ie the potential for waking on the second day). With W as the multiset {H,T,T}, P(W|T) = 2. You can regard this as expectation number of times you see Tails, or the extension of probability to multisets.
The difference in approach is that you have to put the double counting of waking given tails in as a boost to payoffs given Tails, which seems odd as from the point of view of you having just been woken you are being offered immediate take-it-or-leave-it odds. This is made clearer by looking at the twins scenario; each person is offered at most one bet.