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.

Continuity problem is that the 1/2 answer is independent of the ratio of expected number of wakings in the two branches of the experiment

Why is this a problem?

The next clause of the sentence is the problem

unless the ratio is 0 (or infinite) at which point special case logic is invoked to prevent the trivially absurd claim that credence of Heads is 1/2 when you are never woken under Heads.

The problem is special casing out the absurdity, and thus getting credences that are discontinuous in the ratio. On the other hand, you seem to take 1/21in PSB (ie you do let it depend on the ratio) but deviate from 1/21 when multiple runs of PSB aggregate, which is not what I had expected...

D was used in the comment I was replying to as an "event" that was studiously avoiding being W.

http://lesswrong.com/lw/28u/conditioning_on_observers/201l shows multiple ways I get the 1/3 solution; alternatively betting odds taken on awakening or the long run frequentist probability, they all cohere, and yield 1/3.

The problem as I see it with W is that it's not a set of outcomes, it's really a multiset. That's fine in it's way, but it gets confusing because it no longer bounds probabilities to [0,1]. Your approach is to quash multiple membership to get a set back.

No; P(H|W) = 1/21

Multiple ways to see this: 1) Under heads, I expect to be woken 1/10 of the time Under tails, I expect to be woken twice. Hence on the average for every waking after a head I am woken 20 times after a tail. Ergo 1/21.

2) Internally split the game into 2 single constant games, one for k1 and one for k2. We can simply play them sequentially (with the same die roll). When I am woken I do not know which of the two games I am playing. We both agree that in the single constant game P(H|W) = 1/21.

It's reasonably clear that playing two single constant games in series (with the same die roll and coin flip) reproduces the 2 constant game. The correleation between the roll and flip in the two games doesn't affect the expectations, and since you have complete uncertainty over which game you're in (c/o amnesia), the correlation of your current state with a state you have no information on is irrelevant.

P(H|W ∩ game i) = 1/21, so P(H|W) = 1/21, as the union over all i of (W ∩ game i) is W. At some level this is why I introduced PSB, it seems clearer that this should be the case when the number of wakings is bounded to 1.

3) Being woken implies either W1 or W2 (currently being woken for the first time or the second time) has occured. In general note that the expected count of something is a probability (and vice versa) if the number of times the event occurs is in {0,1} (trivial using the frequentist def of probability; under the credence view it's true for betting reasons).

P(W1 | H) = 1/10, P(W2 | H) = 0 P(W1 | T) = 1, P(W2 | T) = 1, from the experimental setup.

Hence P(H|W1) = 1/11, P(H|W2) = 0 You're woken in 11/20 of experiments for the first time and in 1/2 of experiments for the second, so P(W1| I am woken) = 11/21

P(H | I am woken ) = P(H ∩ W1 | I am woken ) + P(H ∩ W2 | I am woken ) = P(H | W1 ∩ I am woken).P(W1 | I am woken) + 0 = 1/11 . 11/21 = 1/21.

The issues you've raised with this is seem to be that you would either: Set P(W1 | I am woken) = 1 or Set P(W1 | T) = P(W2 | T) = 1/2 [ so P(H|W1) = 1/6 ], and set P(W1 | I am woken) = 6/11.

My problem with this is that if P(W1 | I am woken) =/= 11/21, you're poorly calibrated. Your position appears to be that this is because you're being "forced to make the bet twice in some circumstances but not others". Hence what you're doing is clipping the number of times a bet is made to {0,1}, at which point expectation counts of number of outcomes are probabilities of outcomes. I think such an approach is wrong, because the underlying problem is that the counts of event occurences conditional on H or T aren't constrained to be in {0,1} anymore. This is why I'm not concerned about the "probabilities" being over-unity. Indeed you'd expect them to be over-unity, because the long run number of wakings exceeds the long run number of experiments. In the limit you get well defined over unity probability, under the frequentist view. Betting odds aren't constrained in [0,1] either, so again you wouldn't expect credence to stay in [0,1]. It is bounded in [0,2] in SB or your experiment, because the maximum number of winning events in a branch is 2.

As I see it, the 1/21 answer (or 1/3 in SB) is the only plausible answer because it holds when we stack up multiple runs of the experiment in series or equivalently have uncertainty over which constant is being used in PSB. The 1/11 (equiv. 1/2) answer doesn't have this property, as is seen from 1/21 going to 1/11 from nothing but running two experiments of identical expected behaviour in series...

Continuity problem is that the 1/2 answer is independent of the ratio of expected number of wakings in the two branches of the experiment, unless the ratio is 0 (or infinite) at which point special case logic is invoked to prevent the trivially absurd claim that credence of Heads is 1/2 when you are never woken under Heads.

If you are put through multiple sub-experiments in series, or probabilistically through some element of a set of sub-experiments, then the Expected number of times you are woken is linearly dependent on the distribution of sub-experiments. The probability that you are woken ever is not.

So the problem is that it's not immediately clear what D should be. If D is split by total probability to be Heads or Tails, and the numbers worked separately in both cases, then to get 1/2 requires odd conditional probabilities, but 1/3 does not. If you don't split D, and calculate back from 1/3, you get 3/2 as the "probability" of D. It isn't. What's happened is closer to E(H|D) = E(D|H) E(H) / E(D), over one run of the experiment, and this yields 1/3 immediately.

The issue is that certain values of "D" occur multiple times in some branches, and allowing those scenarios to be double counted leads to oddness. I second the observation that caution is generally required.

You're woken with a big sign in front of you saying "the experiment is over now", or however else you wish to allow sleeping beauty to distinguish the experimental wakings from being allowed to go about her normal life.

Failing that, you are never woken; it shouldn't make any difference, as long as waking to leave is clearly distinguished from being woken for the experiment.

No. I assert P(H|W) = 1/21 in this case.

Two ways of seeing this: Either calculate the expected number of wakings conditional on the coin flip (m/20 and m for H and T). [As in SB]

Alternatively consider this as m copies of the single constant game, with uncertainty on each waking as to which one you're playing. All m single constant games are equally likely, and all have P(H|W) = 1/21. [The hoped for PSB intuition-pump]

Before I am woken up, my prior belief is that I spend 24 hours on Monday and 24 on Tuesday regardless of the coin flip. Hence before I condition on waking, my probabilities are 1/4 in each cell.

When I wake, one cell is driven to 0, and the is no information to distinguish the remaining 3. This is the point that the sleeping twins problem was intended to illuminate.

Given awakenings that I know to be on Monday, there are two histories with the same measure. They are equally likely. If I run the experiment and count the number of events Monday ∩ H and Monday ∩ T, I will get the same numbers (mod. epsilon errors). Your assertion that it's H/T with probability 0.5 is false given that you have woken. Hence sleeping twins.

As I see it, initially (as a prior, before considering that I've been woken up), both Heads and Tails are equally likely, and it is equally likely to be either day. Since I've been woken up, I know that it's not (Tuesday ∩ Heads), but I gain no further information.

Hence the 3 remaining probabilities are renormalised to 1/3.

Alternatively: I wake up; I know from the setup that I will be in this subjective state once under Heads and twice under Tails, and they are a priori equally likely. I have no data that can distinguish between the three states of identical subjective state, so my posterior is uniform over them.

If she knows it's Tuesday then it's Tails. If she knows it's Monday then she learns nothing of the coin flip. If she knows the flip was Tails then she is indifferent to Monday and Tuesday. 1/3 drops out as the only consistent answer at that point.

The reason it corresponds to Sleeping Beauty is that in the limit of a large number of trials, we can consider blocks of 20 trials where heads was the flip and all values of the die roll occurred, and similar blocks for tails, and have some epsilon proportion left over. (WLLN)

Each of those blocks corresponds to Sleeping Beauty under heads/tails.

No; between sedation and amnesia you know nothing but the fact that you've been woken up, and that 20 runs of this experiment are to be performed.

Why would an earlier independent trial have any impact on you or your credences, when you can neither remember it nor be influenced by it?