Anthropical Motte and Bailey in two versions of Sleeping Beauty

Ape in the coat

This is the first post in my series on Anthropics. The next one is Conservation of Expected Evidence and Random Sampling in Anthropics

Introduction

There is a curious Motte-and-Bailey dynamic going on with anthropic reasoning. The Motte is that it's just about a specific scoring rule and a corresponding way to define probability estimates. The Bailey is that we have psychic powers to blackmail reality by creating copies of our minds or using amnesia drugs. I'm going to explore these dynamics using two version of the famous Sleeping Beauty mind experiment: classical with the amnesia drug, and incubator, where either one or two Beauties are created. Simultaneously, I'll showcase the dissimilarities between them.

As far as I know people, who are Thirders in classical Sleeping Beauty, are also Thirders in the incubator version for mostly the same reasons. The SIA school of anthropics considers 1/3 to be the correct answer in both problems. And yet these problems are quite different. Adam Elga's proof for thirdism definitely doesn't work for the incubator version. Reasons for answering 1/2 are also different: Double Halfer in one and Halfer in the other.

Classic version of Sleeping Beauty

Let's start from the original version of the experiment

Here is a simple implementation in Python

def classic(heads_chance=0.5):
    days = ['Monday']
    if random.random() >= heads_chance: # result of the coin toss
        coin = 'Tails'
        days += ['Tuesday']
    else:
        coin = 'Heads'
    return days, coin

First, let's clear out what I mean by different scoring rule producing either 1/3 or 1/2.

coin_guess = []
for n in range(100000):
    days, coin = classic()
    for d in days:            
        coin_guess.append(coin == 'Heads')

coin_guess.count(True)/len(coin_guess) # 0.3322852689217815

coin_guess = {}
for i in range(100000):
    days, coin = classic()
    for d in days:            
        coin_guess[i] = (coin == 'Heads')
        
coin_guess = list(coin_guess.values())
coin_guess.count(True)/len(coin_guess) # 0.50167

That is, depending on whether we count per day or per experiment we get different answers, upon running the experiment multiple times. This is completely unsurprising.

Personally, I've always answered 1/2 in Sleeping Beauty and found the Thirders approach silly. Clearly, my credence for Heads should represent the probability of the coin actually landing Heads per toss, not per my guess! For example, in the Monty Hall problem my credence of winning on a door switch is 2/3 not because now I make two guesses instead of one, but because I actually win two out of three times this way in a repeated experiment. If my credence for a coin being Heads is 1/3 and in reality, if we do the same experiment a hundred times, in about fifty of them the coin is heads - then it's an obvious example of a map not representing the territory - which is quite understandable considering the usage of the amnesia drug.

Yet, even I can appreciate that the Thirders position points to something. If we specifically care about subjective experience, than indeed two thirds of them are going to be when the coin is Tails. As long as we do not claim that this fact gives an ability to predict the result of the coin toss better than chance, then we are just using different definitions, while agreeing on everything. We can translate from Thirder language to mine and back without any problem. Whatever betting schema is proposed, all other things being equal, we will agree to the same bets.

This is what I call the Anthropic Motte. The Bailey is the claim that the coin is actually more likely to be Tails when I participate in the experiment myself. That is, my awakening on Monday or Tuesday gives me evidence that lawfully update me to thinking that the coin landed Tails with 2/3 probability in this experiment, not just on average awakening.

This Bailey has originated from anthropic theories, talking about "updating on existence" such as SIA and SSA and will be more evident in the incubator version of the Sleeping Beauty. But first let's have a look at the Adam Elga's paper Self-locating belief and the Sleeping Beauty problem with a proof for Thirdism, which, I believe has also contributed to the Bailey and the confusion between it and the Motte.

On Adam Elga's proof for Thirdism

I find a lot of issues with it, following from what the author acknowledges himself:

Notice that the difference between your being in T1 [Tails and Monday] and your being in T2[Tails and Tuesday] is not a difference in which possible world is actual, but rather a difference in your temporal location within the world

Basically, you can't lawfully use probability theory in this setting assuming that Tails and Monday, Tails and Tuesday and Heads and Monday are three exhaustive and exclusive outcomes which probabilities are supposed to be added to 1. T1 are causally connected to T2, they happen to the same person, even if this person doesn't remember it.

According to Elga:

However

$P (T u e s d a y | T a i l s) = 1 / 2$

And so

$P (T a i l s) = 2 / 3$

Notice that it's just the probability of Tails, not probability of Tails conditional on awakening - the model Adam Elga uses doesn't account for such things at all, so all the justifications based on the updating on awakening are inappropriate here.

This, of course, directly contradicts the initial claim that the coin is fair, which is, among other things, used to prove that

$P (T a i l s | M o n d a y) = P (H e a d s | M o n d a y)$ = 1/2

Now, whether the statement above is true or not, is a big deal. And not only because it's crucial for Adam Elga's proof. Among those who give the answer 1/2 in the Sleeping Beauty experiment there are two distinct positions: Halfers who do not agree with this statement and Double Halfers who do. Thankfully this disagreement is easy to solve.

coin_guess = []
for n in range(100000):
    days, coin = classic()
    beauty_knows_monday = (days[0] == 'Monday')
    if beauty_knows_monday :
        coin_guess.append(coin == 'Heads')
        
print(coin_guess.count(True)/len(coin_guess))  # 0.49958

That is, if the Beauty knows that she is awake on Monday it doesn't give her any information whether the coin landed Heads or Tails. This is expected because Beauty was supposed to be awoken on Monday regardless of the coin toss result, moreover she could've been awaken on Monday even before the coin toss was made. If the credence for Heads in such situation was anything but 1/2 it would mean that participation in the experiment gave the Sleeping Beauty psychic powers to predict the result of the fair coin toss better than chance.

To conclude, in the classic Sleeping Beauty experiment, regardless of what you mean by credence that the coin is Heads, 50% of coin tosses are Heads, Sleeping Beauty doesn't have a better way to determine them than chance and knowing that the current day is Monday doesn't help either.

Incubator version of Sleeping Beauty

Now, let's look at the incubator version of Sleeping Beauty, where there is no sleeping drug or memory loss, but either one or two Beauties are created based on the result of the coin toss.

Here is the implementation of this scenario

def incubator(heads_chance=0.5):
    if random.random() >= heads_chance: # result of the coin toss
        coin = 'Tails'
        my_room = 1 if random.random() >= 0.5 else 2 # room sample
        other_room = 2 if my_room == 1 else 1
        return {'my_room': my_room, 'other_room': other_room}, coin
    else:
        coin = 'Heads'
        my_room = 1
        return {'my_room': my_room}, coin

This version of experiment evades the previous critique. Outcomes T1 and T2 are not causally connected, are randomly selected and do not happen to the same participant in different time.

However, Elga's proof won't work in this situation for a different reason.

Suppose someone left a definitive clue in Room 1 that it's indeed Room 1. This time it will be relevant new information as, finding yourself in Room 1 is twice more likely if the coin is Heads.

$P (H e a d s | R o o m 1) = 2 P (T a i l s | R o o m 1)$

A crucial element of Elga's proof isn't satisfied. In this case Halfers' reasoning is correct and Double Halfers' is wrong. We can see it on the repeated experiment

coin_guess = []
for n in range(100000):
    rooms, coin = incubator()
    beauty_knows_room1 = (rooms['my_room'] == 1)
    if beauty_knows_room1:
        coin_guess.append(coin == 'Heads')
 
 print(coin_guess.count(True)/len(coin_guess))  # 0.6688515435956072

So even if you believe that Adam Elga's proof for Thirdism is sound in the classical version of Sleeping Beauty, it's not applicable for the incubator version. It can't be the justification for answering 1/3 in both cases. Likewise the reasons to answer 1/2 are different in both experiments.

Anthropical Motte here is to claim that Beauties care for each other and invoke total utilitarianism. Being right on Tails means that two Beauties are right, while being right on Heads - that only one. Such attempt to construct the same scoring rule as with the classical version makes it completely clear that we are not talking about probability of coin being Heads or Tails anymore, but about the expected utility of the answer.

The Bailey is that you in particular are going to find yourself in the Tails world more likely if you were created during the experiment. Not that just on average most people created in such experiments are created when the coin landed tails. You in particular can guess that coin landed tails with 2/3 probability if you were part of the experiment due to the update on your existence. That the chance that there is another version of you in the different room is 2/3.

Final words on Anthropical Motte and Bailey

People do not usually hold both the Anthropical Motte and the Anthropical Bailey positions simultaneously. It's easy to see that if your scoring rule counted guesses every day/per every Beauty and it was 2/3 times likely that the coin landed Tails, then you should be not Thirder but Fifther.

coin_guess = []
for n in range(100000):
    days, coin = classic(heads_chance=0.33)
    for d in days:            
        coin_guess.append(coin == 'Heads')

coin_guess.count(True)/len(coin_guess) # 0.19632250462079567

Instead, people just assume that they are the same position, switching between them back and forth, often without noticing. This is understandable because usually they are! Usually if two thirds of your subjective experiences happen when the coin is Tails, Tails outcome of the coin toss are two times more likely. Usually if something is true on average for 2/3 of the population, it is 2/3 likely true for you as well. Sleeping Beauty cases are just specifically designed as adversarial attacks on this heuristic. Of course some people are confused.

Let's get rid of this confusion. Anthropical Motte and Bailey are two different things. And while the Motte is fine, the Bailey is responsible for, or at least, a serious symptom of much of absurdity of anthropic reasoning, which I'm going to resolve in the series of future posts.

The next post in the series is Conservation of Expected Evidence and Random Sampling in Anthropics.

This is the first post in my series on Anthropics. The next one is Conservation of Expected Evidence and Random Sampling in Anthropics

Introduction

Classic version of Sleeping Beauty

Let's start from the original version of the experiment

Here is a simple implementation in Python

def classic(heads_chance=0.5):
    days = ['Monday']
    if random.random() >= heads_chance: # result of the coin toss
        coin = 'Tails'
        days += ['Tuesday']
    else:
        coin = 'Heads'
    return days, coin

First, let's clear out what I mean by different scoring rule producing either 1/3 or 1/2.

coin_guess = []
for n in range(100000):
    days, coin = classic()
    for d in days:            
        coin_guess.append(coin == 'Heads')

coin_guess.count(True)/len(coin_guess) # 0.3322852689217815

coin_guess = {}
for i in range(100000):
    days, coin = classic()
    for d in days:            
        coin_guess[i] = (coin == 'Heads')
        
coin_guess = list(coin_guess.values())
coin_guess.count(True)/len(coin_guess) # 0.50167

That is, depending on whether we count per day or per experiment we get different answers, upon running the experiment multiple times. This is completely unsurprising.

On Adam Elga's proof for Thirdism

I find a lot of issues with it, following from what the author acknowledges himself:

Notice that the difference between your being in T1 [Tails and Monday] and your being in T2[Tails and Tuesday] is not a difference in which possible world is actual, but rather a difference in your temporal location within the world

According to Elga:

However

$P (T u e s d a y | T a i l s) = 1 / 2$

And so

$P (T a i l s) = 2 / 3$

This, of course, directly contradicts the initial claim that the coin is fair, which is, among other things, used to prove that

$P (T a i l s | M o n d a y) = P (H e a d s | M o n d a y)$ = 1/2

coin_guess = []
for n in range(100000):
    days, coin = classic()
    beauty_knows_monday = (days[0] == 'Monday')
    if beauty_knows_monday :
        coin_guess.append(coin == 'Heads')
        
print(coin_guess.count(True)/len(coin_guess))  # 0.49958

Incubator version of Sleeping Beauty

Now, let's look at the incubator version of Sleeping Beauty, where there is no sleeping drug or memory loss, but either one or two Beauties are created based on the result of the coin toss.

Here is the implementation of this scenario

def incubator(heads_chance=0.5):
    if random.random() >= heads_chance: # result of the coin toss
        coin = 'Tails'
        my_room = 1 if random.random() >= 0.5 else 2 # room sample
        other_room = 2 if my_room == 1 else 1
        return {'my_room': my_room, 'other_room': other_room}, coin
    else:
        coin = 'Heads'
        my_room = 1
        return {'my_room': my_room}, coin

This version of experiment evades the previous critique. Outcomes T1 and T2 are not causally connected, are randomly selected and do not happen to the same participant in different time.

However, Elga's proof won't work in this situation for a different reason.

Suppose someone left a definitive clue in Room 1 that it's indeed Room 1. This time it will be relevant new information as, finding yourself in Room 1 is twice more likely if the coin is Heads.

$P (H e a d s | R o o m 1) = 2 P (T a i l s | R o o m 1)$

A crucial element of Elga's proof isn't satisfied. In this case Halfers' reasoning is correct and Double Halfers' is wrong. We can see it on the repeated experiment

coin_guess = []
for n in range(100000):
    rooms, coin = incubator()
    beauty_knows_room1 = (rooms['my_room'] == 1)
    if beauty_knows_room1:
        coin_guess.append(coin == 'Heads')
 
 print(coin_guess.count(True)/len(coin_guess))  # 0.6688515435956072

Final words on Anthropical Motte and Bailey

coin_guess = []
for n in range(100000):
    days, coin = classic(heads_chance=0.33)
    for d in days:            
        coin_guess.append(coin == 'Heads')

coin_guess.count(True)/len(coin_guess) # 0.19632250462079567

The next post in the series is Conservation of Expected Evidence and Random Sampling in Anthropics.

If the Tuesday bet is considered to be "you can take the bet, but it will replace the one you may or may not have given on a previous day if their was one", then things line up to half again.

I can think of two interpretations of the setup you're describing here, but for both interpretations, Beauty does the right thing only if she thinks Heads has probability 1/3, not 1/2.

Note that depending on the context, a probability of 1/2 for something does not necessarily lead one to bet on it at 1:1 odds. For instance, if based on almost no knowledge of baseball, you were to assign probability 1/2 to the Yankees winning their next game, it would be imprudent to offer to bet anyone that they will win, at 1:1 odds. It's likely that the only people to take you up on this bet are the ones who have better knowledge, that leads them to think that the Yankees will not win. But even after having decided not to offer this bet, you should still think the probability of the Yankees winning is 1/2, assuming you have not obtained any new information.

I'll assume that Beauty is always given the option of betting on Heads at 1:1 odds. From an outside view, it is clear that whatever she does, her expected gain is zero, so we hope that that is also her conclusion when analyzing the situation after each awakening. Of course, by shifting the payoffs when betting on Heads a bit (eg, win $0.90, lose $1) we can make betting on Heads either clearly desirable or clearly undesirable, and we'd hope that Beauty correctly decides in such cases.

So... first interpretation: At each awakening, Beauty can bet on Heads at 1:1 odds (winning $1 or losing $1), or decide not to bet. If she decides to bet on Tuesday, any bet she may have made on Monday is cancelled, and if she decides not to bet on Tuesday, any bet she made on Monday is also cancelled. In other words, what she does on Monday is irrelevant if the coin landed Tails, since she will in that case be woken on Tuesday and whatever she decides then will replace whatever decision she made on Monday. (Of course, if the coin landed Heads, her decision on Monday does take effect, since she isn't woken on Tuesday.)

Beauty's computation on awakening of the expected gain from betting on Heads (winning $1 or losing $1) can be written as follows:

P(Heads&Monday)*1 + P(Tails&Monday)*0 + P(Tails&Tuesday)*(-1)

The 0 gain from betting on Monday when the coin landed Tails is because in that situation her decision is irrelevant. The expected gain from not betting is of course zero.

If Beauty thinks that P(Heads)=1/3, and hence P(Heads&Monday)=1/3, P(Tails&Monday)=1/3, and P(Tails&Tuesday)=1/3, and the formula above evaluates to zero, as we would hope. Furthermore, if the payoffs deviate from 1:1, Beauty will decide to bet or not in the correct way.

If Beauty instead thinks that P(Heads)=1/2, then P(Heads&Monday)=1/2. Trying to guess what a Halfer would think, I'll also assume that she thinks that P(Tails&Monday)=1/4 and P(Tails&Tuesday)=1/4. (Those who deny that these probabilities should sum to one, or maintain that the concept of probability is inapplicable to Beauty's situation, are beyond the bounds of rational discussion.) If we plug these into the formula above, we get that Beauty thinks her expected gain from betting on Heads is $0.25, which is contrary to the outside view that it is zero. Furthermore, if we shift the payoffs a little bit, so a win betting on Heads gives $0.90 and a loss betting on Heads still costs $1, then Beauty will compute the expected gain from betting on Heads to be $0.45 minus $0.25, which is $0.20$, so she's still enthusiastic about betting on Heads. But the outside view is that with these payoffs the expected gain is minus $0.05, so betting on Heads is a losing proposition.

Now... second interpretation: At each awakening, Beauty can bet or decide not to bet. If she decides to bet on Tuesday, any bet she may have made on Monday is cancelled, but if she decides not to bet on Tuesday, a bet she made on Monday is still in effect. In other words, she makes a bet if she decides to bet on either Monday or Tuesday, but if she decides to bet on both Monday and Tuesday, it only counts as one bet.

This one is more subtle. I answered essentially the same question in a comment on this blog post, so I'll just copy the relevant portion below, showing how a Thirder will do the right thing (with other than 1:1 payoffs). Note that "invest" means essentially the same as "bet":

- - -

First, to find the optimal strategy in this situation where investing when Heads yields -60 for Beauty, and investing when Tails yields +40 for Beauty, but only once if Beauty decides to invest on both wakenings, we need to consider random strategies in which with probability q, Beauty does not invest, and with probability (1-q) she does invest.

Beauty’s expected gain if she follows such a strategy is (1/2)40(1-q^2)-(1/2)60(1-q). Taking the derivative, we get -40q+30, and solving for this being zero gives q=3/4 as the optimal value. So Beauty should invest with probability 1/4, which works out to giving her an average return of (1/2)40(7/16)-(1/2)60(1/4)=5/4.

That’s all when working out the strategy beforehand. But what if Beauty re-thinks her strategy when she is woken? Will she change her mind, making this strategy inconsistent?

Well, she will think that with probability 1/3, the coin landed Heads, and the investment will lose 60, and with probability 2/3, the coin landed Tails, in which case the investment will gain 40, but only if she didn’t/doesn’t invest in her other wakening. If she follows the strategy worked out above, in her other wakening, she doesn’t invest with probability 3/4, and if Beauty thinks things over with that assumption, she will see her expected gain as (2/3)(3/4)40-(1/3)60=0. With a zero expected gain, Beauty would seem indifferent to investing or not investing, and so has no reason to depart from the randomized strategy worked out beforehand. We might hope that this reasoning would lead to a positive recommendation to randomize, not just acquiescence to doing that, but in Bayesian decision theory randomization is never required, so if there’s a problem here it’s with Bayesian decision theory, not with the Sleeping Beauty problem in particular.

- - -

On the other hand, if Beauty thinks the probability of Heads is 1/2, then she will think that investing has expected return of (1/2)(-60) + (1/2)40 or less (the 40 assumes she didn't/doesn't invest on the other awakening, if she did/does it would be 0). Since this is negative, she will change her mind and not invest, which is a mistake.