K-Bounded Sleeping Beauty Problem

UPD: As mentioned in the comments I’ve misrepresented thirders reasoning, due to a mathematical mistake. So this post isn’t valid. Shame on me! Will keep it till tomorrow as motivation to double check my reasoning in the future

Introduction

I’ve spent quite some time arguing that Thirdism in Sleeping Beauty is incorrect, appealing to theoretical properties of probability function and sample spaces. While it has successfully dissolved the confusion for some people, for many others it didn’t. So in this post I’m taking a different approach

I’ve used some betting arguments before, but even they turned out to be too convoluted. Thankfully, I’ve finally came up with the clearest case. A betting argument that demonstrates, that trying to apply Thirder reasoning, according to which one should assume that they are experiencing a random awakening among all possible awakenings—also known as Self Indexing Assumption—leads to bizarre behavior.

Generalizing Sleeping Beauty

Consider the problem that I’m calling K-Bounded Sleeping Beauty:

You are put to sleep. A fair coin is tossed either until it comes up Heads, or k Tails a row are produced. You receive $2^{n}$ undistinguishable awakenings, where n is the number of times the coin came up Tails. After each awakenings you put back to sleep and receive an amnesia drug which makes you completely forget that you had this awakening. You are awaken during such experiment. What is you credence that in this experiment the coin was tossed k times and the outcome of the k-th toss is Tails?

While the formulation may initially appear complex, it’s nothing more than a certain generalization of Sleeping Beauty.

Specifically, when k=1, K-Bounded Sleeping Beauty reduces to regular Sleeping Beauty problem, where the coin is tossed only once and there are either 1 or 2 awakenings.

As we know, for it Thirders claim that:

$P (T a i l s) = P (T a i l s & M o n d a y) + P (T a i l s & T u e s d a y) = 2 / 3$

Now let’s look at more interesting cases.

When k=2, the possible outcomes of the coin tosses are H, TH and TT, leading to either 1, or 2 or 4 awakenings correspondingly.

A Thirder, reasoning about such problem, would think that the sample space consists of 7 equiprobable outcomes, corresponding to the awakenings states:

$Ω = {H 1; T H 1; T H 2; T T 1; T T 2; T T 3; T T 4}$

$P (H 1) = P (T H 1) = P (T H 2) = P (T T 1) = P (T T 2) = P (T T 3) = P (T T 4) = 1 / 7$

Therefore:

$P (T T) = P (T T 1) + P (T T 2) + P (T T 3) + P (T T 4) = 4 / 7$

When k=3, the possible outcomes of the coin tosses are H, TH, TTH and TTT, leading to either 1, 2, 4 or 8 awakenings.

By the same logic as previously, a Thirders would think that:

$P (T T T T) = 8 / 15$

And so on.

In a general case, according to Thirder reasoning:

$P (T T T . . ._{k - t i m e s} . . . T) = 2^{k} / (2^{k + 1} - 1)$

Which, as we may notice, is more than ¹⁄₂ for every natural k.

$2^{k} / (2^{k + 1} - 1) > 1 / 2$

This means that for any value of k, a Thirders, participating in K-Bounded-Sleeping-Beauty would gladly accept a bet on every awakening with 1:1 odds that the k-th coin toss happened and is Tails.

Which is a terrible idea.

Demonstration

Here is the implementation of K-Bounded Sleeping Beauty in python:

def k_bounded_sleeping_beauty(k):
   coin = None
   coins = []
   num_tails = 0
   
   while coin != 'Heads' and num_tails < k:
       coin = 'Heads' if random() < 0.5 else 'Tails'
       coins.append(coin)
       if coin == 'Tails':
           num_tails = num_tails + 1
           
   num_awakenings = pow(2, num_tails)
   
   return coins, num_tails, num_awakenings

And here is iterated implementation of such betting for k=10

score = 0
k = 10
n = 10000
for i in range(n):
   coins, num_tails, num_awakenings = k_bounded_sleeping_beauty(k)
   
   if num_tails == k:
       score = score + num_awakenings
   else:
       score = score - num_awakenings
       
print(score/n) # -4.01031

As we can see, betting 1$ per awakening at 1:1 odds would, on average, results in 4$ lost per experiment.

The actual odds for this kind of bet are 1:5. They can be calculated using the correct Halfer model for Sleeping Beauty problem in a similar manner I do here.

In a general case:

$U (T T T . . ._{k - t i m e s} . . . T) = 1 / 2 \sum_{i = 1}^{k} U (! T T T . . ._{k - t i m e s} . . . T)$

Which corresponds to 2:k odds.

Unbounded Sleeping Beauty

Even more bizarre behavior happens when we remove the restriction on the number of Tails in the row.

You are put to sleep. A fair coin is tossed either until it comes up Heads. You receive $2^{n}$ undistinguishable awakenings, where n is the number of times the coin came up Tails. After each awakenings you put back to sleep and receive an amnesia drug which makes you completely forget that you had this awakening. You are awaken during such experiment. What is you credence that in this experiment the coin was tossed k times and the outcome of the k-th toss is Tails?

Under such conditions, a thirder would believe with certainty approaching 100% that, for any arbitrary large number, this amount of coin tosses has happened and is Tails, therefore predictably loosing all their money in a single probability experiment.

I was going to initially talk about this case, but it allows to much wiggle room for Thirders to rationalize the absurdity of their behavior, appealing to general weirdness of the infinities. Thankfully, K-Bounded Sleeping Beauty doesn’t present such an opportunity.

Conclusion

I believe, this presents a quite devastating case against Thirdism in Sleeping Beauty, in particular and Self Indexing Assumption, for cases of reasoning about problems involving amnesia, in general. This kind of reasoning simply doesn’t generalize, leading to clearly wrong credence estimates in similar problems.