本文通过引入K-Bounded Sleeping Beauty问题，对Sleeping Beauty问题中三等分观点（Thirder）进行反驳。K-Bounded Sleeping Beauty问题是对原问题的推广，涉及多次抛硬币和多次唤醒，并假设每次唤醒后被试者会失去记忆。作者通过分析Thirder在该问题上的推理方式，发现其会导致荒谬的赌博行为，即无论k取何值，Thirder都会认为硬币抛掷k次且结果为尾的概率大于1/2，并欣然接受1:1赔率的赌注，最终导致亏损。文章通过Python代码模拟了赌博过程，证明了Thirder的推理错误，并对比了正确概率模型（Halfer）的结果，最终得出结论：Thirder的推理方式无法推广到类似问题，导致错误的概率估计。

🤔 **K-Bounded Sleeping Beauty问题:** 该问题是对经典Sleeping Beauty问题的推广，引入了硬币抛掷次数上限k，以及每次抛掷后唤醒次数与抛掷结果的关系，增加了问题的复杂度，但本质上仍是关于概率和条件概率的推理问题。

🎲 **Thirder观点的赌博策略:** 在K-Bounded Sleeping Beauty问题中，Thirder会认为硬币抛掷k次且结果为尾的概率大于1/2，因此会接受1:1赔率的赌注，认为自己能盈利。

💻 **Python模拟实验:** 作者利用Python代码模拟了K-Bounded Sleeping Beauty问题下的赌博过程，结果表明，Thirder的赌博策略会导致平均亏损，而非盈利，证明了Thirder的概率估计存在错误。

📊 **正确概率模型（Halfer）:** 文章指出，正确的概率模型（Halfer）给出的结果与Thirder的估计结果存在差异，Halfer模型下，赌注的赔率应为2:k，而非1:1。

🚫 **Thirder观点的局限性:** 通过K-Bounded Sleeping Beauty问题，作者证明了Thirder的推理方式无法推广到类似问题，其对概率的估计存在根本性错误，尤其是在涉及记忆缺失等情况时，Thirder的推理方式会导致荒谬的结果。

Published on November 26, 2024 5:53 PM GMT

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:

 $P\left(Tails\right)=P\left(Tails\&Monday\right)+P\left(Tails\&Tuesday\right)=2/3$

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:

$\Omega =\{H1;TH1;TH2;TT1;TT2;TT3;TT4\}$

$P\left(H1\right)=P\left(TH1\right)=P\left(TH2\right)=P\left(TT1\right)=P\left(TT2\right)=P\left(TT3\right)=P\left(TT4\right)=1/7$

Therefore:

$P\left(TT\right)=P\left(TT1\right)+P\left(TT2\right)+P\left(TT3\right)+P\left(TT4\right)=4/7$

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:

$P\left(TTTT\right)=8/15$

And so on.

In a general case, according to Thirder reasoning:

$P(TTT..{.}_{k-times}...T)=2^k/(2^{k+1}-1)$

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 = 0k = 10n = 10000for 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:

$U(TTT..{.}_{k-times}...T)=1/2{\sum }_{i=1}^{k}U(!TTT..{.}_{k-times}...T)$

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.

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