Perspective Reasoning’s Counter to The Doomsday Argument

Abstract: 

From a first person perspective, a self-aware observer can inherently identify herself from other individuals. However, from a third person perspective this identity through introspection does not apply. On the other hand, because an observer’s own existence is a prerequisite for her reasoning she would always conclude she exists from a first person perspective. This means observers have to take a third person perspective to meaningfully contemplate her chance of not coming into existence. Combining the above points suggests arguments which utilize identity through introspection and information about one’s chance of coming into existence fails by not keeping a consistent perspective. This helps explaining questions such as doomsday argument and sleeping beauty problem. Furthermore, it highlights the problem with anthropic reasonings such as self-sampling assumption and self-indication assumption.

 

Any observer capable of introspection is able to recognize herself as a separate entity from the rest of the world. Therefore a person can inherently identify herself from other people. However, due to the first-person nature of introspection it cannot be used to identify anybody else. This means from a third-person perspective each individual has to be identified by other means. For ordinary problems this difference between first- and third-person reasoning bears no significance so we can arbitrarily switch perspectives without affecting the conclusion. However this is not always the case.

One notable difference between the perspectives is about the possibility of not existing. Because one’s existence is a prerequisite for her thinking, from a first person perspective an observer would always conclude she exists (cogito ergo sum). It is impossible to imagine what your experiences would be like if you don’t exist because it is self-contradictory. Therefore to envisage scenarios which oneself does not come into existence an observer must take a third person perspective. Consequently any information about her chances of coming into existence is only relevant from a third-person perspective.

Now with the above points in mind let’s consider the following problem as a model for the doomsday argument (taken from Katja Grace’s Anthropic Reasoning in the Great Filter):

 

God’s Coin Toss

Suppose God tosses a fair coin. If it lands on heads, he creates 10 people, each in their own room. If it lands on tails he creates 1000 people each in their own room. The people cannot see or communicate with the other rooms. Now suppose you wake up in a room and was told of the setup. How should you reason the coin fell? Should your reason change if you discover that you are in one of the first ten rooms?

The correct answer to this question is still disputed to this day. One position is that upon waking up you have learned nothing. Therefore you can only be 50% sure the coin landed on heads. After learning you are one of the first ten persons you ought to update to 99% sure the coin landed on heads. Because you would certainly be one of the first ten person if the coin landed on heads and only have 1% chance if tails. This approach follows the self-sampling assumption (SSA).

This answer initially reasons from a first-person perspective. Since from a first-person perspective finding yourself exist is a guaranteed observation it offers no information. You can only say the coin landed with an even chance at awakening. The mistake happens when it updates the probability after learning you are one of the first ten persons. Belonging to a group which would always be created means your chance of existence is one. As discussed above this new information is only relevant to third-person reasoning. It cannot be used to update the probability from first-person perspective. From a first person perspective since you are in one of the first ten rooms and know nothing outside this room you have no evidence about the total number of people. This means you still have to reason the coin landed with even chances.

Another approach to the question is that you should be 99% sure that the coin landed on tails upon waking up, since you have a much higher chance of being created if more people were created. And once learning you are in one of the first ten rooms you should only be 50% sure that the coin landed on heads. This approach follows the self-indication assumption (SIA).

This answer treats your creation as new information, which implies your existence is not guaranteed but by chance. That means it is reasoning from a third-person perspective. However your own identity is not inherent from this perspective. Therefore it is incorrect to say a particular individual or “I” was created, it is only possible to say an unidentified individual or “someone” was created. Again after learning you are one of the first ten people it is only possible to say “someone” from the first ten rooms was created. Since neither of these are new information the probability of heads should remains at 50%.

It doesn’t matter if one choose to think from first- or third-person perspective, if done correctly the conclusions are the same: the probability of coin toss remains at 50% after waking up and after learning you are in one of the first ten rooms. This is summarized in Figure 1.

Figure 1. Summary of Perspective Reasonings for God’s Coin Toss

The two traditional views wrongfully used both inherent self-identification as well as information about chances of existence. This means they switched perspective somewhere while answering the question. For the self-sampling assumption (SSA) view, the switch happened upon learning you are one of the first ten people. For the self-indication assumption (SIA) view, the switch happened after your self-identification immediately following the wake up. Due to these changes of perspective both methods require to defining oneself from a third-person perspective. Since your identity is in fact undefined from third-person perspective, both assumptions had to make up a generic process. As a result SSA states an observer shall reason as if she is randomly selected among all existent observers while SIA states an observer shall reason as if she is randomly selected from all potential observers. These methods are arbitrary and unimaginative. Neither selections is real and even if one actually took place it seems incredibly egocentric to assume you would be the chosen one. However they are necessary compromises for the traditional views.

One related question worth mentioning is after waking up one might ask “what is the probability that I am one of the first ten people?”. As before the answer is still up to debate since SIA and SSA gives different numbers. However, base on perspective reasoning, this probability is actually undefined. In that question “I” – an inherently self identified observer, is defined from the first-person perspective, whereas “one of the first ten people” – a group based on people’s chance of existence is only relevant from the third-person perspective. Due to this switch of perspective in the question it is unanswerable. To make the question meaningful either change the group to something relevant from first-person perspective or change the individual to someone identifiable from third-person perspective. Traditional approaches such as SSA and SIA did the latter by defining “I” in the third person. As mentioned before, this definition is entirely arbitrary. Effectively SSA and SIA are trying to solve two different modified versions of the question. While both calculations are correct under their assumptions, none of them gives the answer to the original question.

A counter argument would be an observer can identify herself in third-person by using some details irrelevant to the coin toss. For example, after waking up in the room you might find you have brown eyes, the room is a bit cold, dust in the air has certain pattern etc. You can define yourself by these characteristics. Then it can be said, from a third-person perspective, it is more likely for a person with such characteristics to exist if they are more persons created. This approach is following full non-indexical conditioning (FNC), first formulated by Professor Radford M.Neal in 2006. In my opinion the most perspicuous use of the idea is by Michael Titelbaum’s technicolor beauty example. Using this example he argued for a third position in the sleeping beauty problem.Therefore I will provide my counter argument while discussing the sleeping beauty problem.

 

The Sleeping Beauty Problem

You are going to take part in the following experiment. A scientist is going to put you to sleep. During the experiment you are going to be briefly woke up either once or twice depending the result of a random coin toss. If the coin landed on heads you would be woken up once, if tails twice. After each awakening your memory of the awakening would be erased. Now supposed you are awakened in the experiment, how confident should you be that the coin landed on heads? How should you change your mind after learning this is the first awakening?

The sleeping beauty problem has been a vigorously debated topic since 2000 when Adam Elga brought it to attention. Following self-indication assumption (SIA) one camp thinks the probability of heads should be 1/3 at wake up and 1/2 after learning it is the first awakening. On the other hand supporters of self-sampling assumption (SSA) think the probability of heads should be 1/2 at wake up and 2/3 after learning it is the first awakening.

Astute readers might already see the parallel between sleeping beauty problem and God’s coin toss problem. Indeed the cause of debate is exactly the same. If we apply perspective reasoning we get the same result – your probability should be 1/2 after waking up and remain at 1/2 after learning it is the first awakening. In first-person perspective you can inherently identify the current awakening from the (possible) other but cannot contemplate what happens if this awakening doesn’t exist. Whereas from third-person perspective you can imagine what happens if you are not awake but cannot justifiably identify this awakening. Therefore no matter from which perspective you chose to reason, the results are the same, aka double halfers are correct.

However, Titelbaum (2008) used the technicolor beauty example arguing for a thirder’s position. Suppose there are two pieces of paper one blue the other red. Before your first awakening the researcher randomly choose one of them and stick it on the wall. You would be able to see the paper’s color when awoke. After you fall back to sleep he would switch the paper so if you wakes up again you would see the opposite color. Now suppose after waking up you saw a piece of blue paper on the wall. You shall reason “there exist a blue awakening” which is more likely to happen if the coin landed tails. A bayesian update base on this information would give us the probability of head to be 1/3. If after waking up you see a piece of red paper you would reach the same conclusion due to symmetry. Since it is absurd to purpose technicolor beauty is fundamentally different from sleeping beauty problem they must have the same answer, aka thirders are correct.

Technicolor beauty is effectively identifying your current awakening from a third-person perspective by using a piece of information irrelevant to the coin toss. I purpose the use of irrelevant information is only justified if it affects the learning of relevant information. In most cases this means the identification must be done before an observation is made. The color of the paper, or any details you experienced after waking up does not satisfy this requirement thus cannot be used. This is best illustrated by an example.

Imagine you are visiting an island with a strange custom. Every family writes their number of children on the door. All children stays at home after sunset. Furthermore only boys are allowed to answer the door after dark. One night you knock on the door of a family with two children . Suppose a boy answered. What is the probability that both children of the family are boys? After talking to the boy you learnt he was born a Thursday. Should you change the probability?

A family with two children is equally likely to have two boys, two girls, a boy and a girl or a girl and a boy. Seeing a boy eliminates the possibility of two girls. Therefore among the other cases both boys has a probability of 1/3. If you knock on the doors of 1000 families with two children about 750 would have a boy answering, out of which about 250 families would have two boys, consistent with the 1/3 answer. Applying the same logic as technicolor beauty, after talking to the boy you shall identify him specifically as “a boy born on Thursday” and reason “the family has a boy born on Thursday”. This statement is more likely to be true if both the children are boys. Without getting into the details of calculation, a bayesian update on this information would give the probability of two boys to be 13/27. Furthermore, it doesn’t matter which day he is actually born on. If the boy is born on, say, a Monday, we get the same answer by symmetry.

This reasoning is obviously wrong and answer should remain at 1/3. This can be checked by repeating the experiment by visiting many families with two children. Due to its length the calculations are omitted here. Interested readers are encouraged to check. 13/27 would be correct if the island’s custom is “only boys born on Thursday can answer the door”. In that case being born on a Thursday is a characteristic specified before your observation. It actually affects the chance of you learning relevant information about whether a boy exists. Only then you can justifiably identify whoever answering the door as “a boy born on Thursday”and reason “the family has a boy born on Thursday”. Since seeing the blue piece of paper happens after you waking up does not affect your chance of awakening it cannot be used to identify you in a third-person perspective. Just as being born on Thursday cannot be used to identify the boy in the initial case.

On a related note, for the same reason using irrelevant information to identify you in the third-person perspective is justified in conventional probability problems. Because the identification happens before observation and the information learned varies depends one which person is specified. That’s why in general we can arbitrarily switch perspectives without changing the answer.

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人称错乱导致的逻辑悖论(II)

在第一部分里已经阐述了我的观点:末日论证(doomsday argument)、睡美人问题(sleeping beauty problem)等悖论的产生原因在于人称错乱。在第一人称思维下,人无法想象自己不存在,存在概率的概念也就不适用。但在第一人称思维里,有自我意识的个体却能通过内省(introspection)来区分自己与其他人,而且这种区分是无需其他信息的。相反的,在第三人称思维里可以做自己不存在的假设,存在概率也有意义,但是却不能通过内省来区分自己和他人,因此任何区分都必须立足于个体之间的不同。如果一个论证即使用了第三人称特有的存在概率又使用了第一人称特有的自我区分那么它必然在思考过程中发生了人称错乱。这个论证就是无效的。末日论证(doomsday arugment)就是这样一个例子。

这里说一个相关的问题。假如你玩俄罗斯转盘,一把六发左轮手枪里放一颗子弹,那么你扣一次扳机被杀的概率是多少?因为这个实验的可能结果之一是自己不再存在(当然这儿不考虑天堂地狱来生等等),而第一人称下无法做这种假设,因此第一人称下这个问题是无意义的,也无法回答。也就是说这个问题也只在第三人称下才有解答(1/6)。如果你觉得这个结论有点不太让人信服的话不妨跳出贝叶概率来看看频率概率。这两种对概率的理解都有同样的人称问题。简单地说频率概率就是要大量重复试验(趋近于无限)然后以结果的相对频率作为概率。第一人称下你要大量重复这个实验找出相对频率是不可行的,因为你只能死一次所以一旦输掉一个实验后面的实验就都做不了。相反的,因为第三人称下的所有人都处于同等的逻辑位置,大量重复这个实验就可以找很多人来玩这个游戏,比如有600个人来玩的话,大概会有100人被杀。因此概率就是1/6。举这个例子主要是想说明一个实验结果和思考者的存在有相互冲突的问题只有在第三人称下才有意义。

说了这么多关于末日论证(doomsday argument)的,下面说说它的姊妹篇睡美人问题(sleeping beauty problem),这个问题可以概括如下:假设你参加了这样一个实验。你先睡去,然后一个科学家根据抛硬币的结果来决定叫醒你几次。如果是正面就叫醒一次,反面就两次。在两次叫醒之间科学家会抹除你第一次醒来的记忆。现在你在实验中醒来了,那么请问硬币落在正面的概率是多少?假设这时科学家告诉你这是你第一次被叫醒,那么此时正面的概率又是多少呢?

一派观点认为在刚刚醒来时正面的概率是1/2,因为你没得到任何新信息。而在被告知这是第一次被叫醒以后正面的概率就该相应的上升到2/3。另一派观点认为在醒来时正面的概率就应该是1/3,因为在背面的情况下你醒来的次数是正面的两倍。也就是说如果硬币落在背面你才更可能在这次被叫醒。而当知道这是第一次被叫醒以后正面的概率再上升到1/2

睡美人问题实际上就是第一部的上帝硬币问题。只是细节上它使用了删除记忆这样一个手段。删除记忆可以看做是一道信息壁垒,这道壁垒使你无法用内省来了解另外一次叫醒时发生的事情。换句话说,在第一人称下两次被叫醒的你其实是两个独立的个体,而被叫醒也意味着自己的意识存在。这样分析的话睡美人问题也就应该有和上帝掷硬币一样的答案。正确的方法是首先确定思考的人称。在第一人称下,在醒来时因为自己没有得到任何新信息,概率停留在1/2。得知这是第一次叫醒以后,因为自己对于是否会被再叫醒一次依然一无所知所以概率仍然停留在1/2。这里不能考虑我处于第一次叫醒而非第二次的条件概率,因为两次叫醒区别在于发生概率不同,即我的意识存在的概率不同,而存在概率在第一人称下是无意义的。你也可以以第三人称回答这个问题。首先第三人称下不能以内省来自我区分,因此也就不能用内省来区分这次叫醒。所以一开始的已知信息只是泛指的睡美人被叫醒了一次。这个已知信息在正面和背面都会出现所以概率不变停留在1/2。而被告知这是第一次叫醒以后已知信息就成为了第一次叫醒发生了,同样的这和正面反面都吻合,概率依然停留在1/2。和上帝掷硬币问题一样,我正处在第一次叫醒的概率是多少?这个问题因为本身发生了人称错乱所以是没有意义也没有答案的。

要说把人称错乱问题表现到极致的必然是尼克博斯滕(Nick Bostrom)的模拟假说(simulation argument)。他在2003年提出了这个观点。内容核心是如果人类的计算能力不断发展最终可以在电脑中虚拟出一个世界来研究他们的祖先,这个祖先模拟将会真实到其中的人也无法得知自己是虚拟的。紧接而来的问题就是我们如何得知自己现在不是生活在一个虚拟世界里呢?可以预见如果人类的计算能力能够做到虚拟世界,那么虚拟世界的数量一定远远大于仅有一个的现实世界数量,虚拟人也必定远远多过现实人。因此假设虚拟世界存在那么我们生活在唯一一个真实世界的概率就趋近于零。至此博斯滕得出了三者任一的结论:要么人类的计算能力永远达不到模拟祖先的水平,要么计算能力能达到但是因为某种原因没人去做类似的模拟,要么我们现在就活在虚拟世界里。这个论证过程的死结依然是人称错乱。我是虚拟人的概率或者我生活在虚拟世界的概率中的都是无需其他信息就不言自明的。这需要读者的自我意识用内省来理解。换言之,这是第一人称下定义的。而在模拟假说(simulation argument)中虚拟和现实的区别就在于存在概率不同。其中现实世界一定存在,而虚拟世界的存在是未知或随机的。所以两者只有在第三人称下才能区分。因此我是虚拟人的概率是一个人称错乱、没有答案的问题。接下来基于这个问题答案的论证自然也都是无效的。(其实接下来的论证里还有一点我反对的,但既然这里已经错了后面的就算全对也没意义,所以就不提了)

综上所述,人称的重要性不言而喻。人择原理(antropic principle)和我们自身存在相关的论证都必须要考虑思维人称的一致性,否则就会陷入诡辩和悖论的循环。出现类似末日论证(doomsday argument)、放肆的哲学家(presumptuous  philosopher )等类似问题。

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人称错乱导致的逻辑悖论

我发现目前比较有名的几个和人择原理(anthropic principle)相关的争议性话题都起源于人称错乱。这其中包括末日论证(doomsday argument),睡美人问题(sleeping beauty problem),模拟假说(simulation argument)等。这里简要的做一下笔记。

先说末日论证(doomsday argument)。它的理论很简单:假设你在所有人类中的出生序列是平均分布的,则根据你的出生序列可得知人类末日可能要更快到来。试想一下这个实验,上帝抛一个硬币。如果正面朝上的话他就创造10间屋子,每个屋子里再创造一个人。如果反面的话他就创造1000间屋子,每个屋子里创造一个人。假设你是被创造的人之一,在屋子里醒来以后上帝把实验过程完全都告诉你,现在要你判断正面的概率是多少。然后上帝再告诉你其实你在前10个屋子里,你再判断一下正面的概率是多少。答案是什么呢?

有一派人认为一开始正面的概率是1/2,因为你除了知道抛了一个硬币以外一无所知。而当得知自己在前10个屋子里以后正面的概率大大提高到接近99% 因为如果硬币落在正面则你必然在前10个房间内,而如果是背面你在前10间的概率只有百分之一。 根据贝叶斯定理的简单推导可得知正面的概率远大于背面。末日论证就是根据这个理论。只要考虑你在所有人类中出生的排序就可以推导出人类总数更可能比较少,也就是说目前尚未出生的人类比根据客观信息判断得出的要少,所以末日要比想象中更早到来。

这个结论看似非常无稽,但是目前为止都没有普遍接受的反驳,这里只介绍一个最多人推崇的反驳观点。这一派认为你一刚开始就应该认为正面的概率极小:接近1%。因为反面会创造1000个人,而正面只会创造出10个。反面的情况下你更可能被创造出来,所以你的存在就是有利于反面的证据。当被告知你在前10间里面以后正面的概率再上升到1/2。简而言之,你的存在和你的房间号两个证据正好互相抵消。也就是说根据自己出生排序的末日论证是错误的。然而这派理论也有很多问题。比如说来我们应该推断宇宙中充满了生命,因为这样的话你更可能存在。再比如在客观证据不能推断出两种理论对错的情况下,我们只需要考虑那种理论下和我们类似的生命体更多就可以下结论了(具体参见Nick Bostrom提出的放肆的哲学家Presumptuous Philosopher) 。总而言之争论还在继续。

我的观点是争论原因是思维中的人称混乱,具体如下。我们在思考问题的时候其实有两种不同的框架。第一种框架姑且称之为第一人称思维。在这个框架里承认我是自己思维的载体,我通过感知和经历获取信息,在头脑里做逻辑推导得出结论。第二种框架姑且称之为第三人称思维,这个框架下推导被看做是不取决于载体的抽象步骤,它假想一个理性的观察者根据已知信息会做何种判断。这两种框架在处理绝大多数问题时逻辑步骤是毫无分别的,因此我们平时并不会刻意区分它们。比如小明等10人参加一个实验,如果硬币落在正面则每个人获得1块钱,如果背面朝上则随机5个人获得1块钱。现在小明拿到了1块,那么硬币落在正面的概率是多少?这里小明可以以第一人称考虑:正面的情况下我更容易拿到钱,现在我拿到了钱,所以正面的概率更大是2/3。而他也可以以第三人称考虑:正面的情况下小明更容易拿到钱,而现在小明拿到了钱,所以正面的概率更大是2/3。两个框架下推导步骤是完全一样的。

但是两种人称并非完全没有差别。首先,因为第一人称思维将我作为思维载体,因此在这个框架下我必定存在,也就是笛卡尔说的我思故我在。在这个框架里也就不能做我不存在的假设。你可以尝试想象一下,如果自己不存在的话,你的感知和思想将会是什么样的。你很快就会发现这是自我矛盾因此无法想象的。而第三人称就没有这个问题,因为它不以为逻辑载体,无论你我他还是隔壁老王,所有人在第三人称思维里的位置都一样。想象自己不存在和想象其他人不存在并没有区别。你可以尝试一下想象自己不存在,必定发现是在以第三人称思考。基于这个区别,任何有关自己存在概率的信息也只能应用于第三人称思维。

两种人称的第二个区别就是关于自我识别。第一人称下的自我识别是固有的,无需其他信息。任何有自我意识的人都可以清楚的区分自己和他人。因为对其他人的了解靠感知,对自己的了解则靠内省。但因为内省只限于本人,第三人称下就不能使用。这也造成所有人在第三人称思维中的地位都一样,因此识别任何人都只能基于个体间的不同。假设有一对双胞胎,第三人称区分他们只能基于他们的不同,如果实在难以区分就人为制造一个。最简单的就是给他们起两个不同的名字。而对于第一人称的双胞胎本人来说,即使他们不知道彼此之间有何不同,甚至也可以没有名字,他们依然不会把另一个人错当成自己。渴了不会让另一个人喝水,累了也不会让另一个人睡觉。

根据以上两种人称思维的区别:1.存在概率仅限于第三人称,2,自我识别仅限于第一人称,我们再来看一下上帝抛硬币的问题。如果以第一人称来考虑的话,当在房间中醒来时我对硬币结果一无所知,而我发现自身的存在也是一种必然,没有新的信息,所以正面的概率保持在1/2。而得知我在前10个房间意味着我对于是否存在10个以的人仍然一无所知,因此正面的概率还是1/2。注意这里不能理解为我属于无论硬币结果如何都会被创造出来的10个人中的一个。这是有关于我存在概率的信息。这个信息只能应用于第三人称思维,在第一人称框架中无意义。而如果以第三人称来考虑这个问题,醒来的时候因为第三人称没有固有的自我识别,因此不能说某个特定的个体被创造出来了。已知信息只是有至少一人被创造出来。这是实验步骤一开始就确定的,无论硬币结果都是如此,因此没有任何新信息,正面的概率停留在1/2。而当被告知我属于前10个房间中的一人则已知信息是前10个房间里至少有一人被创造出来了。这还是无论硬币落在哪一面都会发生的事件,正面的概率还是1/2。综上所述无论是第一人称思维还是第三人称思维结论都是一致的,概率始终停留在1/2。这个问题不同于一般的地方在于两种人称思维的逻辑步骤并不相同,如果解题过程中发生人称错乱,即在第一、第三人称中随意转换就会得出错误答案。前面说过的两个派别就是犯了这个错误。即使用了第三人称下才有意义的存在概率,又使用了第一人称下才可能的自我识别。两者的区别只在于人称转换在何处发生而已。

这里提一句,有人可能纠结于这么一个问题:在实验中醒来以后,自问我身处在前10个房间里的概率是多少。这个问题本身就已经陷入人称错乱了,自然也就没有答案。问题中的利用了第一人称的固有识别。身处前10个房间的人就是存在概率100%的一组人,而存在概率又是第三人称特有的概念。因此不管使用那个人称都没法理解这个问题或是给与回答。如果想要回答这个问题就必须要用第三人称定义第一人称中固有识别出的。这个定义并不是唯一的,根据不同的定义方法答案也就不同。这也是造成很多争论的原因。有兴趣的可以搜索SSA self-sampling assumption)和SIAself-indication assumption)。而尼克博斯滕的模拟假说(simulation argument)就是在试图构建并回答一个这样的无异议的、人称错乱的问题。因此它得出一个反常的结论也就不足为奇了。

细心的读者可能会反问,在之前提到的小明拿钱的例子里在第三人称的已知信息为什么是特指的小明拿到了钱而不仅仅是泛指的某人拿到了钱呢?如果已知信息是泛指的某人拿到了钱这同样不包含任何新信息。硬币的概率岂不是还应该停留在1/2么。同样的,在上帝的硬币实验里在醒来后我也可以随机给自己去一个名字,比如叫二蛋,这样在第三人称里的已知信息就可以变成特指的二蛋被创造了出来而非泛指的有人被创造了出来。如此说来被创造的人越多,就越可能有人随机取名叫二蛋,因此二蛋的存在也就是硬币落在背面的证据了。这里涉及的核心问题就是什么特征可以用来特指某个个体,而什么特征不能。为了说明这个问题请参考以下z这个例子。

假设有这么一个村子,姑且称之为A村。已知村子里所有人家都有两个孩子。村里有个奇怪的习俗,天黑以后所有孩子必须回家,而且有人敲门的话必须是一个孩子去应门。这天晚上你随机敲了一家的门,听见屋里大人喊老大去开门。然后一个男孩打开了门,请问你这家两个孩子都是男孩的概率有多大?离这个村子不远还有一个B村。同样的所有人家都是两个孩子。这个村子的习俗和A村不同,应门的必须是男孩,家里没有男孩的话会无人应门。你又随机敲了一家的门,一个男孩开了门,经过简单的问话你得知这个孩子是家里的老大。请问这家两个孩子都是男孩的概率又有多大?

有的朋友可能认出来这个例子就是男孩女孩悖论(Boy or Girl Paradox)。你对A村和B村两户人家的了解都可以概括为家里的老大是个男孩,然而他们有两个男孩的概率却是不一样的。A村人家的概率是1/2,B村那户人家的概率是1/3。(不相信的可以假设每个村子都有1200户,按照两个步骤演算一下,过程不复杂但是比较繁琐这里就不赘述了)在这个问题里A村的已知信息是特指的家里的老大是个男孩B村的已知信息只是泛指的家里的某个孩子是男孩、或者说家里有个男孩。因为老大这个特征虽然和孩子的性别没有相关性,但是在A村的例子里却先用这个特征指定了一个孩子,它却决定了已知信息里孩子的性别是什么。显然如果当时家长让老二去开门你看到的就未必是一个男孩了。而在B村的例子里,因为习俗的原因已知信息必然是有关于男孩的(家里有或者没有)。只要有人开门,老大这个特征对已知信息就没有任何影响。试想一下如果那个孩子告诉你自己是老二,逻辑上这和他是老大是完全对称的,丝毫不影响概率计算。因此也就不能用老大来特指这个孩子。

再回到小明和二蛋这两个例子。在小明的例子里如果已知信息不是关于他,而是10人中的其他小朋友,比如小李小王等,因为他们未必和小明一样拿到了钱,那么已知信息就会不同。因此特指是正确的。而二蛋只是在被创造出来之后的随机命名,不管是起名叫二蛋还是铁牛、栓柱,都不会改变有某人被创造出来的已知信息,也就是说叫不叫二蛋都没有影响。因此用这个特征特指也自然就是错误的了。同样的,其他特征,比如身高体重或醒来以后的经历等等,都不能用来特指某个个体。已知信息只能是泛指的有一个人被创造出来了,因此概率应该始终停留在1/2

先写到这儿。睡美人问题(sleeping beauty problem)和模拟假说(simulation argument)有时间再说。

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Perspective Importance and Solution to the Sleeping Beauty Problem

This is a simplified argument for perspective disagreement in the sleeping beauty problem. The purpose is to present the idea in a shorter, more structured way than my first post.

Perspective Argument:

Considering the following experiment equivalent to the sleeping beauty problem:

(If you are skeptical about this equivalency please check my first post. It has more detailed explanation why my argument does not dependent on the equivalency)

Duplicating Beauty:

Beauty falls asleep as usual. The experimenter tosses a fair coin before she wakes up. If the coin landed on T then a perfect copy of beauty will be produced. The copy is precise enough that she cannot tell if herself is the clone or the original. If the coin landed on H then no copy will be made . The beauty(ies) will then be randomly put into two identical rooms. At this point another person, let’s call him the selector, randomly chooses one of the two rooms and enters. Suppose he saw a beauty in the chosen room. What should the credence for H be for the two of them? 

For the Selector his probability is easy to calculate. Because he is twice more likely to see a beauty in the room if T, simple bayesian updating gives us his probability for H as 1/3.

For Beauty, her room has the same chance of being chosen regardless if the coin landed on H or T. Therefore seeing the Selector gives her no new information about the coin toss. So her answer should be the same as in the original sleeping beauty problem: if she is a halfer 1/2, if she is a thirder 1/3. 

This means the two of them would give different answers according to halfers and would give the same answer according to thirders. Notice here the Selector and Beauty can freely communicate however they want and it won’t change their answer since they have the same information regarding the coin toss. So halving would give rise to a perspective disagreement.

 

This perspective disagreement is quite unusual (and against Aumann’s Agreement Theorem), so it could be used as an evidence against halving thus supporting Thirdrism and SIA. I would argue why SIA has its own form of perspective disagreement in a separate post. For now I want to argue that this disagreement is logically sound. 

First of all, this perspective disagreement from bayesian reasoning is also mirrored by frequentist interpretation of probability. Let’s take a frequentist’s approach and see what happens if the experiment is repeated, say 1000 times. For the Selector, this simply means someone else go through the coin toss 1000 times and let him chooses a random room after each time. On average there would be 500 H and T. He would see a beauty for all 500 times after T and see a beauty only 250 times after H. Meaning out of all the instances where he sees a beauty in the chosen room, 1/3 of which would be after H. Therefore he is correct in giving 1/3 as his answer.

From beauty’s perspective repeating the experiment simply means she goes through the experiment and wake up in a random room awaiting the Selector’s choice again.  So by her count, taking part in 1000 repetitions means she would recall 1000 coin tosses after waking up.  In those 1000 coin tosses there should be about 500 of H and T each. She would see the Selector about 500 times with equal numbers after T or H. Therefore her answer of 1/2 is also supported by long run frequency.

 

I think this is the easiest way to see the relative frequency from beauty’s perspective: Ignore the clones first, if someone experienced 1000 coin toss she would expect to see 500 Heads. Now consider how the cloning would affect her conclusion. If there is a T somewhere along the tosses thus a clone is created, both the original and the clone would have remembered exactly the same tosses happened earlier. They would also have the same expectation about coin tosses after they split. The relative frequency would be the same for both the original and the clone. Therefore  she does not have to consider whether she is physically a clone or the original. After beauty wakes up remembering a large number of tosses she shall expect about half of those landed in Heads.

Alternatively, a more complicated way would be to think about all the coin tosses happened and her own position. For example a beauty experiencing two coin tosses there are several ways this can happen from a third party’s perspective:

1. H-H: Both coin tossing landed on H. There is a 1/4 chance for this happening. Beauty experience 2 Hs.

2. H-T: First coin landed on H and second coin landed on T. There is a 1/4 chance for this happening and there would be 2 beauties at the end of the experiment. It doesn’t matter if she is any one of the two she experienced 1H and 1T.

3. T-(H,H): First toss landed on T and both resulting beauties experience H at the second toss. There is 1/8 chance for this to happen, doesn’t matter if she is any one of the two beauties at the end, she experienced 1 H and 1T.

4. T-(H,T): First coin landed on T and a H and T happened at the second round of tosses respectively. There is 1/4 chance for this to happen. Depending which coin toss she experienced at the second round, half the time (1/8) she experienced 1H and 1T, while the other half (1/8) she experienced 2 Ts. 

5. T-(T,T): All tosses landed on T. There is 1/8 chance, doesn’t matter which second toss is hers she experienced 2Ts. 

The expected number of H from beauty’s perspective: (1/4)x2+(1/4)x1+(1/8)x1+(1/8)x1=1: half the number of tosses. Obviously this calculation can be generalized to other number of tosses. 

Both the above methods correctly gives the relative frequency from beauty’s perspective. The most common mistake is for a beauty to think she is equally likely to be any one of the beauties at the end of multiple coin tosses. This seemingly intuitive conclusion is valid from a third party’s (the selector’s) perspective, aka if he randomly chooses one from all the ending beauties then all beauties are in symmetrical position. However from beauty’s first person perspective the ending beauties are not in symmetrical position especially since they have experienced different coin toss results.

If we call the creation of a new beauty a “branch off”, here we see that from Selector’s perspective experiments from all branches are considered a repetition. Where as from Beauty’s perspective only experiment from her own branch is counted as a repetition. This difference leads to the disagreement.

This disagreement can also be demonstrated by the difference in betting odds. Imagine the selector and the beauty(ies) enters a bet about the coin toss result whenever they meet. The payoff enabling them to break even in the long run would be different. For the selector, choosing any of the two rooms after T leads to the same observation: he always sees a beauty and enters another bet. However, for the two beauties the selector’s choice leads to different observations: either she can see him and enters another bet or not. Effectively the selector is twice more likely to enter a bet than a beauty if the coin landed on tails. If they bet 1 dollar on H, the longterm breakeven payoff would be 3 dollars for the selector and 2 dollars for beauty respectively. (I will discuss the use of bets as arguments for different positions in the sleeping beauty problem in a separate post in more detail.)

It should be noted by choosing a random room the selector is essentially choosing among all possible observers as stated by SIA. We can also construct the experiment so that the selector is choosing among all actual observers as stated by SSA. For example a third party who knows the content of the rooms can randomly label the two rooms 1 & 2 in his mind. The selector can pick a number as he wish. The third party would then place him into the corresponding room if it is occupied, and place him into the other room if it is not. This way the selector would always see a beauty in the room. Therefore when he sees a beauty his probability would remain at 1/2 as there is no new information. However from beauty’s perspective she is guaranteed to meet the selector in H and only has half the chance in case of T. Therefore seeing the selector is evidence confirming H from her perspective. Assuming she is a halfer, her probability of H would become 2/3. The two of them would still be in disagreement as in the previous case. In essence the two parties interpret the meeting differently which leads to the disagreement. The selector interpret it as a meeting between him and A beauty, any beauty. Whereas beauty interpret it as the meeting between the selector and THE beauty, herself. 

Another important question would be what should beauty’s probability be once learning she is actually the original. If we continue to apply the perspective reasoning then her probability would remain unchanged at 1/2. This can be shown in two ways.

One way is to put the new information into the frequentist approach mentioned above. In Duplicating Beauties, when a beauty wakes up and remembering 1000 repetitions she shall reason there are about 500 of H and T each among those 1000 tosses. The same conclusion would be reached by all beauties without knowing if she is physically the original or created somewhere along the way. Now suppose a beauty learns she is the original. She could ignore the cloning part of the experiment entirely and still conclude those 1000 tosses contain about 500 of H and T each. This means her probability of H would remains unchanged at 1/2.

Another way to see why beauty should not change her probability is to see the agreement/disagreement pattern between her and the selector. It is worth noting that beauty and the selector will be in agreement once knowing she is the original. As discussed above one way to understand the disagreement is they interpret the meeting differently. The selector interpret it as a meeting with ANY beauty whereas beauty interpret it as a meeting with a specific beauty aka herself. However once we distinguish the beauty(ies) by stating which one is the original (and possibly which one is the clone) the selector and the beauty would have to have the same interpretation, that the meeting is between the selector and a specific beauty. We can also use bets mentioned above to demonstrate this. After T, seeing either beauty is the same observation for the selector while it  is different observations for beauties. This in turn causes the selector to enter twice more bets than beauty, leading to the difference in their betting odds. However if a bet is only set between the selector and the original beauty then the selector would no longer be more likely to enter a bet. He and the Beauty would enter the bet with equal chances. Meaning their betting odds ought to be the same, aka they must be in agreement regarding the probability of H. 

Now consider the duplicating beauty problem with the selector choosing a random room. Once the selector saw a beauty in the room his probability of H changes to 1/3 while beauty’s probably remains at 1/2. At this time they are in disagreement. However, if the experimenter tells them the beauty in the room is the original then selector’s probability of H would increase to 1/2 again since he has equal chance of choosing the room with original beauty regardless of coin toss result. Because they ought to be in agreement beauty’s probability should remain unchanged at 1/2.

The same result can be shown if the selector is choosing among existing beauties as SSA suggested using the procedure mentioned previously. As shown above once they meet the beauty’s probability of H rises to 2/3 while the selector’s probability remains at 1/2. If the experimenter tells them the beauty in the room is the original then selector’s would adjust his answer. Since he would always meet with the original beauty if H and meet her half the time if T his probability of H would rises to 2/3. Because they ought to be in agreement beauty’s probability must remain unchanged at 2/3.

The above methods shows beauty should not update her answer after learning she is the original. Beauty’s antibayesianism, just like the perspective disagreement, is quite unusual. This is again due to perspective reasoning. The concept of original or clone is only relevant from a third person’s perspective. From beauty’s perspective by introspection and recollection she is every way the same person who went to sleep last night. Therefore the bayesian update can only be performed from a third person perspective, aka by the selector. 

Traditionally the answer to this problem is deemed to be between self-sampling assumption and self-indication assumption. Self-sampling assumption states beauty should reason as if she is randomly selected among all actual existent (past, present and future) beauties while self-indication assumption states beauty should reasonas if she is randomly selected among all potentially exist beauties. Both methods attempt to answer the question from a third party’s (the selector’s) perspective. As shown above beauty and the selector could disagree with each other therefore neither methods give correct answers. Applying the same logic could help us debunking other problems such as the doomsday argument, the presumptuous philosopher and  the simulation argument. 

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A Solution To The Sleeping Beauty Problem

Edit on 2017-12-03: I suggest new visitors read this post first. It is a newer (and much shorter) entry which better summarize my main argument.

ABSTRACT:I argue perspective disagreements in the setup of Sleeping Beauty Problem(SBP) is logically sound. This suggests double-halving is the correct answer. Counter examples against traditional Halfers and Thirders are also provided. As a result I propose the rejection of both Self-Sampling Assumption(SSA) and Self-Indication Assumption(SIA).  Because an observer reasoning as if she is randomly selected from a reference class switched to an outsider’s perspective which alters the answer. This could be important in anthropic reasoning. 

THE SLEEPING BEAUTY PROBLEM

The problem can be briefly described as follows:

Sleeping Beauty Problem(SBP):

Beauty is undergoing an experiment in which she will be waken up once or twice based on a fair coin toss. If the coin landed on tails (T) she will wake up twice with a memory wipe in between so that the 2 awakenings are indistinguishable. If the coin landed on heads (H) she will not be subjected to any memory wipe so only her first awakening needs to be considered. What should her credence of H be when she wakes up not having memories of a previous awakening.

There are two main camps to the problem. Halfers think the probability should be 1/2 and thirders argue the probability is 1/3. As of right now majority of the literature seems to lean towards the thirders’ view. There are some disagreements among halfers as what would the probability of H be after beauty learns that she is indeed in the first awakening. Here I will simply call those who changes their answer to 2/3 as halfers and call whose answers remains at 1/2 as double-halfers.

It is worth pointing out that halfers agrees with the Self-Sampling Assumption(SSA): that all else equal, an observer should reason as if she is randomly selected from the set of all actually existent observers. Thirders on the other hand agrees with the Self-Indication Assumption(SIA): that all else equal, an observer should reason as if she is randomly selected from the set of all possible observers. With these in mind, let us take a look at the perspective disagreement in SBP.

THE DISAGREEMENT

To see where the disagreement lies I purpose to consider the following problem:

Duplicating Beauty:

Beauty falls asleep as usual. The experimenter tosses a fair coin before she wakes up. If the coin landed on T then a perfect copy of beauty will be produced. The copy is precise enough that she cannot tell if herself is old or new. If the coin landed on H then no copy will be made . The beauty(ies) will then be put into two identical rooms respectively. What should their probability of H be once awake fully knowing the setup.

I imagine many would agree the Duplicating Beauty and the original SBP are equivalent. Both camps have literatures using cloning examples to advance their argument. I will focus on Duplicating Beauty here since it facilitate easier description and discussion on the disagreement. However my argument does not depends on this equivalency. So I will later dedicate a brief section on the procedures leading to the same disagreement in original SBP for anybody suspicious of the equivalency.

Now suppose another person is observing the experiment. He will not be shown the coin toss result nor will he ever be cloned. However after beauty(ies) are put into the two rooms he need to randomly choose one room and enter. Let us call him the selector. Now suppose after entering the random room he sees a beauty inside. Since the chosen room is not empty by very simply bayesian updating his probability of H shall be 1/3.

On the other hand, per halfers, the beauty in the same room should have p(H)=1/2 after waking up. Since her room has equal chance of being selected regardless of H or T the probability remains unchanged when she sees the selector. Now the beauty and the selector has a disagreement about p(H). The disagreement remains when they are free to communicate. Here we have two people giving different probabilities to the same proposition while having the same information, in violation of Aumann’s agreement theorem.

One might use this against halfers since thirders’ interpretation will not lead to such disagreement. I will address the problems with thirders’ reasoning and the pitfall of such agreement later with a counter example. For now I would argue both the selector and beauty are correct in their reasoning and this perspective disagreement is valid.

We can see this perspective disagreement with a frequentist approach. What if the experiment is repeated by a large number, say 1000 times. From the selector’s point of view, as long as someone undergoes the experiment and let him choose one out of two rooms it is counted as a repetition. After 1000 repetitions by his count he should expect to find about 750 non-empty rooms, 250 times following H and 500 following T. Consistent with his p(H) of 1/3. From beauty’s point of view each repetition means undergoing the experiment and wakes up again waiting for the selector’s choice.  So by her count, taking part in 1000 repetitions means she would recall 1000 coin tosses after waking up.  In those 1000 coin tosses there should be about 500 of H and T each, about 500 times she sees the selector half following H the other half following T. This is in accordance with her p(H)=1/2. Note the beauty waking up remembering 1000 coin tosses might be created by any one of those coin tosses just as the beauty disagreeing with the selector after the first toss might be a copy. Not being physically the same person should not affect the answer. To see this we can easily modify the Duplicating Beauty problem so the original beauty is always destroyed after each toss with either one or two copies created depending on H or T. In the modified version all resulting beauties are newly created yet it is hard to argue their answer would be different from the unmodified version. Another point worth noting is if all beauties in the experiment, old and new, each recall 1000 repetitions then by the selector’s count the number of repetition would be way greater. This is unsurprising since multiple beauties would be created in the process and each of them require additional tosses to complete their 1000 repetitions.

The perspective disagreement can also be shown by betting odds. To see this however beauties’ money must be duplicated when she is duplicated. In another word each person’s money follow each one’s perspective. This way beauty will not gain any information simply by checking her wallet meaning her money mimics her epistemic state which determines her answer. For the selector, the fair payout of a 1-dollar bet on H should be 3 dollars. Remember in the 1000 repetitions , he will see a beauty and enter the bet about 750 times. In which 250 times following H. If he pays 1 dollar to enter the bet each time, a 3 dollar payout on H would make his expected return 0. On the other hand the fair payout of a 1-dollar bet on H should be 2 dollars for beauty. In her 1000 repetitions she will be seeing the selector and entering the bet about 500 times, 250 times following H andT each –  a 2 dollar payout will make her expected return 0. The apparent disagreement is because the selector is doubly likely to enter the bet than beauty after T. Since from his perspective seeing any one of the two beauties is the same observation. In both cases he enters the bet. Whereas from beauties’ perspective which exact beauty the selector sees leads to different observations.  Although it gives her no information about the coin toss it decides whether or not she will enter the bet.

The above reasoning can be easily applied to original SBP. Although the execution would be rather cumbersome. In SBP the selector must lose track of the time himself in order to randomly select a day out of two. This can be done by using a pill that makes him sleep through beauty’s first awakening. He must choose between the sleeping pill and a undistinguishable placebo and go to sleep together with beauty when she first enters sleep. This way when he wakes up he can check if beauty is awake as well. Now he is in the same position as if he has just entered a random room in Duplicating Beauty. From beauty’s perspective to repeat the experiment means her subsequent awakenings need to be shorter to fit into her current awakening. For example, if in the first experiment the two possible awakenings happen on different days, then the in the next repetition  the two possible awakening can happen on morning and afternoon of the current day. Further repetitions will keep dividing the available time. It would cause each awakening to be increasingly short, making large number of repetitions impractical. However the methodology is still valid, since the length of the awakenings has no effect on beauty’s answer. To correctly see the effect of betting odds mentioned above her money should follow her epistemic state as well. Therefore any change to her wallet should be reversed when the relating memory is erased. In summary beauty and the selector would also be having disagreements in SBP with valid reasons.

ARGUMENT AGAINST SELF-INDICATION ASSUMPTION

The above arguments are of little importance if beauty should assigns p(H)=1/3 to begin with. In that case although she does not change her belief upon seeing the selector, unlike her counterpart does, both of them would still agree on p(H)=1/3. This can be explained by SIA, where she shall reason finding herself exist is the same as randomly selecting a potential observer(a room in this case) and then find the observer actually exist(the room is not empty). This put beauty in the same position as the selector, causing them to agree. To point out the weakness in this reasoning, consider the following example:

The 81-Day Experiment(81D):

There is a building with a circular corridor connected to 81 rooms with identical doors. At the beginning all rooms have blue walls. Then a painter randomly selects an unknown number of rooms and paint them red. Beauty would be put into a drug induced sleep lasting 81 day, spending one day in each room. An experimenter would wake her up if the room she currently sleeps in is red and let her sleep through the day if the room is blue. Her memory of each awakening would be wiped at the end of the day. Each time after beauty wakes up she is allowed to exit her room and open some other doors in the corridor to check the colour of those rooms. Now suppose one day after opening 8 random doors she sees 2 red rooms and 6 blue rooms. How should beauty estimate the total number of red rooms(R).

For halfers, waking up in a red room does not give beauty any more information except that R>0. Randomly opening 8 doors means she took a simple random sample of size 8 from a population of 80. In the sample 2 rooms (1/4) are red. Therefore the total number of red rooms(R) can be easily estimated as 1/4 of the 80 rooms plus her own room, 21 in total.

For thirders, beauty’s own red room is treated differently. As SIA states, finding herself awake is  as if she chose a random room from all 81 rooms and find out it is red. Therefore her room and the other 8 rooms she checked are all in the same sample. This means she has a simple random sample of size 9 from a population of 81. 3 out of 9 rooms in the sample (1/3) are red. The total number of red rooms can be easily estimated as a third of the 81 rooms, 27 in total.

I believe the above calculations are straightforward. The same numbers would be obtained as the most likely cases in a bayesian analysis with uniform priors. Notice if an outside selector randomly checks 9 rooms and happens to open the exact same 9 rooms beauty knows (her own room and the 8 rooms she checked), he would estimate R=27. Because beauty and the selector has exact same information about the rooms, he would be in disagreement with beauty according to halfers and in agreement with beauty according to thirders even if the two of them are free to communicate. The disagreement/agreement pattern is the same as in SBP.

However there are contradictions in thirders’ answer. First of all, before opening any doors if beauty is told that R=21 she should expect to see 2 reds if she opens 8 doors. According to thirders after opening 8 doors and actually seeing the expected number of 2 red rooms beauty must estimate R=27 instead of 21. This change of mind cannot be explained. Secondly, estimating R=27 means if beauty opens another 8 random doors she should expect to see 24/72×8=2.67 red rooms. This means after beauty saw 2 reds in the first 8 random rooms she would expect to see about 3 reds if she choose another 8 rooms. I fail to see how can this be justified. Last but not least, because beauty believes the 9 rooms beauty knows is a fair sample of all 81 rooms, it means red rooms (and blue rooms) are not systematically over- or under-represented it. Since beauty is always going to wake up in a red room, she has to conclude the other 8 rooms is not a fair sample. Red rooms have to be systematically underrepresent in those 8 rooms. This means even before beauty decides which doors she wants to open we can already predict with certain confidence that those 8 rooms is going to contains less reds than average. This supernatural predicting power is a conclusive evidence against SIA and thirders’ argument.

It is also easy to see why beauty should not estimate R the same way as the selector does. There are about 260 billion distinct combinations to pick 9 rooms out of 81. The selector has a equal chance to see any one of those 260 billion combinations. Beauty on the other hand could only possibility see a subset of the combinations. If a combination does not contains a red room, beauty would never see it. Furthermore, the more red rooms a combination contains the more awakening it has leading to a greater chance for a beauty to select the said combination. Therefore while the same 9 rooms is a unbiased sample for the selector it is a sample biased towards red for beauty.

One might want to argue after the selector learns a beauty has the knowledge of the same 9 rooms he should lower his estimation of R to the same as beauty’s. After all beauty could only know combinations in a subset biased towards red. The selector should also reason his sample is biased towards red. This argument is especially tempting for halfers since if true it means their answer also yields no disagreements. Sadly this notion is wrong, the selector ought to remain his initial estimation. To the selector a beauty knowing the same 9 rooms simply means after waking up in one of the red rooms in his sample, beauty made a particular set of random choices coinciding said sample. It offers him no new information about the other rooms.

To make this point clearer, it might be beneficial to understand how people reach to an agreement in an ordinary problem. Consider the following case:

Balls In Urns(BIU):

Suppose there is a urn filled with either 2 blue balls and 1 red ball(BBR) or 2 red balls and a blue ball(BRR) with equal chances. Andy randomly picked 2 balls from the urn and finds one ball of each colour. He correctly conclude the probability of BBR is 1/2, same as  the probability of BRR. Afterwards Bob asked for a red ball and was given one, he then randomly picked 1 ball from the 2 remaining balls in the urn and saw a blue one. He correctly concluded the probability of BBR is 2/3. It turns out however Andy and Bob actually saw the exact same 2 balls. The two of them are free to communicate and argue. Suppose both of them are rational can they reach to an agreement? Who should change his answer?

Again we can use a frequentist approach to solve this problem. Suppose the experiment is repeated many times with equal number of BBR and BRR. We can count the total number of occurrences when they both see the same 1 red and 1 blue balls. The relative frequency of BBR and BRR among these occurrences would indicate the correct probability. Both Andy and Bob should have no problem agreeing with this method. Here it becomes apparent that the exact procedure of the experiment determines whose initial probability is correct and who need to adjust his answer. More specifically how is the red ball given to Bob determined. Scenario 1: Bob is always given the red ball that has been picked by Andy. In this case Andy is correct, Bob should adjust his answer of BBR from 2/3 to 1/2. This is because for both BBR and BRR Andy has the same chance to pick a red and a blue ball. Given the same red ball Andy had picked Bob would have equal chance to pick the same blue ball again regardless the colour of the last ball left. The relative frequency of BBR and BRR given the occurrences would be about the same. Another case would be Scenario 2: Any red ball in the urn can be given to Bob. In this case Bob’s initial judgement is correct and Andy would have to change his probability for BBR to 2/3. Because all else equal, Bob is twice more likely to have the same red ball as Andy if there is only 1 red ball in the urn. The relative frequency of BBR to BRR with the occurrences would be 2:1. Since only one out of the two Scenarios can be true Andy and Bob must agree with each other as long as there is no ambiguity about he experiment procedure.

However, if we duplicate Bob (either by cloning or memory wiping) in the case of BRR and give each Bob a different red ball suddenly both Scenarios become true. To Andy the red ball he picked will always be given to Bob. To Bob the red ball given to him can be any one from the urn. Neither person would have reason to adjust his own probability fully knowing the experiment procedure. In this case the the two person having exact same information will remain in disagreement even if they are free to communicate. I believe the parallel between BIU, 81D and SBP is obvious enough to show why the selector, just as Andy, shall not adjust his probability.

ARGUMENTS AGAINST SELF-SAMPLING ASSUMPTION

I think the most convincing argument against SSA was presented by Elga(2000). He purpose the coin toss could happen after the first awakening. Beauty’s answer ought to remain the same regardless the timing of the toss. As SSA states, an observer should reason as if she is randomly selected from the set of all actual observers. If the selector randomly choose a day among all waking day(s) he is guaranteed to pick Monday if the coin landed on H but only has half the chance if T. From the selector’s perspective clearly a bayesian updating should be performed upon learning it is Monday. A simple calculation tells us his credence of H must be 1/3. As SSA dictates this is also beauty’s answer. Now beauty is predicting a fair coin toss yet to happen would most likely land on T. This supernatural predicting power is a conclusive evidence against SSA.

However, if we recognize the importance of perspective disagreement then beauty is not bound to give the same answer as the Selector. In fact I would argue she should not perform a bayesian update base on the new information. This can be explained in two ways.

One way is to put the new information into the frequentist approach mentioned above. In Duplicating Beauties, when a beauty wakes up and remembering 1000 repetitions she shall reason there are about 500 of H and T each among those 1000 tosses. The same conclusion would be reached by all beauties without knowing if she is physically the original or created somewhere along the way. Now suppose a beauty learns she is indeed the original. She would simply reason as  the original beauty who wakes up remembering 1000 tosses. These 1000 tosses would still contain about 500 of H and T each. Meaning her answer shall remain at 1/2.

Another way to see why beauty should not perform a bayesian update is to see the agreement/disagreement pattern between her and the selector. It is worth noting that beauty and the selector will be in agreement once knowing she is the original. As stated earlier, one way to understand the disagreement is after T, seeing either beauty is the same observation for the selector while it is different observations for beauties. This in turn causes the selector to enter twice more bets than beauty. However once we distinguish the two beauties by stating which one is the original the selector’s observation would also be different depending on which beauty he sees. To put it in a different way, if a bet is only set between the selector and the original beauty then the selector would no longer be twice more likely to enter a bet in case of T. He and the original Beauty would enter the bet with equal chances. Meaning their betting odds ought to be the same, they must be in agreement regarding the credence of H.

To be specific, the disagreements/agreements pattern can be summarized as follows. If the selector randomly chooses one of the two rooms as described by SIA. Upon seeing a beauty in the room the selector’s probability for H will change from 1/2 to 1/3. Beauty’s probability remains at 1/2 as described above. The two of them would be in disagreement. Once they learns the beauty is the original, the selector’s probability of H increases back to 1/2 from 1/3 by simple bayesian updating while beauty’s probability still remains at 1/2. This way the two would be in agreement. The selector can also randomly chooses one beauty from all existing beauty(ies) as described by SSA (here the total number of beauties should be shielded from the selector to not reveal the coin toss result). In this case seeing a beauty gives the selector no new information so his probability for H would remain unchanged at 1/2. On the other hand, from beauty’s perspective she is twice more likely to be chosen if there exist only one beauty instead of two. Therefore upon seeing the selector her credence of H would increase to 2/3. The two of them would also be in disagreement. Once they learn the beauty is the original the selector’s credence for H would increase to 2/3 by bayesian updating. Again beauty would not update her probability and it remains at 2/3. This way the two would agree with each other again.

As shown above, for the two to reach an agreement beauty must not perform a bayesian update upon the new information. This holds true in both cases regardless how the selection is structured.

Beauty’s antibayesianism, just like the perspective disagreement, is quite unusual. I think this is due to the fact that there is no random event determining which beauty is original/clone. While the coin toss may create a new copy of beauty, nothing could ever turn herself into the copy. The original beauty would eventually be the original beauty. It is simply tautology. There is no random soul jumping between the two bodies. Beauty’s uncertainty is because of the structure of the experiment which is purely due to lack of information. Compare this to the selector’s situation. The event of him choosing a room is random. Therefore learning the beauty in the chosen room is original gives new information about the random event. From his perspective a bayesian update should be performed.Where as from beauty’s perspective, learning she is the original does not give new information about a random event, for the simply fact there is no random event to begin with. It only gives information about her own perspective. So she should not performing a bayesian update as the selector did.

DISCUSSIONS

With the above arguments in mind we can clearly see the importance of perspectivism in SBP. It does not matter whether the selector follows SSA or SIA, his answer could not always correctly reflect beauty’s. Once beauty switch to selector’s perspective her answers would change. Therefore it is important for us to consciously track our reasoning process to make sure it contains no change of perspective. As shown above, beauty’s credence of H should be 1/2 when she wakes up and remains at 1/2 once learned it is Monday, aka double-halfers are correct. In another word, learning she exists does not confirm scenarios with more observers, learning she is the first does not confirm scenarios with less observers. Applying the same logic means we should reject Doomsday Argument while disagree with the Presumptuous Philosopher.

There are more interesting implication for perspective reasonings. One thing worth noting is that disagreements can also arise among beauties themselves. Consider DB with 2 repetitions for all beauties. Beauty wakes up in a room remembering taking part of 2 coin tosses. Now the experimenter tells her that including herself there are currently 3 beauties exist in total. Very soon the beauty figured out to get three beauties after 2 rounds means the first coin toss must be T. One of the resulting beauties would then experience another T while the other resulting beauty would experience another H. For ease of discussion lets randomly label the two beauties after the first toss T1 and T2 since they both experienced a T. Suppose T1 experience a H and T2 experience another T on the second round. We can label the the beauty who experienced a H asT1H1, and randomly label the two beauties experienced another T as T2T1 and T2T2. By indifference principle beauty shall reason she is equally likely to be T1 or T2 at her first awakening. Because T1H1 is a direct product of T1, her probability of being T1H1 at second awakening is the same as being T1 at the first awakening which is 1/2.  Because T2T1 and T2T2 are random labels for the direct products of T2, she shall reason the combine probability of her being either T2T1 or T2T2 is the same as being T2 at the first awakening. So her probability of being T2T1 and T2T2 must be 1/4 each. All three beauties would reach to the same conclusion since they have the same information. Notice here they would be in disagreement regarding their probability of being T1H1. Each of them would think herself is twice more likely to be T1H1 (1/2) than the other two beauties(1/4 each).This disagreement might seems alarming. However it is also valid. As discussed above, in problems involving duplications people with the same information can have different probabilities. The reason this disagreement seems more suspicious is because the resulting beauties appears to be in symmetrical positions thus should not disagree with each other. However this symmetry is only valid from an selector’s perspective. For the selector, any resulting beauty has an equal chance of being picked. In another word T1H1, T2T1 and T2T2 are only different in names during a random selection process. Therefore it is correct a randomly chosen beauty has equal chance of being T1H1, T2T1 or T2T2(1/3). From beauty’s perspective however such symmetry does not exist. The three labels, T1H1, T2T1 and T2T2, are not just different in name. If the coin landed H then she would be labeled H1 for certain. If the coin landed on T she would be randomly labeled T1 or T2. Therefore from beauty’s perspective being labeled as H1 is twice likely to happen than being label T1.

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