The Doomsday Argument

The end of the world and life in a computer simulation The end of the world and life in a computer simulation

phil 20229 Je↵ Speaks February 7, 2008

1 The doomsday argument ...... 1 1.1 Bayes’ theorem ...... 2 1.2 Anthropic reasoning ...... 3 2 Responses to the argument ...... 3 3 Are we living in a computer simulation? ...... 4

1 The doomsday argument

LeslieIn begins the reading his exposition for today, ofLeslie the argument introduces with a familiar an example: sort of reasoning:

This reasoningThe basic idea is, fairly here clearly, is one which a kind we of reasoning employ all upon the time which in our we rely ordinary all of thereasoning time. It about might the world. It be summedmight be up summed like this: ifup we as have follows: two theories, and the first makes a certain event much more likely than the other, and we observe that event, that should lead us to favor the first theory. Note that this doesn’t mean that we shouldThe principle always think of confirmation that the first theory is true; rather, what it means is that, whatever our initial estimate of the probability of the first theory, we should increase that estimateE upon is evidence observing for theT1 over event T2 in if question. the probability of E given T1 > the probability of E given T2. Now consider the application of that line of reasoning to the examples of balls shot at random from aHere lottery E is machine. our evidence, Suppose and thatT1 and you T2 know are thattwo theories the balls between in the machine which we are are numbered trying to decide. When sequentiallywe talk about (with nothe repeats) probability beginning of X given with Y, we 1, but are that talking you about don’t the know probability how many that balls X will there take place, if are in theY also machine. happens. Now we start the machine, and a ball comes out with ‘3’ on it. You’re now asked: do you think that it is more likely that the machine has 10 balls in it, or 10,000 balls in it? The line of reasoning sketched above seems to favor the hypothesis that there are just 10 balls in the machine. (Note that, whichever hypothesis you endorse, you could be wrong; this is not a form of reasoning which delivers results guaranteed to be correct. The question is just which of these hypotheses is most likely, given the evidence.) The basic idea here is one which we employ all the time in our ordinary reasoning about the world. It might be summed up as follows:

The principle of conﬁrmation

E is evidence for T1 over T2 if the probability of E given T1 > the probability of E given T2.

Here E is our evidence, and T1 and T2 are two theories between which we are trying to decide. When we talk about the probability of X given Y, we are talking about the probability that X will take place, if Y also happens. Now consider the application of that line of reasoning to the examples of balls shot at random from a lottery machine. Suppose that you know that the balls in the machine are numbered sequentially (with no repeats) beginning with 1, but that you don't know how many balls there are in the machine. Now we start the machine, and a ball comes out with “3” on it. You're now asked: do you think that it is more likely that the machine has 10 balls in it, or 10,000 balls in it? The principle of conﬁrmation suggests - correctly, it seems - that our piece of evidence favors the hypothesis that there are just 10 balls in the machine. This is because the probability that “3” comes out given that balls 1-10 are in the machine is 10%, whereas the probability that this ball comes out given that balls numbered 1-10,000 are in the machine is only 0.01%.

(Note that, whichever hypothesis you endorse, you could be wrong; this is not a form of reasoning which delivers results guaranteed to be correct. The question is just which of these hypotheses is most likely, given the evidence.)

This gives us some information about how to reason about probabilities, but not very much. It tells us when evidence favors one theory over another, but does not tell us how much. It leaves unanswered questions like: if before I thought that the 10,000 ball hypothesis was 90% likely to be true, should I now think that the 10 ball hypothesis is more than 50% likely to be true? If I assigned each of the two hypotheses prior to the emergence of the “3” ball a probability of 0.5 (50% likely to be true), what probabilities should I assign to the theories after the ball comes out? This gives us some information about how to reason about probabilities, but not very much. It tells us when evidence favors one theory over another, but does not tell us how much. It leaves unanswered questions like: if before I thought that the 10,000 ball hypothesis was 90% likely to be true, should I now think that the 10 ball hypothesis is more than 50% likely to be true? If I assigned each of the two hypotheses prior to the emergence of the “3” ball a probability of 0.5 (50% likely to be true), what probabilities should I assign to the theories after the ball comes out? 1.1 Bayes’ theorem To understand Leslie’s argument, we’ll have to understand how to answer these sorts of questions. In fact, we can do better than just saying that in such cases you should raise the probability you One wayassign to answer to onethese theory. questions Weemploys can, a widely using a widely accepted rule of reasoning called ‘Bayes’ theorem’, accepted rule of reasoning called “Bayes’ theorem,” namedsay after howThomas much Bayes, you an 18th should century raise English your probability assignment. (One reason why this theorem is mathematicianwidely and accepted Presbyterian is that minister. following it enables one to avoid ‘Dutch book’ arguments.)

To arriveTo at arrivethe theorem, at Bayes’we begin with theorem, the following we can begin with the deﬁnition of what is called ‘conditional deﬁnition of conditional probability, where, as is standard,probability’: we abbreviate the“the conditional probability probability of one of x claim, given that another is true. In particular, for arbitrary given y”claims as “Pr(xa | y)”:and b, we can say that

P (a b)= P (a&b) | P (b)

This says,In otherin effect, words, that the probability the probability of a given ofb isa given b is the chance that a and b are both true, divided by the chance that a and b are both true, divided by the chancesthe that chances b is true. that b is true. For example, let a = ‘Obama wins’, and let b = ‘a man wins.’ Suppose that each of Obama, Hilary, and McCain have a 1/3 chance of winning. Then the conditional probability is that Obama wins, given that a man wins, is 1/2. Intuitively, if you found out only that a man would win, you should then (given the initial probability assignments) think that there is a 0.5 probability that Obama will win.

P (h) P (e h) P (h e)= | | P (e)

Consider what this would say about the example of the lottery machine. Suppose for simplicity that you know going in that there are only two options, which are equally likely to be correct: that there are 10 balls in the machine, and that there are 10,000. Let e be the evidence that the ﬁrst ball to come out is #3, and let h be the hypothesis that there are 10 balls in the machine. Then we might say:

P (h)=0.5 P (e h)=0.1 | P (e)=0.5(0.1+0.0001) = 0.05005

0.5 0.1 Then we find, via Bayes’ theorem, that P (h e)= 0.05005 =0.999. So, on the basis of the evidence that the first ball to come out was #3, you should| revise your confidence in the 10-ball hypothesis from 50% to 99.9% certainty.

Bayes’ theorem can be restated in the following way:

2 1.1 Bayes’ theorem

In fact, we can do better than just saying that in such cases you should raise the probability you One way to answer these questions employs a widely acceptedassign rule of to reasoning one theory. called “Bayes’ We can, theorem,” using a widely accepted rule of reasoning called ‘Bayes’ theorem’, namedsay after howThomas much Bayes, you an 18th should century raise English your probability assignment. (One reason why this theorem is mathematicianwidely and accepted Presbyterian is that minister. following it enables one to avoid ‘Dutch book’ arguments.)

P (a b)= P (a&b) | P (b)

This says, in effect, that the probability of a given b is the chanceIn other that a and words, b are both the probabilitytrue, divided by ofthea given b is the chance that a and b are both true, divided by chancesthe that chances b is true. that b is true. For example, let a = ‘Obama wins’, and let b = ‘a man wins.’ Suppose that each of Obama, Hilary, and McCain have a 1/3 chance of winning. Then the conditional Let’s work through an example. Suppose that this is some time before the 2008 election, and let a = ‘Obama wins’, and let b = ‘a man wins.’ Suppose that youprobability think that each is of that Obama, Obama Hilary, and wins, McCain given that a man wins, is 1/2. Intuitively, if you found out only have a 1/3 chance of winning. Then what is the conditionalthat probability a man that would Obama win, wins, given you that should a man then (given the initial probability assignments) think that wins, using the above formula? there is a 0.5 probability that Obama will win.

The conditional probability is that Obama wins, given thatUsing a man thiswins, deﬁnitionis ½, since in ofthis conditional case Pr(a&b)= probability,⅓ we can then argue as follows, assuming that P (b) and Pr(b)=⅔. Intuitively, if you found out only that a man would win, you should then (given the initial probability assignments) think that there is a 0.5 probability= 0:that Obama will win. 6 Using this deﬁnition of conditional probability, we can then argue for Bayes’ theoremP (a& asb) follows, assuming 1. P (a b)= P (b) def. of conditional probability that Pr(b)≠0. | 2. P (b a)= P (a&b) def. of conditional probability | P (a) 3. P (a b) P (b)=P (a&b) (1), multiplication by =’;s 4. P (a&| b)=⇤ P (b a) P (a) (2), multiplication by =’s 5. P (a b) P (b)=| P⇤(b a) P (a) (3),(4) | ⇤ P (b a) P (a)| ⇤ C. P (a b)= | (5), division by =’s | P (b) This conclusion is Bayes’ theorem. Often, what we want to know is, intuitively, the probability of some hypothesis ‘h’ given some evidence ‘e’; then we would write the theorem as:

P (h) P (e h) P (h e)= | | P (e)

P (h)=0.5 P (e h)=0.1 | P (e)=0.5(0.1+0.0001) = 0.05005

Bayes’ theorem can be restated in the following way:

2 1.1 Bayes’ theorem

1.1 Bayes’ theorem In fact, we can do better than just saying that in such cases you should raise the probability you In fact, we can do better than just saying thatassign in such to cases one you theory. should Weraise can, the probability using a widely you accepted rule of reasoning called ‘Bayes’ theorem’, assign to one theory. We can, using a widely acceptedsay how rule much of reasoning you should called raise ‘Bayes’ your theorem’, probability assignment. (One reason why this theorem is say how much you should raise your probabilitywidely assignment. accepted (One is that reason following why this theorem it enables is one to avoid ‘Dutch book’ arguments.) widely accepted is that following it enables one to avoid ‘Dutch book’ arguments.) To arrive at Bayes’ theorem, we can begin with the definition of what is called ‘conditional To arrive at Bayes’ theorem, we can begin with the definition of what is called ‘conditional probability’: the probability of one claim, givenprobability’: that another theis true. probability In particular, of one for arbitrary claim, given that another is true. In particular, for arbitrary claims a and b, we can say that claims a and b, we can say that Definition of conditional probability P (a b)= P (a&b) P (a&b) P (b) P (a b)= | | P (b) In other words, the probability of a given b is the chance that a and b are both true, divided by the chances that b is true. For example, let a =In ‘Obama other wins’, words, and the let probabilityb = ‘a man wins.’ of a Supposegiven b is the chance that a and b are both true, divided by that each of Obama, Hilary, and McCainUsing this have thedefinition a 1/3chances chanceof conditional that of winning.b probability,is true. Then Forwe can the example, then conditional argue let for a = ‘Obama wins’, and let b = ‘a man wins.’ Suppose probability is that Obama wins, givenBayes’ that atheorem manthat wins, as each follows, is of 1/2. assuming Obama, Intuitively, that Hilary, Pr(b) if you≠0. and found McCain out only have a 1/3 chance of winning. Then the conditional that a man would win, you should then (givenprobability the initial is probability that Obama assignments) wins, given think that that a man wins, is 1/2. Intuitively, if you found out only there is a 0.5 probability that Obama will win. that a man would win, you should then (given the initial probability assignments) think that Using this definition of conditional probability,there we can is then a 0.5 argue probability as follows, that assuming Obama that P will(b) win. = 0: 6 Derivation of Bayes’ theorem Using this definition of conditional probability, we can then argue as follows, assuming that P (b) 1. P (a b)= P (a&b) def. of conditional probability | P (b) = 0: 2. P (b a)= P (a&b) 6 def. of conditional probability | P (a) 3. P (a b) P (b)=P (a&b)1. P ( (1),a b)= multiplicationP (a&b) by =’;s def. of conditional probability 4. P (a&| b)=⇤ P (b a) P (a) (2),| multiplicationP (b) by =’s | ⇤ P (a&b) 5. P (a b) P (b)=P (b a) P (a)2. P (b a)= P (a) (3),(4) def. of conditional probability | ⇤ P (b a) P (a)| ⇤ | C. P (a b)= | 3. P (a b) (5),P (b division)=P (a by& =’sb) (1), multiplication by =’;s | P (b) 4. P (a&| b)=⇤ P (b a) P (a) (2), multiplication by =’s This conclusion is Bayes’ theorem. Often, what we want to know is, intuitively, the probability 5. P (a b) P (b)=| P⇤(b a) P (a) (3),(4) Theof conclusion some hypothesis of this argument ‘h’ given is Bayes’ some theorem. evidence Intuitively, ‘e’; then what we would it says| write⇤is that theif we theorem want |to know as:⇤ the probability of some theory given a bit of evidence, what we need to know are three things:P (1)(b thea) probabilityP (a) of the evidence C. P (a b)= |P (b) (5), division by =’s given the theory (i.e.,P how(h) Plikely(e h )the evidence is to happen if the theory |is true), (2) the prior probability of the theory, and (3) theP prior(h e )=probability of |the evidence. | P (e) This conclusion is Bayes’ theorem. Often, what we want to know is, intuitively, the probability Consider what this would say about the exampleof some of the hypothesis lottery machine. ‘h’ given Suppose some for simplicity evidence ‘e’; then we would write the theorem as: that you know going in that there are only two options, which are equally likely to be correct: that there are 10 balls in the machine, and that there are 10,000.P Let(h) ePbe(e h the) evidence that the P (h e)= | first ball to come out is #3, and let h be the hypothesis that| there areP (e) 10 balls in the machine. Then we might say: Consider what this would say about the example of the lottery machine. Suppose for simplicity P (h)=0.5 that you know going in that there are only two options, which are equally likely to be correct: P (e h)=0.1 | that there are 10 balls in the machine, and that there are 10,000. Let e be the evidence that the P (e)=0.5(0.1+0.0001) = 0.05005 first ball to come out is #3, and let h be the hypothesis that there are 10 balls in the machine. Then we might say: 0.5 0.1 Then we find, via Bayes’ theorem, that P (h e)= 0.05005 =0.999. So, on the basis of the evidence that the first ball to come out was #3, you should| revise your confidence in the 10-ball hypothesis from 50% to 99.9% certainty. P (h)=0.5 P (e h)=0.1 Bayes’ theorem can be restated in the following way: | P (e)=0.5(0.1+0.0001) = 0.05005

2 0.5 0.1 Then we find, via Bayes’ theorem, that P (h e)= 0.05005 =0.999. So, on the basis of the evidence that the first ball to come out was #3, you should| revise your confidence in the 10-ball hypothesis from 50% to 99.9% certainty.

Bayes’ theorem can be restated in the following way:

2 1.1 Bayes’ theorem

In fact, we can1.1 do Bayes’ better theorem than just saying that in such cases you should raise the probability you assign to one theory. We can, using a widely accepted rule of reasoning called ‘Bayes’ theorem’, In fact, we can do better than just saying that in such cases you should raise the probability you assign to one theory. Wesay can, how using much a widely you accepted should rule raise of reasoning your probability called ‘Bayes’ assignment. theorem’, (One reason why this theorem is say how much you shouldwidely raise accepted your probability is that assignment. following (One it enables reason why one this to theorem avoid is ‘Dutch book’ arguments.) widely accepted is that following it enables one to avoid ‘Dutch book’ arguments.) To arrive at Bayes’ theorem, we can begin with the deﬁnition of what is called ‘conditional To arrive at Bayes’ theorem, we can begin with the deﬁnition of what is called ‘conditional probability’: the probabilityprobability’: of one claim, the given probability that another of is one true. claim, In particular, given forthat arbitrary another is true. In particular, for arbitrary claims a and b, we canclaims say thata and b, we can say that

P (a&b) P (a b)= P (a&b) | P (b) P (a b)= | P (b) In other words, the probability of a given b is the chance that a and b are both true, divided by the chances that b is true.In other For example, words, let thea = ‘Obama probability wins’, and of a letgivenb = ‘ab manis wins.’the chance Suppose that a and b are both true, divided by that each of Obama, Hilary,the chances and McCain that haveb is a true. 1/3 chance For example, of winning. let Thena = the ‘Obama conditional wins’, and let b = ‘a man wins.’ Suppose probability is that Obama wins, given that a man wins, is 1/2. Intuitively, if you found out only that a man would win,that you each should of then Obama, (given the Hilary, initial and probability McCain assignments) have a think 1/3 chance that of winning. Then the conditional there is a 0.5 probabilityprobability that Obama is will that win. Obama wins, given that a man wins, is 1/2. Intuitively, if you found out only Using this definition ofthat conditional a man probability, would win, we can you then should argue as follows,then (given assuming the that initialP (b) probability assignments) think that = 0: there is a 0.5 probability that Obama will win. 6 Derivation of Bayes’ theorem 1. P (a b)= P (a&b) def. of conditional probability | P (b)Using this definition of conditional probability, we can then argue as follows, assuming that P (b) 2. P (b a)= P (a&b) def. of conditional probability | P (a)= 0: 3. P (a b) P (b)=6 P (a&b) (1), multiplication by =’;s | ⇤ 4. P (a&b)=P (b a) P1.(a)P (a b)= P (a&b) (2), multiplication by =’s def. of conditional probability 5. P (a b) P (b)=| P⇤(b a) P (a)| P (b) (3),(4) | ⇤ P (b a) P (a)| ⇤ P (a&b) C. P (a b)= | 2. P (b a)= (5), division by =’s def. of conditional probability | P (b) | P (a) 3. P (a b) P (b)=P (a&b) (1), multiplication by =’;s This conclusion is Bayes’ theorem. Often, what we want to know is, intuitively, the probability 4. P (a&| b)=⇤ P (b a) P (a) (2), multiplication by =’s Theof conclusion some hypothesis of this argument ‘h’ given is Bayes’ some theorem. evidence Intuitively, ‘e’; then| what we⇤ would it says writeis that theif we theorem want to know as: the probability of some theory given a bit of evidence,5. Pwhat(a web) needP to(b know)= areP ( threeb a) things:P ( a(1)) the probability of the evidence (3),(4) given the theory (i.e.,P how(h) Plikely(e h )the evidence| is to⇤ happenP (b aif )theP theory(a)| is⇤ true), (2) the prior probability of the theory, and (3) theP prior(h e )=probability of |the evidence.C. P (a b)= | (5), division by =’s | P (e) | P (b) If we let “h” stand for the hypothesis to be tested and “e” for the relevant evidence, then the theorem can be rewrittenConsider as follows: what this wouldThis say conclusion about the example is Bayes’ of the theorem. lottery machine. Often, Suppose what for we simplicity want to know is, intuitively, the probability that you know going inof that some there hypothesis are only two ‘h options,’ given which some are evidence equally likely ‘e’; to then be correct: we would write the theorem as: that there are 10 balls in the machine, and that there are 10,000. Let e be the evidence that the Bayes’ theorem first ball to come out is #3, and let h be the hypothesis that there are 10 balls in the machine. Then we might say: P (h) P (e h) P (h e)= | | P (e)

This theoremP (h is)=0 very. 5useful, since often it is easy to figure out the conditional probability of the evidence given the theory, butP very(e h hard)=0 to.1 figureConsider out the conditional what probability this would of the theory say about given the the evidence. example of the lottery machine. Suppose for simplicity | P (e)=0.5(0.1+0that.0001) you = 0 know.05005 going in that there are only two options, which are equally likely to be correct: A good example of this sort ofthat situation there is given are by 10 our balls example in of the the lottery machine, balls. and that there are 10,000. Let e be the evidence that the 0.5 0.1 Then we find, via Bayes’first theorem, ball to that comeP (h e)= out is #3,=0 and.999. let So, onh be the the basis hypothesis of the evidence that there are 10 balls in the machine. | 0.05005 that the first ball to comeThen out waswe #3,might you say:should revise your confidence in the 10-ball hypothesis from 50% to 99.9% certainty.

Bayes’ theorem can be restated in the following way: P (h)=0.5 P (e h)=0.1 | P (e)=0.5(0.1+02 .0001) = 0.05005

Bayes’ theorem can be restated in the following way:

2 1.1 Bayes’ theorem

In fact, we can do better than just saying that in such cases you should raise the probability you assign to one theory. We can, using a widely accepted rule of reasoning called ‘Bayes’ theorem’, say how much you should raise your probability assignment. (One reason why this theorem is widely accepted is that following it enables one to avoid ‘Dutch book’ arguments.)

To arrive at Bayes’ theorem, we can begin with the deﬁnition of what is called ‘conditional probability’: the probability of one claim, given that another is true. In particular, for arbitrary claims a and b, we can say that

P (a b)= P (a&b) | P (b)

In other words, the probability of a given b is the chance that a and b are both true, divided by the chances that b is true. For example, let a = ‘Obama wins’, and let b = ‘a man wins.’ Suppose that each of Obama, Hilary, and McCain have a 1/3 chance of winning. Then the conditional probability is that Obama wins,1.1 given Bayes’ that theorem a man wins, is 1/2. Intuitively, if you found out only that a man would win, you should then (given the initial probability assignments) think that thereIn fact, is we a can 0.5 do probability better than just that saying Obama that will in such win. cases you should raise the probability you assign to one theory. We can, using a widely accepted rule of reasoning called ‘Bayes’ theorem’, say how much you should raise your probability assignment. (One reason why this theorem is Usingwidely accepted this definition is that following of conditional it enables probability, one to avoid ‘Dutch we can book’ then arguments.) argue as follows, assuming that P (b) = 0: 6To arrive at Bayes’ theorem, we can begin with the definition of what is called ‘conditional probability’:1. P (a theb)= probabilityP (a&b) of one claim, given that another is true.def. In of particular, conditional for arbitrary probability claims a and b|, we canP say(b) that 2. P (b a)= P (a&b) def. of conditional probability | P (a) 3. P (a bP)(a&Pb) (b)=P (a&b) (1), multiplication by =’;s P (a b)= P (b) 4. P| (a&| b)=⇤ P (b a) P (a) (2), multiplication by =’s 5. P (a b) P (b)=| P⇤(b a) P (a) (3),(4) In other words,| the⇤ probability of|a given⇤ b is the chance that a and b are both true, divided by the chances that b is true.P (b a For) P ( example,a) let a = ‘Obama wins’, and let b = ‘a man wins.’ Suppose C. P (a b)= |P (b) (5), division by =’s that each of Obama,| Hilary, and McCain have a 1/3 chance of winning. Then the conditional Thisprobability conclusion is that Obama is Bayes’ wins, theorem. given that aOften, man wins, what is 1/2. we want Intuitively, to know if you is, found intuitively, out only the probability that a man would win, you should then (given the initial probability assignments) think that ofthere some is a hypothesis0.5 probability ‘h that’ given Obama some will evidence win. ‘e’; then we would write the theorem as: Bayes’ theorem Using this definition of conditional probability, we can then argue as follows, assuming that P (b) P (h) P (e h) = 0: P (h e)= | 6 | P (e) 1. P (a b)= P (a&b) def. of conditional probability This theorem is very useful, since often it is| easy toP figure(b) out the conditional probability of the evidence given the theory, but very hard to figureConsider out the2. Pconditional(b whata)= probabilityP this(a&b) would of the theory say about given the the evidence. exampledef. of of conditional the lottery probability machine. Suppose for simplicity | P (a) that3. youP (a knowb) P ( goingb)=P ( ina&b that) there are only two (1), options, multiplication which by are =’;s equally likely to be correct: A good example of this sort of situation is given by our example of the lottery balls. that4. thereP (a&| areb)=⇤ 10P (b ballsa) P in(a) the machine, and that there(2), multiplication are 10,000. by Let =’s e be the evidence that the Remember that we have two hypotheses:5. P (a thatb) thereP (b )=are| ballsP⇤(b labeleda) P (1-10a) in the machine, and that there are balls (3),(4) labeled 1-10,000 in the machine.first Let’s ball assume to| come⇤ that,P (b ata out )theP ( start,a is)| #3, we⇤ think and that let theseh hypothesesbe the hypothesis have equal that there are 10 balls in the machine. C. P (a b)= | (5), division by =’s probability: they are each assignedThen probability we might| 0.5. (Maybe say:P (b) which sort of machine we are given is determined by coin flip.) This conclusion is Bayes’ theorem. Often, what we want to know is, intuitively, the probability Now suppose that the first ballof that some comes hypothesis out, again ‘h at’ givenrandom, some is a “3”. evidence Bayes’ theorem ‘e’; then can we tell would us, given write the the theorem as: foregoing information, how likely it is Pthat(h the)=0 machine.5 before us contains 10 rather than 10,000 balls. According to Bayesians, we figure this out by conditionalizing on our evidence, i.e. finding the conditional probability of the P (h) P (e h) theory given the evidence. PP(h(ee)=h)=0.1 | | | P (e) Let “h” be the theory that the machineP contains(e)=0 10. 5(0balls..1+0 Then to.0001) compute = the 0 relevant.05005 conditional probability, we need to know three values. Consider what this would say about the example of the lottery machine. Suppose for simplicity First, we need to know Pr(h).that From you the above, know we going know in that that this there is 0.5. are only two options, which0.5 0 are.1 equally likely to be correct: Then we find, via Bayes’ theorem, that P (h e)= =0.999. So, on the basis of the evidence that there are 10 balls in the machine, and that there| are 10,000.0.05005 Let e be the evidence that the Next, we need to know Pr(e that|first h). Fairly ball the obviously, to first come ball this out tois is 0.1. come #3, and out let wash be #3, the hypothesis you should that revise there your are 10 confidence balls in the machine.in the 10-ball hypothesis Then we might say: But we also need to know thefrom unconditional 50% to probability 99.9% of certainty.e, Pr(e). How should we figure this out?

A natural thought is that sinceBayes’ we knowP (h theorem that)=0 each.5 of can the two be hypotheses restated - 10 in ball the and following 10,000 ball - way:have a probability of 0.5, Pr(e) should be the mean of the conditional probabilities of e given those two hypotheses. In this case: P (e h)=0.1 | P (e)=0.5(0.1+0.0001) = 0.05005

0.5 0.1 2 Then we find, via Bayes’ theorem, that P (h e)= 0.05005 =0.999. So, on the basis of the evidence that the first ball to come out was #3, you should| revise your confidence in the 10-ball hypothesis from 50% to 99.9% certainty.

Bayes’ theorem can be restated in the following way:

2 1.1 Bayes’ theorem

To arrive at Bayes’ theorem, we can begin with the definition of what is called ‘conditional probability’:1.1 Bayes’ the theorem probability of one claim, given that another is true. In particular, for arbitrary claims a and b, we can say that 1.1 Bayes’ theorem In fact, we can do better than just saying that in such cases you should raise the probability you assign to one theory. We can, using a widely accepted rule of reasoning called ‘Bayes’ theorem’, In fact,P we(a canb)= do betterP (a&b than) just saying that in such cases you should raise the probability you say how much you should raise your probability assignment.P (b) (One reason why this theorem is assign to one| theory. We can, using a widely accepted rule of reasoning called ‘Bayes’ theorem’, widely accepted is that following itsay enables how much one you to avoid should ‘Dutch raise your book’ probability arguments.) assignment. (One reason why this theorem is widely accepted is that following it enables one to avoid ‘Dutch book’ arguments.) To arrive at Bayes’ theorem, weIn can other begin words, with the the definition probability of what of a isgiven calledb is ‘conditional the chance that a and b are both true, divided by probability’: the probability of oneTothe claim, arrive chances given at Bayes’ that that another theorem,b is true. is we true. For can example, In begin particular, with let thea for= definition arbitrary ‘Obama of wins’, what is and called let ‘conditionalb = ‘a man wins.’ Suppose claims a and b, we can say that probability’:that each theof Obama, probability Hilary, of one claim, and McCain given that have another a is 1/3 true. chance In particular, of winning. for arbitrary Then the conditional claimsprobabilitya and b, is we that can say Obama that wins, given that a man wins, is 1/2. Intuitively, if you found out only P (a&b) that a man would win, you should then (given the initial probability assignments) think that P (a b)= P (b) | thereP ( isa b a)= 0.5P ( probabilitya&b) that Obama will win. | P (b) In other words, the probability of aUsinggiven thisb is the definition chance that of conditionala and b are probability, both true, divided we can by then argue as follows, assuming that P (b) the chances that b is true. For example,In other let words,a = ‘Obama the probability wins’, and of a letgivenb =b ‘ais man the chance wins.’ that Supposea and b are both true, divided by the= 0:chances that b is true. For example, let a = ‘Obama wins’, and let b = ‘a man wins.’ Suppose that each of Obama, Hilary, and McCain have a 1/3 chance of winning. Then the conditional that6 each of Obama, Hilary, and McCain have a 1/3 chance of winning. Then the conditional probability is that Obama wins, given that a man wins,P ( isa& 1/2.b) Intuitively, if you found out only probability1. P is(a thatb)= Obama wins, given that a man wins, is 1/2. Intuitively,def. of if conditional you found out probability only that a man would win, you should then (given the initialP (b) probability assignments) think that that a man would| win,P (a you&b) should then (given the initial probability assignments) think that there is a 0.5 probability that Obama will2. win.P (b a)= def. of conditional probability there is a 0.5 probability| P ( thata) Obama will win. 3. P (a b) P (b)=P (a&b) (1), multiplication by =’;s Using this definition of conditionalUsing probability, this definition we| can⇤ of then conditional argue as probability, follows, assuming we can then that argueP (b as) follows, assuming that P (b) = 0: = 0: 4. P (a&b)=P (b a) P (a) (2), multiplication by =’s 6 6 5. P (a b) P (b)=| P⇤(b a) P (a) (3),(4) P (a&b) | P⇤(a&Pb)(b a) P (a)| ⇤ 1. P (a b)= 1. P (a b)= P (b)def. of conditional probabilitydef. of conditional probability | P (b) C. P|(a b)= |P (b) (5), division by =’s P (a&b) 2. P (b a)=| P (a&b) def. of conditional probability 2. P (b a)= P (a) | P (a)def. of conditional probability | 3. P (a b) P (b)=P (a&b) (1), multiplication by =’;s 3. P (a b) P (b)=P (a&b)This conclusion| ⇤ is Bayes’ (1), multiplication theorem. Often, by =’;s what we want to know is, intuitively, the probability 4. P (a&| b)=⇤ P (b a) P (a)of some4. P (a hypothesis&b)=P (b a) ‘h (2),P’( givena) multiplication some evidence by =’s ‘e’; (2),then multiplication we would write by =’s the theorem as: | ⇤ 5. P (a b) P (b)=| P⇤(b a) P (a) (3),(4) 5. P (a b) P (b)=P (b a) P (a)| Bayes’⇤ P (theoremb a) P (a)| ⇤ (3),(4) | ⇤ P (b a) P (a)| ⇤ C. P (a b)= |P (b) (5), division by =’s C. P (a b)= | | P (h) P (e h) (5), division by =’s | P (b) P (h e)= | This conclusion| is Bayes’P ( theorem.e) Often, what we want to know is, intuitively, the probability This conclusion is Bayes’ theorem.of Often, some hypothesis what we ‘wanth’ given to know some evidence is, intuitively, ‘e’; then the we probability would write the theorem as: of some hypothesisNow suppose ‘h’ that given the somefirst ball evidence that comes ‘ eout,’; then again weat random, would is write a “3”. Bayes’ the theorem theorem can as: tell us, given the foregoing information, how likelyConsider it is that the what machine this before would us contains say about10 rather thethan 10,000 example balls. ofAccording the lottery to machine. Suppose for simplicity Bayesians, we figure this out by conditionalizingP (h on) P our(e h evidence) , i.e. finding the conditional probability of the P (h e)= | theoryP given(h) P the(e h evidence.) that you| knowP going(e) in that there are only two options, which are equally likely to be correct: P (h e)= | P (e) that there are 10 balls in the machine, and that there are 10,000. Let e be the evidence that the | Let “h” be the theory that the machine contains 10 balls. Then to compute the relevant conditional probability, we need to know three values. Considerfirst ball what to thiscome would out say is #3,about and the example let h be of the the lotteryhypothesis machine. that Suppose there for are simplicity 10 balls in the machine. Consider what this would say aboutthat the you example know going of the in lottery that there machine. are only Suppose two options, for simplicity which are equally likely to be correct: First, we need to know Pr(h). ThenFrom the we above, might we know say: that this is 0.5. that you know going in that therethat are there only are two 10 options, balls in the which machine, are equally and that likely there to are be 10,000. correct: Let e be the evidence that the that there areNext, 10 we balls need into know the machine,Pr(e |first h). Fairly ball and obviously, to that come there this out is are is 0.1. #3, 10,000. and let Leth bee be the the hypothesis evidence that that there the are 10 balls in the machine. Then we might say: first ball toBut come we also out need is #3, to know and the let unconditionalh beP the(h)=0 hypothesisprobability.5 of e, that Pr(e). there How should are 10 we balls figure in this the out? machine. Then we might say: A natural thought is that since we knowP (Ph that)=0(e heach)=0.5 of the.1 two hypotheses - 10 ball and 10,000 ball - have a probability of 0.5, Pr(e) should be the mean| of the conditional probabilities of e given those two hypotheses. In this case: P (Pe (he)=0)=0.1.5(0.1+0.0001) = 0.05005 P (h)=0.5 | P (e)=0.5(0.1+0.0001) = 0.05005 P (e h)=0.1 | 0.5 0.1 Then we find, via Bayes’ theorem, that P (h e)= 0.05005 =0.999. So, on the basis of the evidence P (e)=0Plugging.5(0. these1+0 numbers.0001) =into 0Then .Bayes’05005 we theorem, find, via we Bayes’get: theorem, that P (h e)= 0.5 0|.1 =0.999. So, on the basis of the evidence that the first ball to come out was #3, you0.05005 should revise your confidence in the 10-ball hypothesis that the first ball to come out was #3, you should| revise your confidence in the 10-ball hypothesis fromfrom 50% 50% to 99.9% to0 99.9%.5 certainty.0.1 certainty. Then we find, via Bayes’ theorem, that P (h e)= 0.05005 =0.999. So, on the basis of the evidence that the first ball to come out was #3, you should| revise your confidence in the 10-ball hypothesis Bayes’Bayes’ theorem theorem can canbe restated be restated in the following in the following way: way: from 50% toHence, 99.9% after certainty. you see the “3” ball come out of the lottery machine, you should revise the probability you assign to the 10-ball hypothesis from 0.5 to .999 - that is, you should switch from thinking that the machine has a 50% chance of Bayes’ theorembeing cana 10-ball be restatedmachine to in thinking the following that it has a way: 99.9% chance of being a 10-ball machine. 2 2

2 1.1 Bayes’ theorem

P (a b)= P (a&b) | P (b)

Hence, after you see the “3” ball come out of the lottery machine, you should revise the probability you assign to the 10-ball hypothesis from 0.5 to Consider.999 - that is, you what should this switch would from thinking say about that the themachine example has a 50% of chance the lottery of machine. Suppose for simplicity being a 10-ball machine to thinkingthat that you it has know a 99.9% going chance in of being that a there10-ball machine. are only two options, which are equally likely to be correct: that there are 10 balls in the machine, and that there are 10,000. Let e be the evidence that the An intuitive way to think about what this all means is in terms of what bets you would be willing to accept. If you think that something has a 50%ﬁrst chance ball of happening, to come then out you is should #3, be and willing let to hacceptbe theall bets hypothesis which give you that there are 10 balls in the machine. better than even odds, and willingThen to reject we all might bets which say: give you worse odds. Analogous remarks apply to different probability assignments.

This link between probability assignmentsP ( hand)=0 bets .is5 one way to bring out a strength of the Bayesian approach to belief formation. Following the Bayesian rule of conditionalization is the only way to avoid being subject to a Dutch book. P (e h)=0.1 | A Dutch book is a combination of bets which,P (e)=0 no matter.5(0 what.1+0 the outcome,.0001) =is sure 0.05005 to lose. For example, suppose that we have a 10 horse race, and you have the following views about the probabilities that certain horses will win. 0.5 0.1 Then we find, via Bayes’ theorem, that P (h e)= =0.999. So, on the basis of the evidence #3 has at least a 40% chance of winning | 0.05005 #6that has a the slightly first better ball than to even come chance out of waswinning #3, you should revise your confidence in the 10-ball hypothesis #8from has a 50%better than to 99.9%1 in 3 chance certainty. of winning If these are your probability assignments, then you should be willing to make the following three bets: Bayes’ theorem can be restated in the following way: $2 on #3 to win, at 3-2 odds $3 on #6 to win, at even odds $1.50 on #8 to win, at 3-1 odds 2 1.1 Bayes’ theorem

P (a b)= P (a&b) | P (b)

This link between probability assignmentsConsider and what bets is one this way would to bring say out a about strength theof the example Bayesian approach of the to lottery machine. Suppose for simplicity belief formation. Following the Bayesian rule of conditionalization is the only way to avoid being subject to a Dutch book. that you know going in that there are only two options, which are equally likely to be correct: that there are 10 balls in the machine, and that there are 10,000. Let e be the evidence that the A Dutch book is a combination of bets which, no matter what the outcome, is sure to lose. For example, suppose that we have a 10 horse race, andfirst you ballhave the to following come views out isabout #3, the andprobabilities let h thatbe certain the hypothesishorses will win. that there are 10 balls in the machine. Then we might say: #3 has at least a 40% chance of winning #6 has a slightly better than even chance of winning #8 has a betterP (h than)=0 1 in. 35 chance of winning P (e h)=0.1 If these are your probability assignments, then| you should be willing to make the following three bets: P (e)=0.5(0.1+0.0001) = 0.05005 $2 on #3 to win, at 3-2 odds $3 on #6 to win, at even odds $1.50 on #8 to win, at 3-1 odds 0.5 0.1 Then we find, via Bayes’ theorem, that P (h e)= =0.999. So, on the basis of the evidence | 0.05005 But this is a Dutch book. Can youthat see thewhy? first ball to come out was #3, you should revise your confidence in the 10-ball hypothesis from 50% to 99.9% certainty. One of the main arguments given in favor of updating your beliefs using Bayesian conditionalization, using Bayes’ theorem, is that other ways of updating your beliefs will leave you, in the above sense, subject to a Dutch book. And being subject to a Dutch book seemsBayes’ like theorema paradigm of can irrationality. be restated It seems insort theof like following the probabilistic way: version of having inconsistent beliefs. Hence it is a strength of Bayesianism that it avoids this situation.

2 1.1 Bayes’ theorem

P (a b)= P (a&b) | P (b)

With this Bayesian apparatusConsider in hand, let’s what return to this Leslie’s would argument. say about the example of the lottery machine. Suppose for simplicity that you know going in that there are only two options, which are equally likely to be correct: Leslie asks us to consider twothat hypotheses there about are the 10 future balls course in the of human machine, civilization: and that there are 10,000. Let e be the evidence that the

Doom Soon. The ﬁrsthuman ball race will to go come extinct out by 2150, is #3, with the and total let h be the hypothesis that there are 10 balls in the machine. humans born by theThen time of we such might extinction say: being 500 billion. Doom Delayed. The human race will go on for several thousand centuries, with the total humans born before the race goes extinct being 50 thousand billion. P (h)=0.5

Leslie thinks that we can use a certainP (kinde h of)=0 evidence.1 we have, along with Bayesian conditionalization, to show how likely these two hypotheses are.| To do this, though, we’ll ﬁrst have to think about how likely we think these two hypotheses are to beginP ( ewith.)=0.5(0.1+0.0001) = 0.05005

Doom Soom is pretty grim; it means that the human race will go extinct during the lives of your grandchildren.0.5 0.1 Then we find, via Bayes’ theorem, that P (h e)= =0.999. So, on the basis of the evidence Let’s suppose that we think that this is pretty unlikely - maybe that it has a 1% chance of happening.| 0 .Let’s05005 suppose (we’ll relax this assumptionthat the later) first that Doom ball toDelayed come is the out only was other #3, possibility, you should so that it has revise a 99% your confidence in the 10-ball hypothesis chance of happening. from 50% to 99.9% certainty. What evidence could we possibly have now to help us decide between these hypotheses now? Some obvious candidates spring to mind: theBayes’ proliferation theorem of nuclear can weapons; be restated astronomical in calculations the following of the probabilities way: of large asteroids colliding with the earth; prophecies involving the end of the Mayan calendar; etc. But the evidence that Leslie has in mind is of a different sort.

2 1.1 Bayes’ theorem

P (a b)= P (a&b) | P (b)

Using this definition of conditional probability, we can then argue as follows, assuming that P (b) = 0: 6 1. P (a b)= P (a&b) def. of conditional probability | P (b) 2. P (b a)= P (a&b) def. of conditional probability | P (a) 3. P (a b) P (b)=P (a&b) (1), multiplication by =’;s 4. P (a&| b)=⇤ P (b a) P (a) (2), multiplication by =’s 5. P (a b) P (b)=| P⇤(b a) P (a) (3),(4) | ⇤ P (b a) P (a)| ⇤ C. P (a b)= | (5), division by =’s | P (b) This conclusion is Bayes’ theorem. Often, what we want to know is, intuitively, the probability of some hypothesis ‘h’ given some evidence ‘e’; then we would write the theorem as: Doom Soon. The human race will go extinct by 2150, with the total Bayes’ theorem humans born by the time of such extinction being 500 billion. P (h) P (e h) P (h e)= | Doom Delayed. The human race will go on for several thousand | P (e) centuries, with the total humans born before the race goes extinct being 50 thousand billion. Consider what this would say about the example of the lottery machine. Suppose for simplicity Doom Soom is pretty grim; it means that the human race will go extinct during thethat lives of you your knowgrandchildren. going in that there are only two options, which are equally likely to be correct: Let’s suppose that we think that this is pretty unlikely - maybe that it has a 1% chancethat of there happening. are Let’s 10 balls in the machine, and that there are 10,000. Let e be the evidence that the suppose (we’ll relax this assumption later) that Doom Delayed is the only other possibility, so that it has a 99% chance of happening. first ball to come out is #3, and let h be the hypothesis that there are 10 balls in the machine. Then we might say: What evidence could we possibly have now to help us decide between these hypotheses now? Some obvious candidates spring to mind: the proliferation of nuclear weapons; astronomical calculations of the probabilities of large asteroids colliding with the earth; prophecies involving the end of the Mayan calendar; etc. But the evidence that Leslie has in mind is of a different sort. P (h)=0.5 P (e h)=0.1 That evidence is: each of us is one of the first 50 billion human beings born. Let’s now ask, using| the Bayesian method, what probability, in light of this evidence, we should assign to Doom Soon (DS) andP ( Doome)=0 Delayed.5(0. 1+0.0001) = 0.05005 (DD).

First, what is the probability of our evidence, conditional on the truth of DS? It seems that it should be 0.1; if there 0.5 0.1 will be 500 billion total human beings, then I have a 1 in 10 chance of being amongThen the first we 50 find, billion born. via Bayes’ theorem, that P (h e)= 0.05005 =0.999. So, on the basis of the evidence that the first ball to come out was #3, you should| revise your confidence in the 10-ball hypothesis Second, what is Pr(e | DD)? By analogous reasoning, it seems to be 1 in 1000, orfrom .001. 50% to 99.9% certainty. But what is the probability of e, by itself? On the current assumption that either DS or DD is true, and that DD is 99 times more likely to be true, it seems that Pr(e) should be equal to: Bayes’ theorem can be restated in the following way:

[99 * Pr(e | DD) + 1 * Pr(e | DS)] / 100 = .00199

2 1.1 Bayes’ theorem

P (a b)= P (a&b) | P (b)

0.5 0.1 P (DS) P (e DS) .01 .1 Then we find, via Bayes’ theorem, that P (h e)= =0.999. So, on the basis of the evidence P (DS e)= ⇤ | = ⇤ = .503 0.05005 | P (e) .00199 that the first ball to come out was #3, you should| revise your confidence in the 10-ball hypothesis from 50% to 99.9% certainty. P (DD) P (e DD) .99 .001 P (DD e)= ⇤ | = ⇤ = .497 | P (e) .00199 Bayes’ theorem can be restated in the following way:

P (a b)= P (a&b) | P (b)

So, even if we begin by thinking that the probability of Doom Soon is only 1%, reflection on the simple fact that we are born among the first 50 billion humans shows that we should think that there is a greaterP (h)=0 than 50%.5 chance that the human race will be extinct in the next 150 years. P (e h)=0.1 | Varying our initial assumptions changes the outcome dramatically. Suppose that reflectionP upon(e)=0 the risk.5(0 of .1+0.0001) = 0.05005 climate change, nuclear war, etc. makes us think that we should assign DS and DD roughly equal initial probabilities - suppose we think, before considering the fact that we were born in the first 50 billion people, that the probability of each is approximately 0.5. On this assumption, the probability of Doom Soon after 0.5 0.1 conditionalizing on our evidence is just over 99%. Then we find, via Bayes’ theorem, that P (h e)= 0.05005 =0.999. So, on the basis of the evidence that the first ball to come out was #3, you should| revise your confidence in the 10-ball hypothesis Alternatively, we might think (as Leslie says, p. 202) that if the human race survivesfrom past 50%2050, then to 99.9%it will likely certainty. colonize other planets, making it quite likely that the population of humans will ultimately grow to be at least 50 million billion (50 quadrillion). On this assumption, even if we begin by assigning Doom Soon a chance of only 1%, conditionalizing on the evidence that we were among the first 50 billion born Bayes’gives Doom theorem Soon a probability can be restated in the following way: of 99.9%.

2 1.1 Bayes’ theorem

P (a b)= P (a&b) | P (b)

Using this definition of conditional probability, we can then argue as follows, assuming that P (b) = 0: 6 1. P (a b)= P (a&b) def. of conditional probability | P (b) 2. P (b a)= P (a&b) def. of conditional probability | P (a) 3. P (a b) P (b)=P (a&b) (1), multiplication by =’;s 4. P (a&| b)=⇤ P (b a) P (a) (2), multiplication by =’s 5. P (a b) P (b)=| P⇤(b a) P (a) (3),(4) | ⇤ P (b a) P (a)| ⇤ C. P (a b)= | (5), division by =’s | P (b) This conclusion is Bayes’ theorem. Often, what we want to know is, intuitively, the probability of some hypothesis ‘h’ given some evidence ‘e’; then we would write the theorem as: Doom Soon. The human race will go extinct by 2150, with the total Bayes’ theorem humans born by the time of such extinction being 500 billion. P (h) P (e h) P (h e)= | Doom Delayed. The human race will go on for several thousand | P (e) centuries, with the total humans born before the race goes extinct being 50 thousand billion. Consider what this would say about the example of the lottery machine. Suppose for simplicity So, even if we begin by thinking that the probability of Doom Soon is only 1%, reflectionthat on you the simple know fact going that in that there are only two options, which are equally likely to be correct: we are born among the first 50 billion humans shows that we should think that therethat is a greater there than are 50% 10 balls in the machine, and that there are 10,000. Let e be the evidence that the chance that the human race will be extinct in the next 150 years. first ball to come out is #3, and let h be the hypothesis that there are 10 balls in the machine. Varying our initial assumptions changes the outcome dramatically. Suppose that reflectionThen upon we mightthe risk of say: climate change, nuclear war, etc. makes us think that we should assign DS and DD roughly equal initial probabilities - suppose we think, before considering the fact that we were born in the first 50 billion people, that the probability of each is approximately 0.5. On this assumption, the probability of Doom SoonP (h after)=0 .5 conditionalizing on our evidence is just over 99%. P (e h)=0.1 | Alternatively, we might think (as Leslie says, p. 202) that if the human race survives past 2050,P (e then)=0 it will.5(0 likely.1+0 .0001) = 0.05005 colonize other planets, making it quite likely that the population of humans will ultimately grow to be at least 50 million billion (50 quadrillion). On this assumption, even if we begin by assigning Doom Soon a chance of only 1%, conditionalizing on the evidence that we were among the first 50 billion born gives Doom Soon a probability 0.5 0.1 of 99.9%. Then we find, via Bayes’ theorem, that P (h e)= 0.05005 =0.999. So, on the basis of the evidence that the first ball to come out was #3, you should| revise your confidence in the 10-ball hypothesis The numbers are surprising. But more surprising than the numbers is the way we arrived at them. One thinks of revising one’s view about the likelihood of Doom Soon based on empirical claims aboutfrom nuclear 50% weapons, to 99.9% certainty. climate change, asteroids, etc. It seems crazy that such dramatic changes in our view about the extinction of the human race should result from mere reflection on how many human beings have beenBayes’ born. theorem can be restated in the following way:

This is why this argument deserves to be considered a paradox. We have a plausible argument from Leslie that we should radically change our view of the future based on the number of human beings who have lived, but it seems clear that it is unreasonable to change our views on this topic for this reason. 2 This is why this argument deserves to be considered a paradox. We have a plausible argument from Leslie that we should radically change our view of the future based on the number of human beings who have lived, but it seems clear that it is unreasonable to change our views on this topic for this reason.

We’ll move on shortly to consider some objections to Leslie’s argument. But ﬁrst lets consider another, even more surprising, application of the same sort of reasoning, due to Nick Bostrom (in “Are you living in a computer simulation?”).

Suppose we met some Martians, or other beings from another planet, who were made of entirely different materials from us - suppose, for example, that they were composed almost entirely of silicon. Could such beings be conscious?

It seems clear that they could; beings could be conscious even if made of radically different materials than us. But what makes such beings conscious? Plausibly, what makes them conscious is not what they are made of, but how what they are made of is organized; even something made of silicon could be conscious if its silicon parts were arranged in a structure as complex as a human brain.

If this is true, then it seems clear that computers could be conscious. Indeed, given reasonable assumptions about future increases in the computing power available to us, it seems quite plausible that in the relatively near future computers will be able to run simulations of human life which contain many, many conscious beings.

With continuing increases in computing power, there will presumably come a time in which a vast, vast majority of all conscious beings to have existed will have existed only in computer simulations.

Are you living in a computer simulation right now? It seems that reasoning analogous to that used in the doomsday argument suggests that it is reasonable for you to believe that you are. After all, if the hypothesis that the vast number of conscious beings to have existed exist only in a computer simulation is true, then the probability that you do not exist in a computer simulation is extremely low.

Of course, there is another option: human beings could become extinct before the relevant advances in computing power are realized. So, it seems, it is overwhelmingly likely that either you are living in a computer simulation, or human beings are soon to become extinct. How might one respond to Leslie’s argument? (We’ll leave it an open question whether the responses we consider will also apply to the “computer simulation” argument.)

One could of course respond that the Bayesian apparatus on which it depends is faulty. But that would seem hasty; let’s consider some other possibilities.

One possibility is that the problem begins with the obviously false assumption that Doom Soon and Doom Delayed are the only relevant possibilities. Would Leslie’s argument still work if we relaxed this assumption, and took into account the fact that there are many possible futures for the human species? What would the analogous situation be with the example of the numbered balls and the lottery machine?

Let’s consider a different line of objection:

Look, we know that something must be wrong with this way of arguing. After all, couldn't someone in ancient Rome have used this reasoning to show that the end the world would come before 500 AD? And wouldn't they have been wrong? So mustn't there also be something wrong with our using this reasoning?

Is this a good objection?

A more promising line of objection focuses on the apparent assumption that one’s location in the birth order of human beings is random. Leslie asks you to, in effect, assimilate this case to the lottery ball example. But why think that the fact that I am born in the ﬁrst 50 billion people is relevantly analogous to the number “50 billion” coming out of the lottery machine?

To develop this objection, let’s think more closely about exactly which assumptions are involved in the lottery machine example. (This follows the discussion in Mark Greenberg’s “Apocalypse not just yet.”) Doom Soon. The human race will go extinct by 2150, with Doom Delayed. The human race will go on for several the total humans born by the time of such extinction being thousand centuries, with the total humans born before 500 billion. the race goes extinct being 50 thousand billion.

An analogous case would be a lottery machine which we know to contain either 500 billion or 50 thousand billion balls. To make the numbers smaller, let’s think of a pair of lottery machines with, respectively, 500 and 50,000 balls. Suppose you are confronted with a lottery machine which you know to be of one of the types just described.

Suppose now that the lottery machine spits out a ball which reads “50.” (This is supposed to be analogous to ﬁnding that you are among the ﬁrst 50 billion people born.) Isn’t this evidence, as Leslie says, that the machine has 500 rather than 50,000 balls in it?

It is - but only if we make two assumptions about the machines.

First, we must assume that the ball which comes out is randomly selected by the lottery machine. If, for example, balls with numbers higher than “100” are slightly larger and never come out of the machine, then obviously the fact that a ball labeled “50” came out of the machine would be no evidence at all about how many balls are in the machine.

Second, and just as important, we must assume that the ball with the label “50” on it would still be in the urn if it contained 500 balls. Doom Soon. The human race will go extinct by 2150, with Doom Delayed. The human race will go on for several the total humans born by the time of such extinction being thousand centuries, with the total humans born before 500 billion. the race goes extinct being 50 thousand billion.

It is - but only if we make two assumptions about the machines.

Second, and just as important, we must assume that the ball with the label “50” on it would still be in the urn if it contained 500 balls.

This second condition is easy to miss, since we of course assume that a machine with N balls in it contains those balls labeled 1-N. But of course things don’t have to work this way. Suppose that the 500 balls in the 500-ball machine were randomly selected from the balls in the 50,000 ball machine. Then the “50” ball might not be in the 500-ball machine. In this case, would the emergence of the “50” ball from the machine count in favor of the hypothesis that the machine before us contains 500 rather than 50,000 balls?

The problem is that any way of understanding the analogy between the Doomsday argument and the lottery machine example seems to violate one of the two assumptions needed to legitimate the reasoning used in the lottery machine case.

To see this, think about the question: what is the analogue in the Doomsday argument of the numbers written on the balls in the lottery machine case? Doom Soon. The human race will go extinct by 2150, with Doom Delayed. The human race will go on for several the total humans born by the time of such extinction being thousand centuries, with the total humans born before 500 billion. the race goes extinct being 50 thousand billion.

It is - but only if we make two assumptions about the machines.

Second, and just as important, we must assume that the ball with the label “50” on it would still be in the urn if it contained 500 balls.

To see this, think about the question: what is the analogue in the Doomsday argument of the numbers written on the balls in the lottery machine case?

Here’s one idea: the number “written on you” is the number you happen to be in the birth order of the human species. So (let’s suppose) my number is “50 billion” because I just happened to be the 50 billionth person born.

On this idea, people don’t have “built in” numbers: rather, they are assigned the numbers they get based on when they are born.

But now imagine that the lottery machine worked this way. On this view, there are an undisclosed number of balls in the machine, none of which have numbers written on them. We write the numbers on the balls as they come out of the machine, beginning with “1.” Now suppose we get to “50.” Is the fact that a ball with such a low number came out evidence that we have a 500-ball machine before us? Clearly not, because the ﬁrst assumption - the assumption of random selection - is violated. Doom Soon. The human race will go extinct by 2150, with Doom Delayed. The human race will go on for several the total humans born by the time of such extinction being thousand centuries, with the total humans born before 500 billion. the race goes extinct being 50 thousand billion.

It is - but only if we make two assumptions about the machines.

Second, and just as important, we must assume that the ball with the label “50” on it would still be in the urn if it contained 500 balls.

So let’s suppose instead that every human being comes with a “built in” number, just like the lottery balls have numbers written on them prior to their emergence from the lottery machine.

There are two problems with this suggestion. First, how do we know what anyone’s number is, on this way of viewing the analogy? What does it even mean to say that people have a certain number?

Second, if we can come up with a way of assigning numbers to all the people that could have existed - we violate the second assumption. For there is no guarantee that, if N people exist, they will have the numbers 1-N. This is just like the case in which 500 balls are randomly selected from the 50,000 ball machine - in this case, even if we somehow knew that my number was 50 billion - this would provide us no evidence at all about how many human beings will eventually exist.