Monday, November 5, 2012

Infinity and probability

You are one of a countably infinite number of blindfolded people arranged in a line. You have no idea where in the line you are. Each person tosses a independent fair die, but doesn't get to see or feel the result.

Case 1: What is the probability you tossed a six?

Obviously 1/6.

Case 2: You are informed that, surprisingly, all the even-numbered people tossed sixes, and all the odd numbered people tossed something else. What is the probability you tossed a six?

It sure seems like it's no longer 1/6. For suppose it's still 1/6. Then when you learned that all the even-numbered people tossed sixes and the odd-numbered ones didn't, you thereby received evidence yielding probability 5/6 that you were one of the odd-numbered ones. But surely you didn't. After all, if there were ten people in the line and you learned that all but the tenth tossed a six, that wouldn't give you probability 5/6 that you were the tenth!

But as long as infinitely many people tossed six and infinitely many didn't (and with probability one, this is true), there is always some ordering on the people such that relative to that ordering you can be correctly informed that every second person tossed a six. That puts the judgment in Case 1 into question.

Moreover, why should the alternating ordering in Case 2 carry any weight that isn't already carried by the fact which you know ahead of time (with probability one) that there are equally many sixes as non-sixes?

But if the judgment in Case 1 is wrong, then were we to find out we are in an infinite multiverse, that would undercut our probabilistic reasoning which assumes we can go from intra-universe chances to credences, and hence undercut science. Thus, a scientific argument that we live in an infinite multiverse would be self-defeating.

I don't exactly know what to make of arguments like this.

16 comments:

residentoftartarus said...

I think the correct answer to case 2 is 1/2 via the following reasoning:

A = You throw a six.

B = Your place in line is even-numbered.

C = All the even-numbered people tossed sixes.

Now, P(A|C) = P(B|C) since (A & C) is the same event as (B & C), and P(B|C) = P(B) since B and C are independent events. Ergo, P(A|C) = P(B|C) = P(B) = 1/2.

Similarly, in the case of ten people in line we have the following computation:

A = You throw a six.

B = You're the tenth person in line

C = Everyone but the tenth person tossed a six.

Then P(A|C) = P(B|C) = P(B) = 1/10 via the exact same reasoning. Indeed, it's quite obvious that (A & C) is the same event as (B & C) and we can manually check that B and C are independent events in this case by verifying that P(B & C) = P(B)P(C).

Alexander R Pruss said...

What I am not clear on is whether P(B)=1/2 in case 2.

residentoftartarus said...

I think your confusion stems from not clearly differentiating between two different possibilities.

In case 2 of your entry, the even-numbered people toss sixes subsequent to everyone in line being ordered in some way, in which case I believe the analysis I gave in the previous comment follows.

However, if we change the situation in case 2 so that the blindfolded people are intelligently ordered subsequent to their toss in order to achieve the desired result of having all the even-numbered people toss a six (insofar as this is possible) then I believe the probability in that case is 1/6 via the following reasoning:

A = You throw a six.

B = All the even-numbered people tossed sixes.

C = An infinite number of people tossed a six.

Now, P(B) = P(C) since B and C are the same event, but then it follows that P(B) = 1 since P(C) = 1. Moreover, P(A|B) = P(A) since P(B) = 1, hence P(A|B) = 1/6.

Alexander R Pruss said...

By "ordering" I meant an abstract total ordering relation. Set-theoretically speaking, all of the orderings exist. There is the standard numerical ordering. But there are infinitely many others.

residentoftartarus said...

Ah, I just now saw your reply.

As per my last comment, if B = {your place in line is even-numbered} then P(B) will depend on your setup. However, if the blindfolded people are randomly ordered independent of anything else then it seems clear to me that P(B) should equal 1/2.

residentoftartarus said...

If you just mean some abstract total ordering then I am quite confident that the probability of any individual being even-numbered should is 1/2; however, I would concede that any mathematical demonstration of this would involve some technical details.

residentoftartarus said...

Omega = the set of all total orderings on the positive integers = the set of all bijections from the positive integers to the positive integers.

F = The sigma-algebra generated by the subsets {g in Omega | g(i) is divisible by j}, where i and j are any positive integers.

Now, P({g in Omega | g(i) is divisible by j}) = 1/j induces the natural probability measure on F. But then it follows from the natural probability measure that P({g in Omega | g(i) is divisible by 2}) = 1/2 for any positive integer i.

Thomas Larsen said...

"Thus, a scientific argument that we live in an infinite multiverse would be self-defeating."

There are good arguments to the effect that, on an unrestricted multiverse, it's far more likely than not that I am a Boltzmann brain - which would undermine any scientific basis I might have for an unrestricted multiverse in the first place.

Gill said...

Either I missed something or there isn't any puzzle here.

In case 1, 1/6 is correct only if the die is assumed to be symmetrical. I actually doubt whether we are justified to make that assumption. But maybe in the absence of evidence to the contrary, we are.

In case 2, however, since people who tossed even all got sixes, we do have strong evidence that the die isn't symmetrical. And we have no evidence as to how it isn't, which is crucial, since the different ways in which the die might be asymmetrical would give rise to different probability distributions. So in that case all bets should be off. We simply can't make warranted probability judgments.

Did I miss anything?

Alexander R Pruss said...

Gill:

I am assuming, throughout, that we are certain that it's a fair die, i.e., one that has equal chances of landing on each side.

Mr Larsen:

1. Physicists are trying to find measures on multiverses where it's not likely that we're Boltzmann brains. Maybe they'll succeed.

2. But in any case, if the line of thought in this post pans out, then the Boltzmann brain argument fails--as probabilistic reasoning breaks down entirely in such contexts.

residentoftartarus:

One problem is that in the case where we have the sequence where even numbered items are sixes and odd-numbered ones are non-sixes, there is an alternate ordering, where in the exact same physical situation instead the items whose positions are divisible by 3 are sixes. For instances, the ordering can be given by 0, 1, 3, 2, 5, 7, 4, 9, 11, 6, ....

Let B = your place in line is even in the standard ordering.
B1 = your place in line is divisible by three in the alternate ordering.

If P(B)=1/2, then by parity of reasoning P(B1)=1/3. But B holds if and only if B1 does.

Now maybe the standard ordering is somehow probabilistically special. But why?

residentoftartarus said...

Alex,

If you change B and B1 as follows:

B = {your place in line is even} = {g in Omega | g(1) is divisible by two}

B1 = {your place in line is divisible by three} = {g in Omega | g(1) is divisible by three}

Then P(B) = 1/2 and P(B1) = 1/3 according to the probability measure I gave earlier and it's not the case that B is equivalent to B1 in event space so that there's no contradiction.

The way this works is that you privilege some fixed numbering of the blindfolded people such that you are given the number 1, and then identify the set of all numberings of the blindfolded people as the set of all bijections from the positive integers to itself.

Alexander R Pruss said...

But why should you privilege some ordering?

residentoftartarus said...

You privilege some numbering of the blindfolded people so that you can identify the set of all numberings of the blindfolded people as the set of all bijections from {positive integers} to {positive integers}.

So, consider some function f from {blindfolded people} to {positive integers} such that f(you) = 1 (note that f acts as a numbering of the blindfolded people). Then for every bijection g from {positive integers} to {positive integers} we have a numbering (g o f) from {blindfolded people} to {positive integers}. Indeed, this is how we identify the aforementioned set of all numberings with the relevant bijections.

Of course, it hardly matters what numbering we privilege at the beginning of our analysis.

Alexander R Pruss said...

Let's make this a little more concrete. Suppose that the people are arranged in an infinite two-dimensional field, with clearly defined x and y axes. Each axis defines an ordering. Suppose that ordered by x coordinates the pattern is:
six non-six six non-six six non-six
And ordered by y coordinates the pattern is:
six non-six non-six six non-six non-six non-six non-six ...

You don't know where in the field you are.

What is your probability of six?

Well, going by the reasoning in Case 2 as applied to the x-axis ordering, your probability of six is 1/2. But applying this reasoning to the y-axis ordering, your probability of six is 1/3.

And quite possibly, if you order by Euclidean distance from some origin point, you get an even different probability.

So which is it?

residentoftartarus said...

Alex,

In your field example the x and y axes are giving you two different random variables X and Y and the analysis proceeds as follows:

P({you tossed a six}) = P({X(1) is divisible by two} or {Y(1) is divisible by three}) = P({X(1) is divisible by two}) + P({Y(1) is divisible by three}) - P({X(1) is divisible by two} and {Y(1) is divisible by three}) = 1/2 + 1/3 - 1/6 = 2/3.

Nevertheless, I am beginning to see your point and will think about this more later.

residentoftartarus said...

Alex,

Sorry, the analysis in my last comment is wrong.

Here's the setup for your field example:

A = You throw a six.

B = X(1) is a multiple of two.

C = Y(1) is a multiple of three.

D = X^-1({multiples of two}) = Y^-1({multiples of three}) are precisely the people who throw sixes.

Now, P(A|D) = P(B|D) = P(C|D) since (A and D) = (B and D) = (C and D) just as before. However, B and C are not independent of D since in that case 1/2 = P(B) = P(B|D) = P(C|D) = 1/3. In other words, the lesson of your field example is that the independence assumption I made in my first comment does not necessarily, which is fine by me since I always found that assumption to be somewhat problematic.

The computation of P(A|D) is still an open problem for me at this point.