## Mathematical Conundrum or Not? Number Five

• 404
You forgot Heads and Wednesday, Heads and Thursday, Heads and Friday. . . . . and on forever.

Then do the same with Tails.

I was assuming the two possible interview days in the experiment. But since the probabilities are all equal, it doesn't matter how large the background sample space is (at least in a finite universe). On conditionalization, all the probability will still be distributed equally between H1, T1 and T2.

And in the process don't forget the terms on the days Beauty will be awakened AND interviewed (AKA the sample space), was defined before the experiment started.

Yes. So do you see any problem with my approach in principle?
• 872
. But since the probabilities are all equal, it doesn't matter how large the background sample space is (at least in a finite universe).

Oh really? So you think they are the same thing.

If your distribution is the same as Elga's then they should have the same center.

Let's find out, by comparing the means.

Egla's distribution: (1/3+1/3+1/3)/3=1/3

Your distribution: (1/3+1/3+1/3+0)/4= 1/4

So in your distribution, on average Beauty will get one out of every four attempts correct, because oddly enough you are actually suggesting zero is in that distribution of possible outcomes. When working with sets zero is not the same thing as null.
• 872

Let's look at this from the ground up.

A sample space is the set of all possible outcomes of a random process.

An event is a subset of that sample space.

Let E be the event and let S be the sample space.

Then the Equally Likely Probability Formula is:

P(E) = the number of outcomes in E/ the number of outcomes in S or P(E) = N(E)/N(S)

(N(R) is just the number of elements in R)

Now just to clear it up, in set theory {A,B,C} is not equal to {A,B,C,0}. Let {A,B,C} bet set 1, and let {A,B,C,0} bet set 2.

Consider,

We randomly select one element from each sample space, then our possible outcomes are:

For set 1: A or B or C

For set 2: A or B or C or 0

That means for set 2 there is a one in four chance of 0 being selected. Put that in the context of our problem and that really does not make any sense.

Now conditional probability is the the probability of event K given that event L has already occurred.

---

Consider this argument:

Our sample space is {H,T} with P(H)=N(H)/N(S) or P(H) = 1/2.

Let H equal the set {M1} and let T equal the set {M2, Tu}

Where,

M1 equals Monday and Heads
M2 equals Monday and Tails
Tu equals Tails and Tuesday.

So now our possible events are sets. Set H has one element and set T has two elements, each with a 50% chance of being selected. Then P(M2) or P(Tu), given the event T, by our Equally Likely Probability Formula is P(M2) = 1/4 and P(Tu) = 1/4. Given Tails she could be in P(M2) or P(Tu), so 1/4 + 1/4 = 1/2 therefore P(M2) + P(Tu) = 1/2 = P(T) = P(H).
• 404
Oh really? So you think they are the same thing.

I was referring to the background sample space in my earlier post that included all combinations of day outcomes and coin toss outcomes and that are assigned the same probability. I'm saying that it doesn't matter what size the background sample space is, as long as it is finite and the elements have equal probabilities.

I'm then conditionalizing on Beauty being awake (and interviewed) to produce a second sample space with only three elements {H1, T1, T2} with 1/3 probability each. That second sample space is the relevant sample space and matches Elgas.

The second probability distribution from my earlier post should have been:

       Mon  Tue
Heads  1/3
Tails  1/3  1/3

• 1.6k

Thanks. I think I finally understand the halfer position. (The one thing I'm not completely clear on is how the Monday interview is retroactively determined to be a single or half of a double in the variant where the coin is tossed after the first interview.)

What puzzles me is why Beauty would reason this way.

My Beauty reasons this way, as I've said before:
(1) If I knew it was Monday, I'd know it could be heads or tails, even chance.
(2) If I knew it was Tuesday, I'd know it was tails.
(3) I know I'll be interviewed on Monday, but interviewed on Tuesday only half the time.
(4) Therefore my weighted expectation of heads is 2/3(1/2) + 1/3(0) = 1/3

The halfer Beauty reasons this way:
(1) If I knew I was in the single interview track, I'd know it was heads.
(2) If I knew I was in the double interview track, I'd know it was tails.
(3) I'm in the first track half the time and in the second half the time.
(4) Therefore my weighted expectation is 1/2(1) + 1/2(0) = 1/2

But this is just pretend reasoning.

It's like "working out" your expectation of heads in a simple coin toss this way:
(1) If I knew it was heads, I'd know it was heads.
(2) If I knew it was tails, I'd know it was tails.
(3) It's heads half the time and tails half the time.
(4) Therefore my weighted expectation is 1/2(1) + 1/2(0) = 1/2

What's the point of that?

And indeed, Lewis's "proof" has but a single step.

(No argument in this post, just clearing my head.)
• 1.6k
Continuing:

How many times does Beauty expect to be asked for her credence?
(1) If I knew it was heads, I'd know I'll be asked once.
(2) If I knew it was tails, I'd know I'll be asked twice.
(3) It's heads half the time and tails half the time.
(4) Therefore my weighted expectation is 1/2(1) + 1/2(2) = 3/2.

If I'm in the single interview track, and I am half the time, I get 2/3 of my expected interviews.
If I'm in the double interview track, and I am half the time, I get 4/3 of my expected interviews.

My expectation for getting to say "heads" and be right, because I'm in the single interview track, is 2/3(50%) = 1/3.
My expectation for getting to say "tails" and be right, because I'm in the double interview track, is 4/3(50%) = 2/3.
• 1.6k
H has one element and set T has two elements, each with a 50% chance of being selected.

When does anyone ever make a random selection from among only the tails interviews?
• 872
The 50% each applies to H and T; not the elements in the sets, but the sets themselves.
• 1.6k

I misunderstood.

But you are going to say exactly this about M2 and Tu (or T1 and T2), so the question stands.
• 872

In the event of Tails, Beauty will be awakened on Monday and Tuesday, but due to the nature of the experiment she will not be able to tell the difference, either one is equally likely when interviewed. Hence P(M2) = P(Tu) =1/4. It is 1/4 as only a total of 50% was allotted to T.

This temporal uncertainty, is actually where the 1/3 argument is placed. If the uncertainty is about her location in time, and probability is the measure of uncertainty then shouldn't her sample space be {M, T1, T2}?

It all depended on which uncertainty Beauty decides to consider.
• 1.6k
In the event of Tails, Beauty will be awakened on Monday and Tuesday, but due to the nature of the experiment she will not be able to tell the difference, either one is equally likely when interviewed

So what? It's not a situation that arises. Neither she not the experimenters are ever in the position of knowing that the coin landed tails but wondering what day it is. Beauty only wonders what day it is to figure out how the coin landed.

Suppose there was another coin toss to determine whether heads was the single interview or the double this time around. Then half the time heads would be 1/3 of the interviews, and half the time 2/3, so heads would on average be half the interviews and same for tails.

But that is not the case here. The interviews are not randomly distributed.

Beauty knows that when she is asked for her credence, 1/3 of the time the coin has landed (or will land?) heads and 2/3 of the time the coin has landed (or will land?) tails.

Therefore her credence that the coin has landed (or will land?) heads must be 1/3.
• 872
So what? It's not a situation that arises. Neither she not the experimenters are ever in the position of knowing that the coin landed tails but wondering what day it is. Beauty only wonders what day it is to figure out how the coin landed.

If my position was that she somehow magically knew it was tails, then why would I claim it has a 50% uncertainty? If she knew it was tails it would be a 100% certainty and M2 = 50% = Tu with temporal uncertainty. The 50% is the uncertainty of T or H and the 25% is the uncertainty of M2 or Tu.

The conditional probability of tails given that it is M2 or Tu is P(T|M2)= P(.5|.25) = .125/.25 = 1/2 = P(T|Tu) which is equal to P(H). Both days still have the same 50% uncertainty when considering H or T. In fact from that direction all three days have the same uncertainty when considering H or T, which is where we get the 1/3 argument.

The interviews are not randomly distributed.

Randomly just means equal probability. Figured I should clear that up. In the technical sense when talking about random, it means each element in the sample space has an equal chance of being selected. So in the event of tails, on any given consideration of the interview between M2 and Tu Beauty, has a 25% uncertainty of being in either of them. It is 25% and not 50% because of the uncertainty in it being Tails. We are stacking uncertainties.

How it comes out all depends on the considerations of the uncertainties.
• 1.6k
The conditional probability of tails given that it is M2 or Tu is P(T|M2)= P(.5|.25) = .125/.25 = 1/2 = P(T|Tu) which is equal to P(H).

This is still slightly puzzling to me.

P(H | M1) = 1, right? And this is the thing about the double interview track: both them happen if and only if the coin lands tails. From your calculation, P(T | M2 v Tu) = 1, yes? But it should be P(T | M2) = P(T | Tu) = 1, and P(T) = P(M2) = P(Tu) = 1/2. You always get both on tails. You get them one at a time, but we don't necessarily care.

That space of three possibilities, {M1, M2, Tu} has three elements each of which has an unconditional probability of 50%. Conditioned on the whole space, they'll each be 33%.
• 872
P(H | M1) = 1, right? And this is the thing about the double interview track: both them happen if and only if the coin lands tails. From your calculation, P(T | M2 v Tu) = 1, yes? But it should be P(T | M2) = P(T | Tu) = 1, and P(T) = P(M2) = P(Tu) = 1/2. You always get both on tails. You get them one at a time, but we don't necessarily care.

That space of three possibilities, {M1, M2, Tu} has three elements each of which has an unconditional probability of 50%. Conditioned on the whole space, they'll each be 33%.

This is why I keep saying it depends on how Beauty decides to consider her uncertainties.

Remember conditional probability is the the probability of event K given that event L has already occurred. Order matters.

So what is T given M2 or Tu? 50%

Hence, .50 +.50 +.50 = 1.5 > 1. The sum of probability cannot be greater than 1 and since P(H|M1)=P(T|M2)=P(T|Tu) we reallocate the credibility to 1/3 each.

That is when Beauty is considering the uncertainty of her location in time.

Now what if Beauty considers instead the uncertainty of H or T?

Then, purely for demonstration, what is M2 given the event T?

P(M2|T) = P(.25|.5) = .125/.5 = 1/4.

However, if Beauty was considering the uncertainty of H or T, and not her location in time, the only reason to consider the conditional probability of M2 would be for completion; practically she could end at the uncertainty of H or T.

The real issue here is not that we get two different yet seemly reasonable answers; this is not about 1/2 vs 1/3. What the Sleeping Beauty Problem demonstrates is that decision affects the outcome of her solution.

What I find interesting is that this decision also is very likely an unconscious decision. Which may be why we get people who are convinced it is 1/3 and people who are convinced it is 1/2. The unconscious mind made a decision for them on how to consider the probability, a decision that consciously they were never aware of.
• 1.6k

Something I don't remember us talking about: should Beauty, knowing the rules of the experiment, subject her expectation of a tails interview to a discount? It occurs to me that this may be the regime Lewis is describing.

Here's a physical version. You decide to test if a coin is fair by throwing a red marble in an urn on heads, and a blue marble if tails. After a bunch of flips, you'll count the marbles, expecting them to be about equal. Drawing a marble randomly will have the same distribution as the coin itself.

Suppose instead on tails you throw in two blue marbles. Then you'd expect a 2:1 ratio if the coin is fair. A randomly selected marble is now twice as likely to be blue, but each blue is discounted, and is only half the total available evidence of a tails flip, unlike the reds each of which is all the evidence of a heads. Each blue does represent a tails, certainly, and only got in the urn because a tails was tossed, but there's another blue that's evidence of the same toss.

Now suppose the marbles are all white. Still true that you're twice as likely to draw a marble representing a tails toss, but you have to discount.
• 1.6k
@Michael I think I'm a halfer now. (Still some things I'd like to be clearer on.)

@Andrew M, @andrewk, @JeffJo: do you find this argument as convincing as I do?

@Jeremiah
each blue is discounted, and is only half the total available evidence of a tails flip, unlike the reds each of which is all the evidence of a heads

Is this what the conditional probabilities we've been talking about are trying to express? I'm still not clear on how this idea is formalized.
• 1.6k
@Jeremiah
It's as we were discussing: each marble represents an interview event. To count coin toss outcomes you only need one red marble, but two blues, to make up the entire double interview event. Each blue marble is one kind of event, but that event is half of the kind of event we want to count.
• 404
A randomly selected marble is now twice as likely to be blue

That's all we're asking Beauty about. That is, the probability that the next marble to be drawn (or the interview that is being conducted) will be associated with a heads outcome. Which is 1/3.

but each blue is discounted, and is only half the total available evidence of a tails flip, unlike the reds each of which is all the evidence of a heads.

Discounting doesn't help Beauty. Suppose she is a halfer. Should she discount the next drawn marble (or current interview) by 1/2 or not? Well, she should 2/3 of the time. So 2/3 * (1/2 * 1/2) + 1/3 * 1/2 = 1/3. The thirder says that is the real probability of each state for her.
• 1.6k
Discounting doesn't help Beauty. Suppose she is a halfer. Should she discount the next drawn marble (or current interview) by 1/2 or not? Well, she should 2/3 of the time. So 2/3 * (1/2 * 1/2) + 1/3 * 1/2 = 1/3. The thirder says that is the real probability of each state for her.

This is the wrong model. This table
    Mon   Tue
H   1/2
T   1/4   1/4

is right, and here's why.

Suppose you have a machine set up like this: there's a hopper full of red marbles and a hopper with twice as many blue marbles; you push one button and it transfers a single red marble or two blue marbles to another hopper, one you can't see; you push a different button and it dispenses one of the marbles from the small hopper. What are your odds of getting a red marble? 1/2. Half the time only a single red marble goes into the small hopper and then gets dispensed in the second step. (Half the time, two blue marbles go in, and then one of those two is dispensed, so the chance of blue -- a blue, any blue, one of the two in the small hopper -- is also 1/2, despite the fact that twice as many marbles were dispensed at the stage you don't see.)

Now do it this way: you have a hopper full of white marbles; push one button and half the time a single marble is moved to the small hopper, half the time two; you push the second button and get a single marble. How many marbles are left in the small hopper? Dunno. Half the time there's still one there, and half the time there isn't.

You could randomize. Any number of marbles could be dispensed to the small hopper. Getting one tells you exactly nothing. You could have it transfer a random number of reds to the small hopper half the time and a random number of blues half the time. When you push the second button to get your marble, the chances will still be 50:50 of getting a red or a blue.
• 404
What are your odds of getting a red marble? 1/2.

Agreed. But that scenario is equivalent to randomly waking Beauty on either Monday or Tuesday if tails, but not both days. To be analogous to the Sleeping Beauty scenario, the second blue marble has to be dispensed in a separate event (with amnesia in between). That is an additional possible state that Beauty could be in which, for the thirder, decreases the probability for the Heads and Monday (or red marble) state to 1/3.

All that means is that if Beauty is asked to guess which state she is in and she guesses Heads and Monday, then she will be correct 1/3 of the time. Similarly for Tails and Monday and Tails and Tuesday. On the thirder view, probability is about the state she is in, not the coin toss (or day) outcome itself.

The halfer view, while seemingly just representing a fair coin toss as coming up heads half the time, has the consequence that P(Heads|Monday) = 2/3 instead of 1/2. I think that is a reductio of the halfer view.
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