Couldn't she also look at it from the perspective of H or T? Suppose she reasons the coin landed H with p=.5 and T with p=.5. If it landed H it is Monday. If it landed tails it is either Tue or Mon with equal probability. Thus probability it is Monday is .75 and Tuesday is .25.

It seems to me that this is more of a philosophical observer question than it is a math question. anyway thanks to all who have posted great responses, I lurk a lot and almost never post, but I found this question to be very interesting.

It is a math problem, not a philosophical problem.

The probability of heads is .5, agreed. If it lands heads it is Monday which will happen (0.5)x where x is the number of flips. If it lands tails it will either be Monday or Tuesday but it is not a case of either being awakened Monday or Tuesday, but Monday and Tuesday so that Monday happens (0.5)x and Tuesday (0.5)x. Thats where you get Monday equals 2/3 and Tuesday equals 1/3. In your post you described the case where on a tails she is awakened on Monday or Tuesday randomly (but only one of those) so that Monday equals 3 times Tuesday.

I'm going to re-post the jason1990 post that PTB linked to. The fact that he disagrees should at least give pause to the people that are sure one way or another.

As far as I can tell, P(H|M or T) = P(H), since the condition exhausts all possibilities. Is there something wrong with my reasoning here?

I have another way to put it. On Sunday, SB is asked what the probability of heads is before the coin is flipped. Obviously 1/2. She already knows that after she goes to sleep, she will wake up with no knowledge that she doesn't already have. I think the 1/3 crowd is arguing that her answer should change between the time she goes to sleep and the time she wakes up. What new information has she learned that allows her to change her answer?

The trick I figured out is how to keep all of these... kind of. It's well-known and not controversial AFAIK that in monty hall, if you are told that monty opens a random remaining goat, you win 1/3 by staying. If he opens a random remaining door, and it happens to be a goat, you win 1/2 by staying. And if he just opens a goat, and you have no idea what his procedure was, then you can't come up with an answer. It isn't enough to just see a goat, you have to know how you came to see a goat.

And (4) here is the same kind of deal, and I dealt with it in post 35. If you know that when you're told monday, it couldn't have been tuesday, then you get P(h|mon)=.5. But if it could have been tuesday and just wasn't, then knowing it's monday is informative and changes you to P(heads|mon)=2/3. And if you have no idea how you came to know it's monday, then it's indeterminate (many valid parameterizations of my space lead to different numbers).

You can't deal with the different selection procedures in monty hall in one simple probability space (the normal treatment there is just to make a different one), and nobody seems to complain... but you could embed two copies like I did here and handle both answers out of one space if you felt like it.

She does get information. She is being awakened, which happens twice as often on a tails as it does on a heads.

Someone else already did this but consider 1000 repetitions of the experiment with 1000 different people. They will be asked the question "What was the coin toss result?" 1500 times. If they all say "Heads" they will be correct 500 times. If they all say "Tails" they will be correct 1000 times. The coin is 50/50, but the experimental design creates a bias because "Tails" results in asking the question twice. It is a poorly designed experiment for determining the odds of "Heads" vs. "Tails" because the result of the measurement changes the experiment. To properly test the coin performance you must keep the measurement independent of the result of the measurement.

Consider an extreme example. They put her to sleep and flip a coin. If it comes up tails they wake her and ask the coin result. If it is heads, they shoot her in her sleep and she never wakes up. The coin flip is 50/50 but she must answer tails. In this case she gets asked the question 1.0x if it is tails and 0x if it is heads. The coin is still 50/50 but the experimental design creates a bias in the answer because the coin flip modifies the experiment. This case is easy because the bias is infinite, but the math is the same.

Nothing changes.

"What's % chance coin will be heads?" - 50%

If I wake you up, with:
1/3 chance it's Monday and coin is heads
1/3 chance it's Monday and coin is tails
1/3 chance it's Tuesday and coin is tails,
but you don't know which, what's the % chance then from your perspective that the coin was heads? - 33%

There's no changing of information here. This is how she would answer before the experiment as well.

People are getting caught up in the fact that coins are 50/50. It's irrelevant to the question being asked. The question asked has nothing to do with the probability of a coin landing on heads.



The other thing that confuses people is the whole 100% thing. 100% is not meaningful when we are talking about multiples of 1. Yes, it is accurate that she will always be woken up and asked the question. The question is how many times will that happen.

Still waiting for an answer on this from the 1/2ers. I can't imagine how one would refute this logic.

In Bostrom's hybrid model for the case where the experiment is only done once he is a 1/2'er. (For numerous runs his hybrid model tends toward 1/3). In the single trial 1/2'er case he handles the problem of P(Heads|monday) by claiming that if she's told it's Monday then another agency-part has been created for her and so another probability model is required rather than appeal to the conditional probability calculated from the original model. So maybe similar to your approach.

The thing is, with your 50-25-25 model (the same as I proposed in 2007), the model produces a standard P(heads|Monday) = 2/3 conditional probability. Just like in Monty Hall where the player picks door 1, the player's original 1/3,1/3,1/3 model produces a standard conditional probability P(door 1 a car | door 3 a goat) = 1/2. In Monty, the simplest way that conditional probability would apply in action would be if the player simply asks what's behind door 3, is always told when he asks, and this particular time door 3 has a goat. Similarly, if Beauty asks what day it is, is always told when she asks, and it happens to be Monday, then her standard conditional probability calculation P(heads|Monday) should apply. Equivalently, in Monty the player upon choosing door 1 could just always be told what's behind door 3 - and he knows this. If it happens to be a goat he can compute his standard conditional probability. Similarly, like in your calculation, if Beauty is always told the day - and she knows this - and it happens to be Monday she can compute P(heads | Monday) = 2/3. I think you agree with this from your other post:




Whether she is always told the day and this day happens to be Monday, or she just thinks to herself what her probability model tells her if she happened to know this is Monday, her conclusion is the standard conditional probability produced by her 50-25-25 model:

P(heads | Monday) = 2/3

But this is precisely the "Monday Problem" for the 1/2 argument. It makes no difference whether the coin is flipped Sunday night or Monday night. She will always be awakened Monday regardless of the coin's outcome. And if it is flipped Monday night, with her 50-25-25 probability/credence model, and she is always told the day and it happens to be Monday (or just ponders what her credence should be if she knew the day was Monday), she calculates a credence for the outcome of a coin toss yet to be made of 2/3 for heads.

I believe jason1990 tries to avoid this problem by reducing Beauty's credence model and not giving her a Credence for what day it is when told the coin landed Tails. However, as I look at it again I'm wondering what probability model he would propose to do that. I hope he is checking this thread so I'll state the problem to him in my next post.


PairTheBoard

It's not equivalent to the coin being flipped monday night though. It's "equivalent" in that somebody watching a replay of the awakenings, and looking at the coin after all the awakenings are finished, but with no idea what's going on in the meantime will come up with the same counting stats (as will showing the coin to beauty and doing away with amnesia, so obviously the information you're aware of is everything.. in my version the information beauty had when she awoke was the information she had before she went down which was the same as the observer had before she went down), but it obviously can't be represented by my space because the propositions in my space involve a determined coin.

In this version, right before you go to sleep, you are 100% that the coin has not been flipped, and when you wake up, you believe "there is some nonzero chance the coin hasn't been flipped, and some nonzero chance that the coin has been flipped and is tails". Your information has changed, and is asymmetrically weighted to tails no matter what.

In the original version, right before you go to sleep, you are 100% that the coin has been flipped, and when you wake up, you are 100% that the coin has been flipped. Your information is the same.

They just aren't equivalent.

So you're saying P(H | awake) = 1/3
I think we agree that P(awake) = 1
This implies that unconditionally, P(H) = 1/3

I don't have a refutation for your example other than what I've written above. The point of this problem is that both sides have issues which is why there is no consensus.

From 2007 thread - last page

From 2007 thread - page 8 (10 post pages)

I see this model showing posterior 1/2 and prior 1/2 are consistent which I think is all it was intended to do at the time. What I'm wondering now is what probability model is possible to show the consistency of (1)-(4) bolded above?

Thanks,


PairTheBoard

I am in the 50% camp.

To make the numbers easier lets say there are 50 days, and she can either be in interview H1 or T1, T2, T3,... T50.

The flaw in the reasoning is assuming it is equally likely she is in any interview. But she knows the experimental design, so it isn't. Even though the interviews are all similar, she knows that 50% of the time the coin is heads. So she can ascertain that 50% of the time, she is in H1. 1% of the time, she is in T1, 1% she is in T2, etc. So the answer is still 50%.

To make this more clear, imagine the coin is switched with some rigged coin that comes up heads 99.9999999999% of the time. So if it is heads (almost always), she is interviewed once. If it is tails (almost never) she is interviewed twice. Would you reason the following? :

No. Instead you would reason as above, and correctly see that 99.9999999999% you are in H1, 0.00000000005% you are in T1, and 0.00000000005% you are in T2; making the chance of heads in your answer to the interview question still 99.9999999999%.

Wrong.

99.9...8% she is in H1

0.0...1% she is in T1
0.0...1% she is in T2

Tails is still twice as likely as the odds of the coin would indicate.

This is just incorrectly using a mathematical formula to disprove something that is obviously correct.

But it is equally likely!!!

If we do N trails, on average she will be in T50 just as often as H1. If the coin is 50/50 this has to be true.

Every time it's heads, she is in H1. Every time it's tails, she is in T50 (and T49, and...).




Ok, take the original experiment. If heads, we NEVER wake her up. If tails, we wake her up X times, X>0.

Do you still think it's 50/50? Clearly it's now 100% tails. If nothing else, this shows the frequency with which we wake her up is relevant.

There is a tricky part though:

How do I deal with the fact that HM isnt TM? After all, if you tell her its monday, then shouldnt heads be as likely as tails? But in the model I propose, p(H | M) = 2/3, so whats wrong?

Okay, so (*):
HM = 0.5
TM = 0.25
TT = 0.25

You can imagine that they have different rooms labelled HM,TM,TT that are designated for each of the 3 possibilities. These probabilities indicate the probability she will be in the indicated room if you choose a random experiment, then choose a random room visit. Lets call this underlined choice a "clean choice."

If you restrict your sample to clean choices that are also Monday, then yes the chance of heads is 2/3. But that is because of how the sampling is done. For any specific experiment in this sample, if we walk up to her any time it is Monday and tell her it is Monday, she will say the chance that heads is 1/2. BOTH are correct.

Here is a picture to help illustrate this.



In the picture is an atypical clean choice of 4 interviews-in-progress from experiments. (Our choices are marked in blue). Luck evened out and the choices were HM HM TM TT as expected above (see *). In particular, in Experiment 4, a Monday happened, but by luck we randomly picked a time when we were Tuesday interviewing her. When we discarded Tuesday, we were left with a chance of 2/3 heads from our point of view, also as expected (since HM/(HM+TM) = .5/.75 = 2/3)

But when we ask her, the result is that the chance of heads is 1/2, also as expected. The red circled times are the times in the experiment we walk up to her and tell her its monday.

I hope this picture clears it up. If it doesn't, just stare at it from the point of view that Beauty cant distinguish between Experiment 1 and 2, and also cant distinguish between Experiment 3 and 4.

1. Yes you are technically correct. But from the viewpoint of clear choice it isnt equally likely (see my post above.) Meaning if you chose an experiment at random, then chose a room at random, my probabilities would hold.

2. Yes I also agree she should answer 100% tails, but this is a degenerate case not related to the problem. First because the bias of the coin doesnt influence the result. Also the number of times she is woken up doesnt influence the result. So you cant show anything from this.

Your whole premise is a bit unclear, but if I'm to assume when TT comes up, we don't wake her up on Monday, and that she will always be asked the above question, then the above is incorrect.

This is starting to seem like a religious argument where the theists (I.e. the people that are wrong), use way too many words, and very convoluted arguments, and refuse to acknowledge the short, blatantly true posts.

Ok how about this.

Joe runs into Q in a bar.

Joe says, "I haven't seen you in 20 years! I remember that experiment you were talking about, where you flip a coin, if it's heads, you wake SB once. If tails, you wake her a million times over a million days. Every time you wake her you ask her if she thinks the coin was more likely to be heads or tails. You were going to be the interviewer for this experiment."

Q responds, "Yes, it took a long time, but we did get it up and running. I just finished my interview for today in fact."

Joe responds, "Ah, so the coin came up tails then! You've got your work cut out for you."


Is this experiment significantly different from the million SB I proposed above? Is there any flaw with Joe's confidence here? I think Joe can be extremely confident it was tails.

Here is another way to understand the problem.

You are asking her how often she thinks its heads (and she is perfectly rational). Since every waking is indistinguishable, presumably she will give the same answer p. So what answer should she give?

Well, half the time it is heads, and she gives answer p.
The other half of the time it is tails, and she gives answer p a zillion times.
Either way her best answer is to give p = 0.5 .

Of course if she is betting she should take into account that if it is tails she bets many times the amount she initially bets.

Yes it is significantly different (but your story is a bit unclear). But if you mean what I think you mean then yes, it is significantly different, and Joe would be correct to be very confident it was tails.

The reason is that Joe has very good reason to believe that he can observe the contradiction -- that is, P( observed Q had no interview | heads) = .999999 . Here SB can make no such observation.

Nope.

"I'm awake, therefore tails is a zillion times more likely to have been flipped."




What if instead of p, we asked her "What credence do you give to today's date?"

The odds of it being that first Monday are extremely slim. The odds of it being that Monday are exactly equal to it being the 7th Tuesday.

Same question, but the answer clearly isn't 50/50.

Edit: I said "exactly equal", but that should say, "only twice as likely".

Here is another thought for you ZJ. You are in a futuristic capsule. I flip a fair coin secretly to decide whether you either stay in the capsule for 1 hour or 2 hours, but you cant tell the difference because time slows down in the 2 hour mode (making 2h feel like 1h)

You get in the capsule at noon. When you get out, what % of the time would you say that it is 1PM?

This was a rhetorical question but hopefully you can be persuaded by the similarities (Beauty cant tell the diff between 2 interviews and 1 interview -- they collapse into a single unified interview with the same answer)

Say there are 100 days. She would reply 101/200 Day 1, 1/200 Day 2, 1/200 Day 3, etc.