Just do the experiment multiple times and see what the limit approaches as experiments approaches infinity.

http://en.wikipedia.org/wiki/Frequency_probability

I know I said I was done, but I guess not. I was running errands and this comment popped into my mind. I realized that this is a very strong analogy to the SB problem.

I will illustrate.

Experiment 1-

You walk into a room and the examiner asks the following question: "A family has two children. Assuming that boy/girl is exactly 50/50 what is your credence for the proposition that the younger child is a girl." You reason that there are four possibilities listing the older child first: boy/girl, boy/boy, girl/boy, girl/girl. They are equal and two of them have the younger a girl, therefore you answer 1/2. The examiner now says: "Nothing has changed but you have this information, it is not boy/boy. Now what is your credence that the youngest is a girl." You now have 3 equal possibilities and 2 of them have the youngest a girl, so you answer 2/3.

Experiment 2-

You are told you will be put to sleep as in the OP SB problem but that you will be wakened on both Monday and Tuesday and asked the result of a coin flip that was done Sunday night after you were put to sleep. The experiment begins. You awaken and the examiner asks you: "What is your credence for the proposition that the coin was heads." You reason that there are four possibilities, Mon/Heads, Mon/Tails, Tues/Heads, Tues/Tails. They are equal and two of them have the coin as Heads so you answer 1/2. The examiner now says: "Nothing has changed but you have this information, it is not Tues/Heads." (This is the OP SB situation). You now have 3 equal possibilities and only 1 of them has heads so 1/3.

The first example is well known to produce 2/3 even though the original odds of the younger child being a girl was 1/2. The second is point for point exactly the same.

Now I am done. At least I hope so.

Its 2/3rds. The waking up a million times illustrates this perfectly, well done.

jason1990 presented this same model in the 2007 thread. Perhaps you can follow his notation better. He does the same calculation.

2007 Thread - Page 3 (10 posts per page)
jason1990 follows up with the next post

I had been thinking of posting this because on reviewing that thread I noticed that when jason's p (White's c) is close to 0 the conditional probability approaches 1/3. What's going on is that when Tails, the probability of "at least one awakening" is 2p-p^2. So for p close to 1 that's about p ie. about the same as the probability of "at least one awakening" when Heads. But for p small, 2p-p^2 is about 2p or about twice the probability of at "least one awakening" when Heads.

I thought my observation for p close to 0 was interesting but I didn't really see what to make of it since it 's describing approximately one and only one awakening for Tails - except with twice the probability as the one for Heads.

I believe jason1990's point here was that for p<1 there clearly is information being given by an awakening, namely the Event W = "at least one awakening". With that infomation clear we can form a probability space wherein we can condition on that information P(H|W) and where p is close to 1 get P(H|W)= about 1/2. And for p=1 the model still works with P(H|W)=1/2. In this case P(W) = 1 so W doesn't really convey any information. But the important point is how this contrasts with the 1/3er's who claim an awakening conveys information but cannot prove it based on a mathematical model (with the prior P(H)=1/2) where conditioning on the Event of "an awakening" gives P(H|"an awakening")=1/3. What the 1/3'ers do is assert "an awakening" provides information and then assert a (1/3,1/3,1/3) "awakening" model based on arguments appealing to the indifference principle.

I think White makes basically the same point. The Event "at least one awakening" can be modeled and conditioned on in a straightforward way with the Event clearly conveying information for his c<1. While for c<1 the 1/3'ers must overide this clear information and instead build an ad hoc model based on their assertion that there is even more information to be had by "being in an awakening" based on arguably spurious arguments. I don't think White can be so easily dismissed. I wonder when he wrote that paper and if he reads 2+2.



PairTheBoard

Why can't we get in on some of this.


That's pretty bad. Those who argue 1/2 (a) are not under any misaprehension that 1/3er's are suggesting Beauty change her belief about the fairness of the coin. Nor are they unaware that the circumstances in which she is asked her credence are while experiencing an awakening. 1/2'ers are perfectly aware and understand what a Bayes' inference conditioned on her being in an awakening means and that it's plausible it might apply here. They just don't think it does. Groisman deflates nothing here and is just making another 1/3er argument.

And his "argument" is basically one by assertion hidden within an analogy. He tacitly asserts that all awakenings are equally likely by exhibiting their equal likelihood in a large number of trials of an "analogous" experiment set up to show equal likelihood.

1/2ers can just as easily exhibit an "analogous" experiment set up to show that not all awakenings are equally likely and that "this awakening" has 1/2 probability being the result of Heads. In fact several such exhibitions have been given in this thread - Funded by the Sklansky Institute for the Advancement of Smartness.


PairTheBoard

Yea, I think that's in agreement with jason1990's last post in the 2007 thread. The only thing I'm wondering about is, if your model only has atomic outcomes of {Hm} and {Tm+Tt} with equal probability, How can you compute the conditional probability P(H|moday)? Normally you would compute that as,

P(H|monday) = P(H and monday)/P(monday)

But you don't have the detail in the model for P(monday).

At least the credence, P(H|monday)=1/2 is not contradicted by the model, but I don't see how it's implied by it either. I asked this in a previous post as well hoping jason1990 was following the thread. Maybe it's something easy I should see but I'm afraid I don't.

PairTheBoard

You can't do that if you don't have well defined space first. You try to push for MH = 0.5, MT = 0.5 and TT = 0.5. That just doesn't fly as all of them are separate events (at least you treat them as such later) and add up to more than 1.
This is why I asked the question what would happen if SB is only awakened once (in the form of spawning creatures question). You first answered 1/3 (thus pushing for 1/3, 1/3, 1/3 distribution) but then when I clarified what I meant RLK jumped in and said "now it's 1/2" and you didn't respond.
If you assume 1/3,1/3,1/3 distribution in only awakened once version you will face very arguments you are pushing now (just run it multiple times and "calculate EV").

One more example of what you try to do and why you can't:
Imagine you are rolling a dice and record the following events:
a)one of the {1,2,3,4} rolled
b)one of the {4,5,6} rolled
c)exactly 4 rolled

Now you say "run it multiple times and count how often 4 appears in the outcomes". You will end up with probability more than 1.

There is no way around it. You either define space where:
a)MH = 0.5
b)MT
c)TT
and apply indifference principle to b) and c) or:

a)MH
b)MT
c)TT

And apply indifference principle to all of them (arriving at 1/3,1/3,1/3) which will bite you in the ass using the same arguments you are pushing now.

This is misunderstanding of indifference principle.
Remember that there is nothing in original SB problem which tells you that MT = TT. You assume that knowing what experiment is because it seems plausible to you.
Now you try change the experiment in the meantime leaving your application of indifference principle in place. That doesn't fly. (you wouldn't be so fast to apply indifference principle to MH and TH if you knew that on TH sometimes disappear thus making MH and TH asymmetrical events).
To see why again consider experiment with only one awakening (where you conceded MH = 1/2 and MT + TT = 1/2). Following your reasoning you would arrive at 1/3,1/3,1/3 distribution which you don't want.

Saying that distribution is 1/2,1/4,1/4 in experiment with one awakening but 1/3, 1/3, 1/3 (or really 0.5/0.5/0.5 which you magically scale down) with the one with 2 awakenings is exactly the same as my example with rolling a dice and counting 4's. (you arrive at probability more than 1 and then somehow scale it down back to 1)

The children problem (the one with boy/boy, boy/girl, girl/boy , girl/girl) is different because you are specifically told that probability of getting every pair is 0.25 so you don't need indifference principle at all.

OK. He may be off on what 1/2's think. I have no idea how anyone gets to 1/2 in the OP so I cannot imagine what they think. No one has been able to demonstrate anything approaching a mathematical justification for it. All I have seen is a lot of hand waving bs. Suit yourselves. I really do not care. The answer is 1/3. Keep thinking about the problem and maybe some day you will understand why.

In #107 you made the same mistake with indifference principle.
You first describe experiment with 4 awakenings possible and apply indifference principle to get MH = TH but then you change experiment to "but if the coin is heads she is told it's not Tuesday" and expect your previous application of indifference principle to hold in this completely new experiment.
That doesn't work. Again, apply to SB with one awakening to see why. (unless you are advocating 1/3, 1/3, 1/3 distribution in experiment with 1 awakening; if so read ZeeJustin's posts concerning "EV" and "counting outcomes" to see why it doesn't work).

ROTFL.
The experiment have 2 outcomes. Those are MH and {MT, TT}. As outcomes are determined by coinflip we should be comfortable assigning MH = 1/2 and {MT,TT} = 1/2.
SB is awakened within the experiment. She knows she is either in MH world or in {MT, TT} world.
What is her credence she is in MH world ? 1/2 of course.
What is her credence she is in {MT, TT} world ? 1/2 again. Where exactly ? She has no clue but it seems plausible to apply indifference principle here and arrive at MT = 1/4, TT = 1/4.
Why should she bet on tails when being awakened ? Because in {MT, MT} world bets are for twice the stakes.

This meets all the requirements for a mathematical explanation and doesn't require implying probability space with independent events in it adding up to more than 1.

What 1/3'ers have to offer instead ?
Let's run the experiment 100 times. We arrive at 50MH, 50MT, 50TT. As definition of probability used be frequency school is "number of times event occurs divided by number of times it could have occured" they arrive at p(MH) = 1/2, p(MT) = 1/2, p(TT) = 1/2 which should give them pause and reconsider probability space they used to describe the experiment. Instead they assume it has to be 1/3,1/3,1/3 after all somehow.
No matter the fact that MH is independent event to both MT and TT which implies that MT and TT are the same event (because they occupy the same 1/2 in the space left after removing 1/2 occupied by MH). So MT and TT are the same event in that model which doesn't stop 1/3'ers to count it twice (and claim independence there sometimes blind to the fact that it would make size of the space they used to be more than 1).

If you lump MT and TT together because they are indistinguishable, then you need to lump MH into that group as well. From SB's perspective, they are all the same.

Or you could do it from an outside perspective, and have 3 different groups. Having 2 groups is nonsensical.

50/150 = 1/3, not 1/2.

By using the number 100, you are looking at only the coinflip and nothing else. There is data beyond 100 experiments that you are undeniably ignoring by using a sample size of 100 flips.

That leads to the question about distribution in experiment with one awakening. What is your answer there ? 1/2 or 1/3 ? I need to know that before I respond.

Definition of probability though: "Number of times event occured divided by number of times it could have occurred". The number of times it could have occurred is number of times you run the experiment. You can't claim you run experiment 100 times and then use 150 to determine probability. It just doesn't compute. (the event can't occur more times than experiment was run).

By your thinking if I flip a coin and write "BASTARD!" two times every time the coin is heads you could conclude that probability of writing BASTARD is 100%.
(We run experiment 100 times and see 100 "BASTARD"s so 100/100 = 1).
You want to use space:
-p(not writing anything) = 0.5
-p(writing first bastard) = 0.5
-p(writing 2nd bastard) = 0.5

And then treat them all as independent events which doesn't work because sum of events in probability space is never more than 1.

Now feel free to propose different probability space which models this experiment.
If you are accept frequency probability and accept that MH and MT are independent as well as MH and TT are independent you arrive at p(MH) = 1/2 automatically.

Here is how "boy/girl" experiment differs from SB:


(I made a typo, last 1/3 should be 2/3 of course)

As before red line separates independent events. Do you see why you can't draw red line between MT and TT in SB experiment ?

One last example:
We roll a dice and observe the following events:
a)the result is in {1,2,3} ( {} denotes a set of outcomes)
b)the result is in {4,5,6}
c)the result is in {6,5,4}

Once you are awakened during the experiment without seeing the roll first, what is your credence one of the 4,5,6 rolled ?
1/3'ers want to apply the following thinking to answer this question:
"We run experiment 100 times and arrive 50 a)'s, 50 b)'s and 50 c)'s, therefore 50/150 = 1/3 so your credence should be 2/3 for one of the 4,5,6 rolled as it occurred 100 times out of 150".

Do you see how hopeless this thinking is ?

This has been referred to by several people as a convincing argument for 1/3. I think it is a fair argument for comparison to the (.25,.25,.5) model because that model appeals to the indifference principle for Beauty's credence on what day it is if the coin is Tails. It's a fair argument for comparison but I think it fails to rebut for the same reason it's fair for comparison. That reason being it also appeals to the indifference principle for Beauty's credence for what day it is. What it shows is that an appeal to the indifference principle for Beauty's credence on the day can produce both a 1/3 and a 1/2 model - with the 1/3 model possibly more intuitive to some people. But that doesn't really solve the problem of which use of the indifference principle is the "right" one. If anything, it argues that appeal to the indifference principle for credence on the day cannot deflate the disagreement.

What your model does not recognize is the alternate approach recommended by jason1990 in his last post of the 2007 thread and the one TomCowley has now come around to. Specifically, that Beauty does not have adequate information to form any credence for what day it is. Not for monday-tuesday when Tails in the original problem. Nor for monday-tuesday unconditionally in your settup. Your settup simply asserts she should form a 50-50 credence for monday-tuesday. The argument jason1990 made and TomCowley has now come to says that assertion is unwarrented. The probability model for your settup should contain the atomic outcomes {Hm+Ht} and {Tm+Tt} with equal probability and no credence for the day.


PairTheBoard

Or you can introduce 2nd coin to determine which day it is if there are two days to choose from (simulating indifference principle). If you are awake and told you are still in the experiment you can conclude something about 2nd coin toss (that it indicated Monday if first coin was heads) but you can't conclude anything new about 1st coin so you stick to your a priori.

Also notice that if we flip two coins, awake SB once and ask her about her credence about first coin in cases of HH, RH and RR but don't ask her at all in the event of HR the correct answer for her once she is asked is 1/3 for H. This is also consistent with frequency argument. If we run it 100 times we will arrive at:
25HH, 25HR, 25RR, 25RH comfortably giving probabilities of 25/100, 25/100, 25/100 and 25/100 respectively. This is because our space now contains an event "we won't ask you anything", so being asked is in fact new information.
What we have in original SB is H, RH, RR events and those as I showed have probabilities of 1/2, 1/2, 1/2 respectively forcing a conclusion that RH = RR (as expected because 2nd coin is not used in original SB problem)

From OP:


In 100 flips, this situation will come up 150 times. To determine a probability space, you cannot look at a point in time when more information will be added later.

If your response to this is "no information is added later" then read this spoiler:




Wait, lol. Just read your definition of probability again. Did you read that off a crackerjack box? Pretty sure if we use that definition, every coinflip is 100%.

You still refuse to apply definition of probability according to frequentists.

"Number of time event occurred divided by number of times it could have occurred"
If you flip a coin 1000 times and try to determine probability of it being heads you use the definition as follows:
-how often heads occurred ? answer : 500 times
-how often heads could have occurred ? : answer 1000 times

What is p(heads) ? answer : 500/1000

You confuse trials with outcomes. Probability is not (your desired outcome)/(all outcomes) but (your desired outcome)/(how many times it could have occurred).
The difference is crucial. One run of experiment can have 1000's outcomes (ie satisfy 1000's of events).

We won't get any further if you deny the very definition of probability used by frequency school and then try to use frequency argument. You can find the definition here:
http://en.wikipedia.org/wiki/Frequency_probability or here:
http://www.stat.wvu.edu/srs/modules/...f/probdef.html

Where nx is number of times your event occured and nt is number of times you run the experiment.

According to this definition you have P(MH) = P(MT) = P(TT) = 1/2 in original SB experiment.

Yeah. If you roll a dice and focus your attention on the following events:
a)roll in {1,2,3}
b)roll in {4,5,6}
c)roll in {6,5,4}

The "situation" will also come up 150 times during 100 rolls. So what ? p(a) is not 50/150 but 50/100. Don't confuse outcomes with trials.

I loled

Punter. You keep making the same mistake.

The question in the OP is not, "What is the probability the coin will land heads?".

The question has very little to do with the odds of a coinflip.

Every single post you make is centered around the flip.

The event is the awakening/interview, not the flip.

Yes of course. We have probability space with the following events: MH, MT, TT.
We now want to determine p(MH).
To do that we open wikipedia and read the definition. It tells us that we need to know how often our event occurs during n trials of our experiment.
Let's try it with 1000 trials. We run 1000 experiments and see how often we encountered MH. We arrive at 500 times (more or less), noting in the meantime that MT and TT also occured 500 times each.

What is now p(MH) according to that definition ?
C'mon...

Conditional probability doesn't mean "and now it's completely different world and completely different experiment" it only means: "now out of all the trials you run focus only on those which satisfy given event, if you only takes those trials into account what is the ratio of those trials where event occurred to number of trials you are focusing on ?" In our example an event "she is awakened" has probability 1 so it doesn't reduce space we are focusing on.

It could be easier to see for you if you stop denoting events MH, MT and TT and starting denoting them as AMH, AMT, ATT (as awakened and monday and heads etc.).
If you do that you will see that focusing on trials where awakened happened didn't reduce probability space. If you want to see an example where focusing on given event shrinks probability space see my drawings about boy/girl experiment.

I am still waiting for you (or any other 1/3'er who used frequency argument) how the hell they arrived at 150 in "50/150" expression.

To be clear about conditional probability I will write it carefully:
T = set of all the trials we run, there are n such trials
E = event which is our "information" so for example "not BB" in boy/girl example or "awakened" in SB example
NE = event we want to know probability of

We are now interested in both p(NE) and p(NE | E).
Let's start with p(NE). It's simple, we run experiment N times and use wikipedia definition:
p(NE) = number of trials where NE occured / total number of trials

if n is 1000 we conclude:
p(NE) = 500/1000 (more or less)

Now someone tells us: "Focus only on those trials out of 1000 you just run, where awakening happened, what is now p(NE) ?"
To anwer that question you open wikipedia again and notice you need to determine the following things:
a)number of trials out of 1000 we run before where awakening occurred
b)number of trials out of 1000 we run before where awakening occurred and P(NE) happened

a) is clearly 1000 again, as awakening happened in every run of our experiment
b)is clearly 500 as MH happened 500 times during those 1000 trials

Therefore:
P(NE | E) = 500/1000 = 50%.

Notice how this work in boy/girl experiment. If you run it 1000 times you will come up with 250 for every pair. Now someone tells you: "focus only on those trials where either GG, BG or GB occured".
You will see that out of 1000 trials only 750 interests you now and out of those 750 there are 500 resulting in GG or GB giving probability of girl being the first child as 2/3.
There is no contradiction anywhere. Just read definitions and apply them carefully.

If someone still can't see it, try to answer SB question but assuming the coin is not fair but skewed towards heads (2/3 of the time resulting heads, 1/3 of the time resulting in tails).
Write down your reasoning and how you apply definition of probability.