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Sleeping Beauty Problem Sleeping Beauty Problem

11-08-2011 , 07:17 PM
So far we have ZJ comparing 1/2ers to theists, and RLK telling 1/2ers they should not gamble for any significant amount of money.
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11-08-2011 , 07:23 PM
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Originally Posted by ZeeJustin
Thanks coffee for clearing up what Justin A was saying.

1 is not unconditionally true. It is only true under these conditions. Therefore, you have to take the word unconditionally out of your conclusion.
Those conditions have to have a probability < 1 for them to be different. So far the only condition I can see is that SB has woken up. Since this is always the case (she is never asked the question without having just woken up), then P(awake) = 1. What I wrote before follows from the laws of probability. If P(B) = 1, then P(A|B) = P(A).

If you want to show that your conclusion escapes this, you need to show that the conditions under which she awakes does not have a 100% chance of being the case. Either that or accept the seemingly absurd conclusion that she should think P(H) = 1/3 before the experiment starts.

Edit: Or I guess you could try to show why "If P(B) = 1, then P(A|B) = P(A)" is wrong if that's what you think.
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11-08-2011 , 08:24 PM
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Originally Posted by Justin A
Those conditions have to have a probability < 1 for them to be different. So far the only condition I can see is that SB has woken up. Since this is always the case (she is never asked the question without having just woken up), then P(awake) = 1. What I wrote before follows from the laws of probability. If P(B) = 1, then P(A|B) = P(A).

If you want to show that your conclusion escapes this, you need to show that the conditions under which she awakes does not have a 100% chance of being the case. Either that or accept the seemingly absurd conclusion that she should think P(H) = 1/3 before the experiment starts.

Edit: Or I guess you could try to show why "If P(B) = 1, then P(A|B) = P(A)" is wrong if that's what you think.
You're not differentiating between the two questions.

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or accept the seemingly absurd conclusion that she should think P(H) = 1/3 before the experiment starts.
If on Sunday you ask her, "What is the probability the coin will land heads?" - she will answer 1/2.

If on Sunday you ask her, "Given that you are woken up and interviewed as outlined, upon an interview/awakening, what is the probability the coin landed heads?" - she will answer 1/3

The problem with what you said in post 104 is that you don't distinguish between these two very different questions.

Step 1 is only correct if change it to P ( H | Awake | lots of words)

Edit: if you want to argue lots of words are implied, then take the word "unconditionally" out of step 3, and put in the same implication, and now your conclusion is fine.

Last edited by ZeeJustin; 11-08-2011 at 08:43 PM.
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11-08-2011 , 09:08 PM
Ok, I am in the 1/3* camp.. may post later about all the tangley issues, too tired now.
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11-08-2011 , 09:42 PM
A lot of the arguments ITT seem based on a misconception of what probability means and how it behaves. I subscribe to the logical probability view that the probability of a proposition is a number assigned to represent the state of information about the proposition. This will coincide with the frequency and propensity views in some cases (you are about to roll a fair die. What do you think the probability of the proposition "The die will land on 1" is? will give the same answer 1/6 under any of those views), but this view allows discussion of things in which the other views really don't make any sense. If I flip a fair coin behind a wall where you can't see it, under the logical probability view, you assign 50% to "the coin is heads" and 50% to "the coin is tails" (because you're told the coin is fair), even though you know the answer is already physically determined and you know it no longer has a physical propensity to come up 50-50.

Another thing that seems commonly mistaken (or ignored) is the probability space. I'm not going to do it justice here, but for people who aren't going to read a technical wiki page, it's enough to basically treat it a set of disjoint propositions, each with a real number 0<=x<=1 attached, whose numbers add to 1 and behave like you'd expect for the logical operations of and, or, not, and if (if what you're if-ing has P>0). The real number attached a proposition is called the probability of the proposition. A probability space is just a mathematical object- for you to take it seriously as a description of something empirical, the propositions and probabilities need to correspond to what's going on- the propositions need to span all possible outcomes, and you need to justify your choice of their associated numbers. With simple cases, people are used to letting the selection process define the space (roll a die and the space becomes {1,2,3,4,5,6}) and using the indifference principle to fill out the numbers that aren't explicitly defined (in the die case, assigning 1/6 to each number). The indifference principle is mangled in many ways, so formulating it in a coherent way is important, namely "Equal states of knowledge about propositions in different problems should be given equal probability assignments." (which, implies the better-known version if and only if you can transform one proposition into another by symmetry/label exchange methods) It's not enough to have some number of propositions of unknown probability and say "we don't know the exact numbers for these yet, so 1/nth each of what's remaining." You have to have equal states of knowledge about each to justify doing that.

This is most obviously mangled in the posts that argue along the line of "choose a day, then choose an awakening, tails>>>>heads". First of all, choosing a day with some probability is something that requires a probability space to have any meaning. So if you just start off with "it is monday", "it is tuesday" as your propositions (to avoid making the explanation to follow any more difficult), that spans the possibilities of when you wake up. But when you go to assign numbers, you can't use the indifference principle because you know monday and tuesday follow different rules, so you have different information about them, so without more argumentation, you'd just be pulling numbers out of your butt with no principle that justifies them. So that line of reasoning doesn't get you to a justified answer.

What's tricky about this problem is that there is no selection process defined by the problem for the awakening you're looking at, and people are so used to using the selection process to populate the space that they just invent a selection process out of thin air. Days were covered above, but the other popular one ITT is sampling by awakenings. Again, this requires a probability space, the indifference principle doesn't apply because your knowledge about how the awakenings come about is different, so you can't justify the 1/3,1/3,1/3 assignment that's implicit in "twice as likely" argument forms. It's a consistent probability space, but so is 1/5, 2/5, 2/5, and nobody thinks that's a justified answer, so you need a reason beyond "it's what the indifference principle would say if it applied here" to argue why 1/3,1/3,1/3 is even better than 1/5,2/5,2/5, much less a justified answer for the whole scenario. I have yet to see one, and I don't think one exists.

So if there's no physical sampling procedure defined, and the indifference principle doesn't apply to the ones pulled out of thin air anyway, are there any other tools to use to justify an answer? Yes. The principle of reflection. Well, more precisely, the slightly restated version of the principle of reflection that goes "Given a probability for a proposition now, you should have the same probability for that proposition later if your information about its likelihood hasn't changed." Which is like the trivial version of the indifference principle, talking about the same proposition in the same problem.

So now, the real problem. You are told the rules and a coin is flipped. Let's recreate my space from post 35 in equivalent but reworded form, based on my future experience from the 5 possible points in time. There are 5 propositions. The first two from before I go to sleep:

a="I haven't slept. I'll go through some stuff and be told the coin is heads"
b="I haven't slept. I'll go through some stuff and be told the coin is tails"

The next 3 from after I go to sleep
c="I will be awakened and have amnesia, be asked my credence, told the coin is heads, and let go" (HM)
d="I will be awakened and have amnesia, be asked my credence, be put back to sleep, be awakened the next day, be asked my credence, told the coin is tails, and let go" (TM)
e="I will be awakened and have amnesia, be asked my credence, told the coin is tails, and be let go." (TT)

each with associated P(a)..P(e)

These are clearly disjoint propositions about my future experience that span all possible outcomes. Because the coin is fair, I assign a and b equal probability 0<P(a)<1/2. Now, in the most general sense, without trying to assign actual numbers yet, I know that P(c),P(d),P(e) are >0 and that P(c)+P(d)+P(e)=1-P(a)-P(b)=1-2P(a). I believe that if it's not a or b, I will be in c with probability P(c)/(1-2P(a)), d with probability P(d)/(1-2P(a)), and e with probability P(e)/(1-2P(a)). Now, I wake up with amnesia. This means that I am in c/d/e and not a/b. So, since this is exactly the same state of knowledge as two sentences ago, by the trivial indifference principle, I must still believe that I am in c with probability P(c)/(1-2P(a)), d with probability P(d)/(1-2P(a)), and e with probability P(e)/(1-2P(a)). This is where the 2/3 crowd makes a critical and obvious mistake. They say "you're twice as likely to be awakened in a tails world (d or e) than in a heads world (c)!!!". This is a probabilistic statement being made without a reference to a probability space. It is true in this space if and only if P(d)+P(e)=2*P(c) and false otherwise.

So if you just decide to go P(c)=P(d)=P(e), ignorning the fact that the indifference principle doesn't apply, then you run into immediate problems. At a/b, you believe that p=1 you'll go through c/d/e and p=.5 that you'll be told the coin is heads at the end. Then, only upon learning that you're in c/d/e- when you knew there was a 100% chance that you would learn you're in c/d/e, your probability is 2/3 tails at the end, violating the trivial indifference principle. Also, at a/b, you believe there's a 50% chance you'll go through heads stuff and a 50% chance you'll go through tails stuff. Again, upon only learning that you're in c/d/e- going through stuff- your probability is 2/3 tails stuff, again violating the trivial indifference principle. Not only is the P(c)=P(d)=P(e) assignment unjustified, it's actively contraindicated.

And P(c)=P(d)+P(e), P(d)=P(e) (aka 1/2,1/4,1/4) works perfectly and obeys the trivial indifference principle in both cases, as well as the actual indifference principle P(d)=P(e) (given tails, P(monday)=P(tuesday)). Those appear to be the unique c/d/e relations for this problem that obey everything. And, in what is exactly 0 coincidence IMO, this corresponds to what you would get if you sampled old experiments with the procedure 1. Choose an experiment (week) at random. 2. Choose a random awakening from that week. 3. See if it ends in H/T. This would give the 1/2,1/4,1/4 distribution of awakenings selected and a 50/50 H/T ratio in experiments (which is the ratio you get if you just 1. Choose an experiment at random. 2. Look at the coin)

The various monday variants with this experiment aren't worth rehashing because I have nothing to add to post 35. PtB's recent variant, coin being flipped on monday night, is worth looking at in a little more detail. I was wrong before- this variant does NOT have the same counting stats as the original when sampled by the method in the previous paragraph. It gives 1/3,1/3,1/3 with 2/3 tails experiments (experiment-look at coin is still 50/50 obviously). Whether PtB wants to believe it or not, and whether anybody else in the world realizes it, these are clearly, demonstrably (completely independent of asking beauty questions, amnesia, credences, etc) not the same experiment. The resulting experiments, sampled in the same way, give different numbers. And I already explained why with a space and with words and I'm still satisfied with it.
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11-08-2011 , 11:09 PM
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Originally Posted by TomCowley
A lot of the arguments ITT seem based on a misconception of what probability means and how it behaves.
Well, I don't think the probability here from SB's perspective is the same probability of the coin toss. If we told her before she went to sleep that a coin would be flipped every day, but she would only be awoken when the coin landed tails, then asked her her credence as to which side the coin landed, her answer would obviously be 100% tails, despite the fact that it was a 50% chance the coin would land tails when the experimenter threw it.

Seems the problem from the OP is a simple extension of that problem, and her credence as to which side the coin landed is dependent on how she is awoken.
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11-09-2011 , 03:23 AM
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Originally Posted by TomCowley
A lot of the arguments ITT seem based on a misconception of what probability means and how it behaves. I subscribe to the logical probability view that the probability of a proposition is a number assigned to represent the state of information about the proposition. This will coincide with the frequency and propensity views in some cases (you are about to roll a fair die. What do you think the probability of the proposition "The die will land on 1" is? will give the same answer 1/6 under any of those views), but this view allows discussion of things in which the other views really don't make any sense. If I flip a fair coin behind a wall where you can't see it, under the logical probability view, you assign 50% to "the coin is heads" and 50% to "the coin is tails" (because you're told the coin is fair), even though you know the answer is already physically determined and you know it no longer has a physical propensity to come up 50-50.
So how does all of this apply when the coin is flipped monday night? What information is added? Previously you said something about the line of, 'there is now a non-zero chance that the coin has not been flipped yet'. So you think it makes a difference between a past coin that has been flipped but I don't know the outcome for, and I believe it to be 50/50, and a fair coin to be flipped in the future which I know to be 50/50. Where p(flipped in the past)+P(yet to be flipped)=1.

Also, I don't believe the rest of your post applies to my post #107, in which the knowledge about the different propositions ARE identical.
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11-09-2011 , 05:55 AM
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Originally Posted by TomCowley
Lots of words
The 1/3ers don't need to rely on the indifference principle, so your whole argument is flawed.



If we did an iterated version of this problem, we all agree that interviews would be conducted at equal frequency across MH, TH, and TT.

Furthermore, if SB's memory were returned, and she were to look back, she would clearly say, "Now that I can remember all the iterations of this experiment, I can clearly see that tails was much more likely to have been flipped across the sample space of interviews (which is the sample space defined in the problem) ."

Doing the experiment multiple times doesn't change the odds.



There's no indifference principle needed it. It's clear as day to see at what frequency interviews are conducted.
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11-09-2011 , 06:21 AM
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The 1/3ers don't need to rely on the indifference principle
Isn't the core of your argument: "She don't know in which out of 3 states she is in so she assumes she is equally likely in any of them" ?

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If we did an iterated version of this problem, we all agree that interviews would be conducted at equal frequency across MH, TH, and TT.
So I flip a coin two times and it comes heads once and tails once. In 50% of cases SB is waken up on Monday and the coin is heads and other 50% of the time SB is waken up on both Monday and Tuesday and it's tails. If we do this multiple times it's clear as the day that (Monday AND tails) is as frequent as (Tuesday AND tails) and both of them combined are as frequent as (Monday & heads) so it's clear overall distribution is 50-25-25 and not 33-33-33.
That's the "clear war" for you

Last edited by punter11235; 11-09-2011 at 06:30 AM.
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11-09-2011 , 06:33 AM
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Originally Posted by punter11235
Isn't the core of your argument: "She don't know in which out of 3 states she is in so she assumes she is equally likely in any of them" ?
Yes, but the word assumes implies a lack of certainty which I don't think is accurate. You're skipping the step of how she makes that conclusion.

See the iterated argument for how she knows they are equally probable, or any number of other arguments in this thread.

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Frequency probability is the interpretation of probability that defines an event's probability as the limit of its relative frequency in a large number of trials.
That's what I'm invoking.
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11-09-2011 , 06:35 AM
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Originally Posted by punter11235
Isn't the core of your argument: "She don't know in which out of 3 states she is in so she assumes she is equally likely in any of them" ?



So I flip a coin two times and it comes heads once and tails once. In 50% of cases SB is waken up on Monday and the coin is heads and other 50% of the time SB is waken up on both Monday and Tuesday and it's tails. If we do this multiple times it's clear as the day that (Monday AND tails) is as frequent as (Tuesday AND tails) and both of them combined are as frequent as (Monday & heads) so it's clear overall distribution is 50-25-25 and not 33-33-33.
That's the "clear war" for you
Bolded part is wrong. If you flip two coins MH, TH and TT all happen once each.
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11-09-2011 , 06:39 AM
FWIW, I'm not saying the indifference principle doesn't work. I'm just saying 1/3ers don't need to rely on it.

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What's tricky about this problem is that there is no selection process defined by the problem for the awakening you're looking at, and people are so used to using the selection process to populate the space that they just invent a selection process out of thin air. Days were covered above, but the other popular one ITT is sampling by awakenings. Again, this requires a probability space, the indifference principle doesn't apply because your knowledge about how the awakenings come about is different
That flaws in Tom's arguments appear there if we want to invoke the indifference people.

A) The selection process is defined by the problem
B) Our knowledge about how the awakenings come about is the same.
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11-09-2011 , 06:50 AM
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Bolded part is wrong. If you flip two coins MH, TH and TT all happen once each.
I will try different wording.
Let's assume we flip a coin every Sunday whole year. We do our experiment and SB wakes up, she now reasons: "I don't know which week it is, 26 times it's heads week and 26 times it's tails week", "If I am in heads week, it's Monday but if I am in tails week it's either Monday or Tuesday with equal probability (indifference principle)" so now if SB is asked about her credence she concludes it's 50% for H&Monday and 25% for both T&Monday and T&Tuesday, where is the flow in her thinking ?

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FWIW, I'm not saying the indifference principle doesn't work. I'm just saying 1/3ers don't need to rely on it.
So you are doing experiment x times and count MH, MT and TT's coming up with some big number, now you are assuming that SB is equally likely in any of them, aren't you ?
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11-09-2011 , 06:57 AM
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Originally Posted by punter11235
26 times it's heads week and 26 times it's tails week
This is where the flaw lies. The sample includes 78 awakenings, not 52.

Any math that has 26/52 is ignoring 1/3 of the awakenings.
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11-09-2011 , 07:00 AM
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Originally Posted by ZeeJustin
This is where the flaw lies. The sample includes 78 awakenings, not 52.
I am saying it includes 52 weeks not awakenings. You want it be a flow, because you want to choose sampling method and then apply indifference principle for your outcomes.
You can't rebut my argument that it's 52 weeks with "but it's 78 states".
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11-09-2011 , 07:08 AM
Let's focus on your argument for a bit. You flip a coin 100 times and come up with 50 MH's 50 MT's and 50 TT's. What is your argument now that it's 1/3 for MH ?
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11-09-2011 , 07:13 AM
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Originally Posted by punter11235
Let's focus on your argument for a bit. You flip a coin 100 times and come up with 50 MH's 50 MT's and 50 TT's. What is your argument now that it's 1/3 for MH ?
50/150 = 1/3
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11-09-2011 , 07:16 AM
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Originally Posted by punter11235
I will try different wording.
Let's assume we flip a coin every Sunday whole year. We do our experiment and SB wakes up, she now reasons: "I don't know which week it is, 26 times it's heads week and 26 times it's tails week", "If I am in heads week, it's Monday but if I am in tails week it's either Monday or Tuesday with equal probability (indifference principle)" so now if SB is asked about her credence she concludes it's 50% for H&Monday and 25% for both T&Monday and T&Tuesday, where is the flow in her thinking ?
She is asked the question twice as often in the tails weeks, therefore we can infer that when she is asked the question she is more likely to be in a tails week.
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11-09-2011 , 08:11 AM
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50/150 = 1/3
Let's see how this reasoning applies to some other situation.
We go allin preflop AA vs KK playing on Stars. You have AA. If you win Stars display WIN in the middle of the screen. If you lose though Stars display LOSE, wait 1 milisecond, display LOSE again, wait 1 milisecond etc. for 4 "LOSE" displays overall.
Are you comfortable accepting that if we go allin, you close your eyes and your only information is that something just got displayed your credence should be that AA won with 50% probability ?

Last edited by punter11235; 11-09-2011 at 08:34 AM.
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11-09-2011 , 08:32 AM
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Originally Posted by punter11235
Let's see how this reasoning applies to some other situation.
We go allin preflop AA vs KK playing on Stars. You have AA. If you win Stars display WIN in the middle of the screen. If you lose though Stars display LOSE, wait 1 milisecond, display LOSE again, wait 1 milisecond etc. for 4 "LOSE" displays overall.
Are you comfortable accepting that if we go allin, you close your eyes and your only information is that something just got displayed it's equally likely to be WIN and LOSE ?
Let's make it more fair. 4 textboxes appear. After the hand you can check ONLY ONE. If you win, one of them shows WIN, if you lose all four show LOSE. You check a textbox, but you are somehow unable to read what it says, but you can determine there was some text. In this scenario, the text is approximately equally likely to be WIN as LOSE.

Why? Because if you win, 3 out of 4 times you would check an empty textbox.
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11-09-2011 , 09:15 AM
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Let's make it more fair
I just counted outcomes and applied "50/150 = 1/3" principle. If you accept that principle you should have no problems accepting result of my experiment. I wouldn't be surprised 1/3'er do I accept it, I just want to make sure.
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11-09-2011 , 09:35 AM
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Originally Posted by punter11235
I just counted outcomes and applied "50/150 = 1/3" principle. If you accept that principle you should have no problems accepting result of my experiment. I wouldn't be surprised 1/3'er do I accept it, I just want to make sure.
No your example sucks because you would only be able to observe the last 'LOSE' or the whole cycle of 'LOSE''s. Any principle can be misapplied. If you have a big bag, and for every win you put in a white ball, for every loss you put in 4 black balls, now at random grab a ball.

Now, what was your point again?
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11-09-2011 , 10:24 AM
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If you have a big bag, and for every win you put in a white ball, for every loss you put in 4 black balls, now at random grab a ball.

Now, what was your point again?
My point is that you should answer "80% for AA" if you grab the ball without looking at it.
If you apply your balls in the bag example to coinflipping (one white ball for heads and 2 black balls for tails) you should still say "50% for heads" if you grab a ball without looking at it.
What 1/3 camp wants is that you should say "33% for heads" if you hold a ball in your hand, it after all is twice as likely a black ball (MT or TT).
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11-09-2011 , 11:02 AM
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Originally Posted by punter11235
Let's focus on your argument for a bit. You flip a coin 100 times and come up with 50 MH's 50 MT's and 50 TT's. What is your argument now that it's 1/3 for MH ?
Imagine that the question is changed in a minor way. Each time she wakes up she pays $0.50 and wins $1 if she guess correctly. Examining your numbers for the outcome above, in that scenario if she guesses heads each time her EV is 0. If she guesses tails each time her EV is +$50. Her credence for heads cannot be equal to her credence for tails.

/thread
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11-09-2011 , 11:24 AM
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Her credence for heads cannot be equal to her credence for tails.
It doesn't follow.
Imagine this experiment: we are flipping a coin and putting 1 white ball in the bag if heads and 2 black balls in the bag if tails (those are numbered 1 and 2). You are now drawing a ball at random. If you draw white ball or black ball with number 2 you bet and experiment ends.
If you draw black ball number 1 though you are forced to repeat your bet but this time instead of randomly drawing a ball we will put the other black ball in your hand.

What is your credence of coin being heads once you draw a ball without looking at it ?
The answer: 50%
Should you bet with 1-1 odds ?
The answer: surely not

I am also waiting for comments about "50/150 = 1/3" principle in the following experiment:
We are flipping a coin and putting one white ball in the bag if heads and 2 black balls if tails. You now draw one ball at random, what is your credence the coin was heads without looking at the ball you hold in our hand ?
Answer according to "50/150 = 1/3" principle: there are twice as many black balls as white balls so drawing a black ball is twice as likely therefore P(heads) = 1/3. What am I missing here ?

Last edited by punter11235; 11-09-2011 at 11:29 AM.
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