Indistinguishable: I prefer the view of problem 2 to problem 1 for the egalitarian reason that I stated earlier. If you prefer problem 1, you have to explain what makes you so special that you get created every time, as opposed to the (at least) 9990 other folks who only get created sometimes.
I prefer problem 2 to problems 3-infinity because I have not yet even solved problem 2. Trying to solve both problem 2 and problems 3-infinity is just more work than I’m willing to commit to at this point. (I have a sneaking suspiscion that problems 2-infinity all have the same probabilities, but I’d have to look a bit more closely to be sure.)
Tyrrell McAllister: Let’s examine what information each of them have.
SB[sub]before[/sub]:
[ol]
[li]Knows the experimental procedure[/li][li]Knows that she has not yet be woken up by the experimenters[/li][li]Knows that SB[sub]during[/sub] does not know whether she has been woken up by the experimenter once or twice.[/li][/ol]
SB[sub]during[/sub]:
[ol]
[li]Knows the experimental procedure[/li][li]Knows that SB[sub]before[/sub] has not yet be woken up by the experimenters[/li][li]Does not know whether she has been woken up by the experimenter once or twice.[/li][/ol]
They have exactly the same information, they just aren’t in the same situation as each other.
Off hand, I can’t think of another situation that produces results like this.
1 H B* .25
2 H C* .25
3 T B* C .125
4 T B C* .125
5 T C* B .125
6 T C B* .125
Okay, I think I’ve got it figured out. What’s going on here is that you’re not accounting for the odds that C is in the experiment, but SB doesn’t know it. Referring to the above chart, we see that it is indeed the case that if you look at the cases where C is in the experiment (cases 2-5, specifically), across those the odds of the coin being heads was indeed 1/3. However, SB doesn’t always know if C is in the experiment! She only knows that C is in the experiment if she sees it on the day she’s being interrogated, which only happens in cases 2, 4, and 5. And in those cases, it’s heads precisely 50% of the time.
I’ll make another comment on this, in case you’re interested. I agree that there are different perspectives, depending on what you know and what information you’re looking for. Way back in the Sleeping Beauty problem, the experimenter has credence of 1/2, while Beauty has credence of 1/3. As other posters have said, this difference comes from the fact that Beauty is trying to guess right on each time she wakes up, while the experimenter is trying to guess right on each experiment. Similarly, the the men and women have different credences for the coin flip than each other, and both have different credences than God. It seems to me that this difference in credences comes from the fact that (a) the fact that any given person exists is not certain, (b) the sex of any given person is not certain and (c) the total number of each gender depends on the result of the coin flip.
I rather enjoy these classical probability problems, so I spent some time with Problem 2 (as I outlined above) and I think I’ve got the answer. It also turns out that it’s easy to extend this to problems 3 through infinity. Like the pennies analogy I mentioned before, the answer depends only on the distribution of men through the two universes. I get that, given that you wake up a man, you’re credence for a coin flip of heads should be 1/11 (the number of men in the heads universe, divided by the total number of men in either universe), and this is true regardless of the number of “potential people” that exist up in heaven. If you’re interested in any of the details, I can also post those.
Excepting of course for the fact that she actually has a credence of 1/2*. Which is not to say that differing perspectives don’t lead to different credences; they certainly can and do when the different perspectives grant differing information. Such as when in the 1000-to-1 gender scenario, both the man and woman, despite meeting, each know that only they themselves is a random member of the complete set of people, and that the other person was selected for the purpose of the meeting from the smaller set of the people of the necessary gender. And of course for the experimenter or god, the credence is a certain 0 or 1 by the time the subjects of the test are asking the question, due to having complete information. This difference in information, resulting from differing perspectives, unsurprisingly results in different credences.
The fact that beauty’s correct credence is actually 1/2** is more useful in removing the conundrum you presented in post #141; that is, the odd case where it appeared that her credence changed without any change in her available information, which is handily resolved by the fact that, when the correct credence is calculated, her credence remains the same as expected.
** Once you get past the large error I was making of incorrectly giving all wakings equal probability, and also past the extremely subtle error Tyrrell McAllister was making of not accounting for Beauty’s partial ignorance of the outcomes, anyway.
I differ with you on this. I think it’s certain that any person who so much a has a credence is certain to exist, as is any person they meet, which dissolves point a. I think, similarly, that (unless we specifically frame the problem otherwise, such as with false genders written on foreheads), that any person may be certain of their own gender as well. And I don’t actually think that the differing quantities per coin flip is the cause of the confusion either; after all, that information is known equally to all the men and women as well.
I’m fairly sure the difference in credence actually comes from the fact that each person knows that they are a random member of the complete set of sentient people, and from there they can tack on their known gender and theorize about the whole group’s gender distribution based on their random sample of one. (Sleeping Beauty can’t do something similar because she doesn’t have access to any information that has a positive or negative correllation to result of the coin flip.)
This *almost *got me to change my mind, until I realized that you accidentally changed the paradox in a meaningful way. Drawing one ball from the bag tells you nothing about the coin flip, but if you know that you’re going to *continue *to draw balls from the bag until it’s empty (while suffering amnesia prior to each drawing, naturally), then the fact that you’re (still?) drawing balls from the bag instead of doing anything else is relevant information.
To make it easier to see, expand the time frame of the original question, so instead of the experiment lasting either 1 or 2 days it lasts either 1 or 14,610 days (40 years), which figures to be the majority of the remainder of your lifespan. You wake up and are interviewed – you don’t know what day of your life this is, but if the coin had come up Heads then the interview would be a once-in-a-lifetime event and you should expect to be doing something else, whereas if it came up Tails the interview is not extraordinary.
Or, coming at it from a slightly different direction, reconfigure the hypothetical so that the experiment will be conducted on you 10 separate times, always starting on the same day of the week, and if the coin comes up Tails then the interview is repeated the rest of the week (Tuesday through Sunday). So, Heads means one interview (on Monday), Tails means 7 interviews. As always, you begin each interview with no recollection of any previous interviews.
Assuming you get 5 Heads and 5 Tails, then there will be 10 interviews on Mondays, and **30 **interviews on non-Mondays. Therefore, you should wake up to each interview knowing that today is probably a non-Monday, which means you must also wake up to each interview knowing that the coin probably landed on tails, since if you’re being interviewed on a non-Monday (which you know you probably are) then the coin can’t have come up Heads.
Was it? You wrote: “(1) Considering some other simpler paradoxes such as the Boy or Girl paradox, you might think that SB shouldn’t condition merely on “C appears during the experiment”. Instead, you might argue, she should condition on the stronger fact that “C appears today”. I tried to argue in posts 30 and 31 for why I don’t think that that reasoning applies in this case.”
However, I’m not saying she should condition on “C appears today”*; I’m saying she should condition on “I know that C appears in the experiment”. That is all that she can be sure of based on the information she has, and as I demonstrated, the chance of that is 50%.
Think of it this way: rather than conditioning on the fact “C appears during the experiment”, we could calculate the odds based on the fact “There are two days in the experiment”. Obviously, that fact tells you with 100% certainty that the coin landed on tails. However, you also have to take into account whether you’ll know that fact is true before you can make determinations based on it. In some of the cases where “C appears during the experiment” is true, SB will not know it - she’ll know that “B appears during the experiment” instead.
I disagree that “today” is without referent without a calendar; “now” is as self-defining and self-evident to a sentient observer as “I” is. However, that’s not really relevent to the point; her knowing that “today” is in the experiment won’t lead you to an erroneous thirder conclusion, becase there are no cases where that is true and SB doesn’t know it.
You’ll never wake up in the interview room and “expect to be doing something else” - the obvious fact that you are waking up in the experiment room remembering nothing prior to going to sleep there at the start of the experiment rules out the possibility that it’s day 6 of the one-day version of the experiment. Similarly, on every day of the experiment, it will be extraordinary; you will not remember any prior wakings regardless of whether any occured, so to you it will always seem to be the first an only waking.
Clearly, if SB had some way to sense that she’d awoken a thousand times prior, that would drastically change the experiment - she’d be certain it landed on tails, becuase she’d know that in the heads case she wouldn’t be feeling this boredom in her bones. But by the terms of the experiment she has no way at all to tell that this isn’t day 1, so every day feels as fresh and new as any other.
The thing is, though, that the odds aren’t the same of each of those interviews occurring. (Which is the error I was making myself for the longest time in this thread.) In actuality the odds of a given day on a tails coin flip occuring is (1/6 * 1/2), and the odds of it being a monday on a heads coin flip is (1/1 * 1/2). Which means that the odds that it’s monday en total are 7/12, and the odds that is isn’t is 5/12.
So, despite non-mondays occuring three times as often over the course of the ten experiments, if you wake up, and had to guess what day it was, Monday is the clear winner, being seven times likelier than any single other day. Which kind of puts the kibosh on being more certain that it landed on tails as a result.
The thing to remember is that more often doesn’t necessarily mean more likely. Going back to the ‘two balls in the bag’ scenaro, suppose that depending on the coin flip, you either put one blue ball in the bag, or you put one blue ball and one red ball in the bag. Now reach in and pull out a single ball, but don’t look at it.
Now: what are the odds that the ball you are holding is blue, and based on the mere fact you are holding the ball (but don’t know what color it is), what is the probability that there is a second ball in the bag?
You grant below that non-Mondays will occur three times as often as Mondays over the course of the experiments, but here you hold that any given interview day is more likely to be a Monday than not. It is not possible that both of these things are true.
It actually doesn’t. Even if we were to use your numbers (which I’m pretty sure are wrong), then tails is still more likely, since half of the Mondays come about as a result of a tails, hence tails would be correct 17/24 of the time.
To restate, we know that we can expect there to be an average of 10 Mondays and 30 non-Mondays over the course of the experiments, and we know that today is one of the experiment-days in that group. Since we don’t know anything else, we *do *know that it’s more likely (from our perspective) to be a non-Monday, and hence tails.
(Watch, I’ll make an embarrassing elementary mistake here). Odds that you’re holding a blue ball = 75%. Since (presumably) you know that you’ve reached in for a ball just once so far, the odds that there is a second ball in the bag are 50/50, same as the coin flip.
My numbers were wrong - I got it in my head we were dealing with six days somehow. (I’m non-jewish; to me a week ends on saturday. That is my only defense.)
The actual numbers are 7/14 for monday 6/14 for the rest.
Of course it’s possible - you’re mixing up the probability that the current day will be a given day with the probability a given day will happen.
The probability the current day will be a given day is reduced by there being more days to pick from, right? If the options are monday through sunday, then you have a 1/7 chance that the current day is a given day. If the options are monday only, its 1/1 aka 7/7 for monday.
This means that the odds of the current day being a given day are precisely like if you were rolling a single 14-sided dice, with a ‘monday’ on eight sides of the dice, and each of the other six sides having a different day. With seven of the sides with ‘monday’ also saying ‘heads’, and one monday and each of the seven non-monday sides saying ‘tails’.
When you roll this dice, what is the chance of it being Monday? 8/14. What is the chance of it being heads? 7/14 = 1/2.
When you count the times actual days came up, you aren’t counting the odds of the current day being a given day. You’re counting the number of times the days come up. This is like taking the dice above and anytime it rolls ‘heads’ you count 1 monday, and anytime it comes up ‘tails’ you count one each of each day. (And you still count one heads, or one tails). This hasn’t got anything to do with the odds of the current day being a given day, because it ignores the ‘current day’ aspect of it completely. (And even though you count the other six days each half as often as you counted monday, so that they total to far more together, that doesn’t change that half the dice rolls were still ‘heads’ dice.)
I maintain that there’s an (N+1)/(N+N) chance that it’s the first day, for any trial with 1 or N days happening with equal probability. That’s how odds work. That’s how you got a 75% chance of the ball drawn being blue (with blue of course meaning ‘the current day is the first day’).
No, this math is good. It correctly reflects the odds of a random ‘current’ day being the first day (blue:75%) or the second day (red:25%), and the fact that whatever you draw, the odds are still 50% that the bag started with two balls.
Now, take that same bag, and if there were two balls in it, suppose that before you reach into the bag that if there was a second ball in there to start with, somebody may or may not have taken the blue ball out (with a 50% chance of having done so). If there was only one ball in there, then they didn’t and the blue ball remains there.
Does this change the odds as to what your ball color might be? Does this change the odds of whether there was originally one ball or two in the bag?
Then ask yourself whether it matters if the person asked themselves these same questions when they took the first ball out of the bag (if they did so - and of course presuming they didn’t know they drew the blue ball).
Huh? Is there an eighth day accounting for the other 1/14?
Ok, I *think *I see where you’re making your mistake. You’re counting Monday every time it comes up, but you’re only counting days Tuesday-Sunday once, even though each one comes up half as often as Monday, and hence the set of [Tuesday-Sunday] as a whole comes up three times as often as Monday (in the restated scenario from my post #146).
Yes, half of the coin flips are heads, but when you’re asked the question, 7/8 of the time the correct answer is tails (3/4 of the time it’s a non-Monday, 1/8 of the time it’s a Monday following a tails), which is identical to the proposition that if you’re being interviewed, the preceding coin flip probably came up tails.
Now, I’m still not entirely sure what you’re trying to do with the numbers, as I don’t get why even the numbers you’re giving don’t imply an equal probability for heads and tails, and you never responded to my point that your numbers (even your revised numbers in your last post) lead to more tails, since only half of the Mondays are heads.
No, it’s only 75% likely to be blue if you know that it’s the first time you reached into the bag, which is not analogous the the SB paradox. To make it the same as the original question, you have to posit that, given you’re holding a ball, there’s a 33% chance it’s the second second time you’ve reached into the bag, which leads us right back to the original conclusion.
I can sort of see why you’d think the way you do: 50% of the time it’s Heads-Day1, 25% of the time it’s Tails-Day1, and a 25% of the time it’s Tails-Day2. If you say it like that it sounds so reasonable. Unfortunately, thinking about it this way runs into two big problems.
First, these probabilities do not correspond to reality, as we see when we run the experiment over and over and wind up with *twice *as many Day 1s as Day 2s, and not three times as many. You can’t populate a sample with a process where each individual *X *has a 75% chance of being Y, but then wind up with only 66% of the sample being composed of Ys. Odds don’t work *that *way. If reality isn’t bearing out your odds, then you’ve made some mistake with the math.
Second, if what you say is true, then there’s a 75% chance that any given interview is being conducted on a Monday. However, half of all Monday interviews are subsequent to a tails flip. (Just because tails-Monday is followed by tails-Tuesday does not make it half as common as heads-Monday; how could it? Tails leads to two interviews, not one interview split in half, which is what you’re proposing.) So, you then get a 25% chance of a tails-Tuesday, and a 32.5% chance of a tails-Monday, which works out to a 62.5 % (5/8) chance of tails. Which is nuts.
(The bolded part, I think, is the crux of the confusion, hence this post is longer than it needs to be. Sorry.)
begbert2, let me answer your footnote first, to give context to the rest of my reply.
My position at the time that I wrote the OP was that indexical words such as “I”, “now”, “today” and “here” need to be translatable into non-indexical terms to have meaning. Furthermore, I held that, to carry out this translation, you needed to have something equivalent to a random calendar (at least) that was active for the longest-possible duration of the experiment. I have since realized that insisting on translatability has too many bizarre consequences. However, I don’t yet see how to attribute meaning (and, in particular, credence) in general to untranslated indexicals. Because my thoughts on this are still unsettled, I haven’t yet posted on them. But the upshot is that, if indexicals are “self-defining and self-evident to a sentient observer”, I don’t yet see how.
Your footnote goes on to say
This is precisely why my earlier post responded to your objection. You wrote:
Now, like I said, I don’t yet know how to make perfect sense of indexicals. But you yourself observe in your footnote that “C appears today” implies that “SB knows it”. Wouldn’t you also grant that the converse is true? If so, then you grant that “C appears today” and “SB knows it” are logically equivalent. Hence, to condition on one is to condition on the other. That is why you did in fact condition on “C appears today”, and why my response did anticipate your objection.
Demonstrably not. The odds that the current day is a given day decrease as the number of days in the set increases. The odds that a given day will occur in the set does not.
To recap. You (not SB, just you, in real life) wake up and have forgotten what day it is. What are the odds that it’s sunday? One in seven. What are the odds that sunday will occur sometime in the week? 100%. When one in seven = 100%, then they’ll be identical for our purposes.
Blast and damn and damn and blast, I cannot get the notion that it’s a six-day experiment out of my head! That was of course supposed to be 8/14 for monday and 6/14 for the total of the rest. Uh, as could be handily deduced from everything else I said.
You have my official persmission to correct me when in the future I claim to never make errors.
It’s not a mistake - it’s deliberate, because I’m trying to illustrate the difference between the the probability that the current day will be a given day, and the probability a given day will happen at all. (Which, as previously illustrated, are not identical for our purposes.)
Look, we’re doing probability here. The probabilitity of an event occuring if there are separate subconditions with separate probabilities for the event is as follows: the odds of the event occuring in a subcondition times the odds of that subcondition happening, summed up over all the subconditions.
So, when there is one condition with one blue ball, and one condition with a blue and a red ball, here is how you calculate the odds:
Case 1 = 1 blue; Case 2 = 1 blue and 1 red
Case 1: Chance of drawing a blue = 100%
Case 2: Chance of drawing a blue = 50%
Case 1: Chance of drawing a red = 0%
Case 2: Chance of drawing a red = 50%
Chance of Case 1: 50%
Chance of Case 2: 50%
Chance of drawing a blue: (100% * 50%) + (50% * 50%) = 50% + 25% = 75%.
Chance of drawing a red: (0% * 50%) + (50% * 50%) = 0% + 25% = 25%.
You know this, you did it yourself and got the correct result.
Now let’s do this again with the seven (repeat to myself, seven, not six :smack:) day scenario. (Using fractions this time to make the math more readable.)
Case 1 = Monday only; Case 2 = Monday through Sun:smack:day
Case 1: Chance of the current day being Monday = 1
Case 2: Chance of the current day being Monday = 1/7
Case 1: Chance of the current day being Tuesday = 0
Case 2: Chance of the current day being Tuesday = 1/7
Case 1: Chance of the current day NOT being Monday = 0
Case 2: Chance of the current day NOT being Monday = 6/7
Chance of Case 1: 1/2
Chance of Case 2: 1/2
Chance of the current day being Monday = 1: (1 * 1/2) + (1/7 * 1/2) = 1/2 + 1/14 = 8/14
Chance of the current day being Tuesday = 1: (0 * 1/2) + (1/7 * 1/2) = 0 + 1/14 = 1/14
Chance of the current day NOT being Monday = 1: (0 * 1/2) + (6/7 * 1/2) = 0 + 6/14 = 6/14
So, yes, most definitely, the odds of it being a monday are lots better that the odds that it’s any other specific day, and also better by a little than the odds of all the other days put together!
Here’s another scenario: Suppose we flip a coin. If it’s heads, you are struck with cluebat once. If it’s tails, you are struck with a cluebat seven times. Suppose you only know that you were struck at least once with the cluebat. Based on that, is it more likely that the coin was heads, or that the coin was tails?
You’re struck seven times with the cluebat if it’s tails. Does that mean that that it was seven times as likely that the coin was tails? If you get struck more times, does the coin become more unfair?
I’m curious what the difference is between the cluebat scenario is and the SB scenario. I don’t see one, myself.
When you understand what I’m trying to do with the numbers, you’ll know why my numbers lead to believing that there’s an equal probability of heads and tails.
Nope. Though I admit that rather than adding to the prior 75% blue ball scenario, I should have rewritten it; the result of adding to it was cryptic. I shall rewrite it now, and make it completely analogous to the SB scenario. (Which I was trying to do with the addendum but didn’t make it clear.)
Okay. You have a bag. Based on a 50/50 coin flip, either a blue ball, or a blue ball and a red ball is put into a bag. However, they’re not mixed up - the blue ball is in front of the red ball.
IF there is a red ball in the bag, there is a 50% chance that somebody will reach into the bag before you and extract the blue ball. If this does NOT occur, when you reach into the bag, you will get the blue ball.
So, you reach in and pull out a ball. What are the odds that the ball is blue, and without looking at the color of the ball in your hand, what does merely having it tell you about the odds of the coin flip?
(I’ll help you out with the first question: regarding the casewise-type analysis like I did earlier in this post, there are three cases: one with a 25% chance of occuring and 100% of the ball being blue; one with a 25% chance of occuring and a 0% chance of the ball being blue, and one with a 50% chance of occuring and a 100% chance of the ball being blue.)
The problem is, in the seven day week outcome, each day is not an outcome of the coin flip. It’s a seventh of an outcome, which you are comparing to an entire outcome in the case of a heads coin flip.
If we were cutting up a cake and giving SB half on one day if it was heads, and a fourteenth of it on each of seven days if it was tails, you would understand how 8/14ths of the cake would be “Monday cake”, how 1/2 of the cake would be “heads cake”, how the mere fact there were more slices on one half doesn’t imply that half is seven times bigger, and how the mere fact that SB gets a slice of cake doesn’t give her a clue which of the eight peices it is.
Well, “I” is easy, you just look down at yourself. But that’s cheating. So negate that.
You solve this by looking at Descartes. “I think therefore I am”. The fact that you’re an observer means that you exist and the fact that you’re sentient means that you know it. Your sentience itself is your referent for “I”. In real-life scenarios like this, where the sentience is specifically attached to an individual, then that individual is a referent for “I” too, since we assume for the sake of argument there’s not a trickster god messing with your ability to tell which body you’re wearing.
I’ll do “here” next. Any sentient individual who finds themselves existing, will find themselves doing so in a place. That place is ‘here’. Other places may be defined relative to here, or, you may find out where ‘here’ lies relative to some objective standard, and then refer to ‘here’ relative to that standard - but that isn’t a necessary step. You can simply define “here” as “the place I am”.
“Now” is directly analogous to “here”. Any sentient individual who finds themself existing will find themself doing so at a point in time. That time is “now”. The point of reference is again the sentience itself; “now” is “the moment of time currently experiencing”. Now, this one is a little tricker to visualise, what with time being noncorporeal and the moment of “now” continually changing, but the principle is the same.
In other words, the terms work just the same as they do in real life. You know what I mean when I say “I am here now”, right? Even if I was lost and didn’t have a clock on me.
The thing is that “C appears today” doesn’t mean the same thing as “C appears in the experiment”. The first implies the second, but the second doesn’t imply the first - there are cases when the second is true and the first is not. Which means you would reasonably expect the second to be more probably true that the first, correct? And it is. There is a 1/2 chance that the first is true, and a 3/4 chance that the second is true. (With a 1/2 and 1/3 chance of the coin behing heads if they’re true, respectively.)
SB only knows the first statement. She doesn’t know the second statement, and has no way to learn it. Which is why she’s forced to use the odds connected to the information she has, which are halfer odds.
I’m going to leave the philosophy aside for now. Suffice to say that I’m now more sympathetic than I was at the time I wrote the OP that you should be able to use indexicals in all the circumstances that you’d intuitively think that you could. However, what’s still unclear to me is how to set up the semantics of indexicals so that their credences are determined and nonparadoxical. Indexicals still confuse me to some degree. I can envy the fact that they don’t seem to confuse you, but that alone doesn’t free me of my own confusion :).
Right, but at the time I was writing what you’re criticizing, I was insisting that only “C appears in the experiment” is meaningful. I was asserting that “C appears today” is not meaningful except as the tautology that C appears on the day that it appears (because the random calendar’s display is the only non-indexical means by which SB can indicate specific days, and I was insisting that all meaningful indexical statements must be translatable into non-indexical statements.)
Your criticism was that I should have conditioned instead on “I know that C appeared”. But we saw that “I know that C appeared” could only mean the same thing as “C appeared today”, and according to my position at the time, the latter is meaningless, so former is also. Hence, it just wasn’t available as something that you could condition on.
The upshot is that I wasn’t making a probability error; I was making a philosophy error by being too strict about what kinds of indexicals are acceptable.
Forgive me for not responding to most of your post, but I’m pretty sure I can pinpoint the source of the disagreement, and a longer post here would just be noise. My last post was definitely too long; sorry.
Ok, this analogy demonstrates exactly what we’re not doing. We’re not giving her one interview on a Heads and 1/7 of an interview, seven times, on a tails. We’re giving her one interview on a Heads and 7 interviews on Tails. We’re not breaking a static quantity of interviews into smaller pieces to give to her gradually, we’re generating *more *interviews. Most of the interviews she experiences will be subsequent to a Tails result.
Look, you just can’t get around this: In my scenario from post #146, we know (and agree) that SB will experience 10 interviews on Mondays and 30 interviews on non-Mondays, which is simply not possible if each individual interview probably takes place on a Monday! In order to convince me you have to show how a 57% chance of a Monday 40 times leads to only 10 Mondays (without its being a fluke), and that can’t be done, because .57*40 = something other than 10.
Would a new hypothetical help? Probably quite the opposite, but I want to try it.
Say you’re napping in front of the TV on a Sunday afternoon, with two football games about to start consecutively: the first involves two very low scoring teams, the second involves two very high scoring teams. You ask your wife to wake you whenever any team scores a Touchdown (assume that both games are guaranteed to include at least one Touchdown). Being a very heavy sleeper, when first awoken you have no recollection any recent real-world events (such as being woken up before).
You’re wife wakes you up and says that someone scored a Touchdown. You have no idea how many times (if any) she’s woken you up already, nor what time it is. Before you glance over at the TV, shouldn’t you expect that you’re more likely to see the game involving the high-scoring teams? The high-scoring game will generate more awakenings than the low scoring game (not more *fractions *of awakenings).
In exactly the same way, SB should expect tails, because tails will result in more awakenings.
Then what percentage of Mondays are subsequent to a Heads result? Aren’t you maintaining 3/4? Because even it was 3/4, that still only gets Heads up to 6/14 of the flips. In order for Heads to be as likely as tails according to you, then a full 7/8 of Mondays must follow a Heads. How could that be?
Well, not to twist the knife, but my criticism was that you were erroneously assuming that SB knew the truth value of “C appears in the experiment”, when in fact that can’t be extrapolated from “I’m seeing C on the calendar”. (You can be sure it’s true if you see C, but you can’t be sure it’s false if you don’t.) I figure this to be an analysis error, and a subtle one at that. There’s not much shame in overlooking it.
If you presume that SB can’t make statements about “today” without a concrete referent, that just removes a peice of knowledge from the set of knowledge at SB’s disposal. Removing such knowledge just brings her closer to defaulting to the fact that she knew the coin flip was fair, a halfer conclusion; to diverge from the halfer conclusion she has to actually be sure of a fact that leads her to a different conclusion.
Don’t worry about long posts; I make them all the time. Look at this one!
Your new hypothetical underscores the mistake you’re making in your interpretation of the problem; you are not viewing a high-scoring and low-scoring game consecutively, you are viewing a high-scoring or a low scoring game, not both.
Consider these two different scenarios.
Suppose we know that one game (Game 1) has one touchdown, and that the other (Game 2) has seven touchdowns. They’re pre-recorded, so there’s no doubt.
Scenario 1: the two games are on the same tape, one after the other. When played, all eight touchdowns will each occur once, with equal probability that any given awakening is happening during any of the eight touchdowns, with a 1 in 8 chance that it’s the Game 1 touchdown.
Scenario 2: the two games are on different tapes; you put one of those tapes in at random (perhaps based on a coin flip), and set aside the other. In that case, if the tape is the Game 1 tape, all the touchdowns will be game 1 touchdowns; if the tape is the Game 2 tape, then all the touchdowns will be game 2 touchdowns. It doesn’t matter how many times you awaken in either case.
The thing to note about scenario 2 is that it doesn’t matter how many touchdowns there are per tape; it still doesn’t change which tape is in the machine, or the odds that that tape was the one you put in the machine. That was determined in advance, and nothing you do afterward will change it.
SB is in scenario 2.
50% of the mondays are Heads Mondays, and 50% of the mondays are tails mondays. The error you’re making is assuming they’re comparable.
Heads occurs on all of the flips - I don’t know where you’re getting 6/14.
Now, despite the fact that 50% of mondays happen on tails, if you awoke on a random day, looked at the calendar, and learned that the day was monday, that wouldn’t give you a 50% chance that the coin was heads. If the coin was tails, then odds are good you wouldn’t have awoken on a monday, wereas on a heads roll you will always awaken on mondays. This makes it much more likely to be a heads roll, if you happened to know it was monday - despite that the same number of mondays occur regardless of how the coin landed.
In addition to being intuitive, we can also calculate this. In a given single experiment, the odds that a given awakening is on a given day are as follows:
7/14: Heads Monday
1/14: Tails Monday
1/14: Tails Tuesday
1/14: Tails Wednesday
1/14: Tails Thurdsay
1/14: Tails Friday
1/14: Tails Saturday
1/14: Tails Sunday
These are trivially arrived at by taking the odds of the coin result and multiplying by the odds of a random awakening being on a given day given that coin result. (I normalized the denominators for convenience.)
Probabilities of the coin being heads assuming you knew it was a given day of the week can be arrived at by taking the ratio of heads occuring on that day and tails occuring on that day. So, if you knew it was monday, there would be a 7 to 1 chance that the coin was heads. If you knew it was not monday, there would be a 0 to 6 chance the coin was heads. Since we don’t know whether it’s monday or not, you scale the proportions appropriately (which I helpfully did in advance) and add them together. Which results in 7 to 7: a 50-50 chance.
Nonetheless, the question isn’t how many times monday occurs over a large number of experiments - it’s what the odds are that in the current experiment, the coin is heads. Getting there by asking what the odds are of it being monday and then the odds of the coin being heads if it’s monday is a roundabout way of getting there - which appears to be prone to error. But done right, it does give the correct result.
I’m afraid that you’re not grasping the subtleties (<he said with self-deprecating irony>) of the position that I held, which is fine because I’ve abandoned it myself.
I was not assuming that “C appears in the experiment” and “I’m seeing C on the calendar” are logically equivalent. I was asserting that “I’m seeing C on the calendar” was not a meaningful statement unless you had some objective means of specifying which “I” you’re talking about. There are potentially several SB’s (one for each day of the experiment). To my mind, she can’t just say “I” and be done with it. To give the statement “I’m seeing C on the calendar” meaning, she needs to have some means of indicating which of the SB’s is the one she is asserting sees C.
If you replied, “She can just point at herself. That’s the SB she’s talking about”, I would have replied, “Ah, but who’s pointing is the pointing that determines the SB that’s being talking about? She can’t indicate who’s pointing has this power unless she can already indicate herself, so this ‘definition’ would be circular. After all, the other SBs (on the other days) may very well point to themselves and utter the very same sentence. How is she to distinguish one of these ‘pointing-acts’ from the rest and imbue it with the power to determine the unique referent of her utterance?”
To reiterate, the claim was not that “I’m seeing C on the calendar” means the same thing as “C appears in the experiment”. The claim was that “I’m seeing C on the calendar” is just a meaningless string of words (or, at best, a tautology if you define the “I” to refer to the SB who sees C on the calendar).
As I said, I don’t yet understand how the statement “C appears on the calendar today” has meaning. At the time of the OP, I was inclined to just insist that it can’t be given meaning. This position would block your attempt to condition on “I’m seeing C on the calendar”. You would just not have that statement available to be conditioned on. I’ve since realized that I can’t maintain this position, but I don’t yet have a satisfactory position to replace it.
Wellll, however abstract you want to make it, you simply can’t (and never could) get to “C appears in the experiment” from where SB is standing, and being less able to define who is standing where doesn’t help matters. And without getting there (or making the grosser error of forgetting that the tails-days don’t have the same weight as the heads-day that I was making earlier in the thread), you can’t become a thirder.