Basically, an agent is put to sleep and told they will be woken up once or twice, depending on the results of a fair coin flip, without the ability to remember other awakenings.
What probability does the agent assign to the event that the coin landed heads?
The intuitive response is 1/3, but this poses obvious epistemological problems. The agent has, ostensibly, no new information at all, and their prior is surely 1/2. Hope someone else finds this as interesting as I do!
Instead of P(Monday | Heads) = P(Monday | Tails) = P(Tuesday | Tails) it is really P(Monday | Heads&Awake) = P(Monday | Tails&Awake) = P(Tuesday| Tails&Awake) or something like that. But the interviewer isn't asking about that, they are asking for the probability of the coin. The 3 positions are only exhaustive given that you are awake to be interviewed about them, not exhaustive of possible states (it's missing P(Tuesday | Heads&Asleep)). Since you're always awakened at least once, I find the argument that being awake has 'given you information that it is not tuesday AND heads' is pretty weak. While true, both heads and tails expect to be awoken while it is not both tuesday AND heads.
She hasn't been given new information by waking up, she also knows as she goes into the experiment - "most of the outcomes where I am being interviewed involve the coin toss coming up tails".
Half of them end up in the heads group. They wake up on Monday and are questioned. Then they sleep until Wednesday and are released.
The other half end up in the tails group, and so are questioned twice (Monday and Tuesday) then released on Wednesday.
Because we gain no information during the experiment, we can make our decision before the experiment.
Let's count. There will be 3N/2 interviews conducted. N/2 of the will be 'heads' interviews and N will be 'tails' interviews. So going in, we can see that when someone experiences the event 'being asked about the coin', 1/3 of the time the coin will be heads and 2/3 o the time it will be tails. Hence, our credence in the coin being heads should be 1/3.
Here is a counterargument. Imagine a slightly different experiment. The people are not asked what their credence in the coin being heads is. They are asked to guess if it is heads or tails. If they are right, the experiment continues and they are eventually released. If they are wrong, this is noted, and the experiment continues until Wednesday, and then they are killed and their home planet is destroyed.
As before, we gain no information during the experiment, and so can decide our answer beforehand. No matter what strategy one picks for making that decision, there is a 50/50 chance that one ends up with a destroyed planet. That indicates that our credence in heads should be 1/2.
Here's a game-theoretic perspective. In general, when an event has a 1/3 chance of happening, an idealized gambler would be indifferent between the following two bets or lottery tickets: (A) win $2 if the event happens; (B) win $1 if the event doesn't happen. (Notice her average payoff is 2/3 no matter which bet she takes.)
Now in the sleeping beauty problem where tails is two awakenings and heads is one, a gambler would be indifferent between (A) winning $2 every time she wakes up and the coin is heads, and (B) winning $1 every time she wakes up and the coin is tails. This suggests that her "belief" is 1/3.
Another way to put it might be that for a risk-neutral agent, doubling the payoff in one state of the world is equivalent to doubling its "perceived probability". In the sleeping beauty problem, doubling the payoff is like experiencing everything twice.
"Mr. Jones has 2 children. What is the probability he has a girl if he has a boy born on Tuesday?" Somehow knowing the day of the week the boy was born changes the result. It's completely bizarre.
Suppose that you ask Mr. Jones weather he has a boy and he says yes. Then the probability that he also has a girl is 2/3.
Suppose that you asked Mr. Jones weather he had a boy born on a Tuesday, and he says yes. Then the probability that he has a girl is less than 2/3, because having two boys gives (about) double the chance for one of them to have been born on a Tuesday.
However, suppose that you asked Mr. Jones weather he has a boy, and if so what day his eldest boy was born on, and he says "yes, and on Tuesday". Then the probability that he also has a girl is again exactly 2/3.
Wikipedia has a detailed explanation: https://en.wikipedia.org/wiki/Boy_or_Girl_paradox
No one said anything about "Mr. Jones has told you…", here. There was nothing about asking Mr. Jones a question and him providing an answer according to some process.
Rather, the question was simply "Mr. Jones has two children. What is the probability he has a girl if he has a boy born on Tuesday?".
There are implicit conventions involved in reading this, but not particularly problematic ones. This implicitly means "Out of all families with two children, at least one of which is a boy born on Tuesday, what proportion have a girl? [Presuming that out of those families, birth gender and day of the week for the two children are all independently uniformly distributed]". And this is a straightforward counting problem.
So the wording seems fine and the problem well-posed to me.
:-O
If the question were "There's Kid 1 and Kid 2, each independently selected with random gender and birth-day-of-the-week. Out of those cases where Kid 1 is a boy born on Tuesday, what proportion are cases where Kid 2 is a girl?", then the answer would indeed be a straightforward 50%; the status of Kid 1 is entirely independent of the status of Kid 2.
But that's not the question. The question is "There's Kid 1 and Kid 2, each independently selected with random gender and birth-day-of-the-week. Out of those cases where at least one (either one, and possibly both) of Kid 1 and Kid 2 is a boy born on Tuesday, what proportion are cases where at least one of Kid 1 and Kid 2 is a girl?".
This is very different, and of course just drawing out the possibilities (all 2 * 7 * 2 * 7 equiprobable-by-stipulation choices of gender and birth-day-of-the-week for Kid 1 and Kid 2) and circling which pairs of subsets are the relevant ones for the two questions reveals the difference, the probabilities for either question elementarily calculable in this way by basic counting.
Also, I notice you said "Somehow knowing the day of the week the boy was born changes the result. It's completely bizarre."
Remember, though, there's no "the boy". The question "On which day of the week was the boy born? Tell me, I need to know!" does not always have a well-defined answer.
Indeed, you'd get the same 14/27 answer even if "Tuesday" in the question "What proportion of two-children families with at least one Tuesday boy have a girl?" was replaced by any other day. And if this seems paradoxically in conflict with the fact that simply asking "What proportion of two-children families with at least one boy have a girl?" has instead the answer 2/3, reflect again upon the fact that some families have two boys born on different days, so that there's no single answer to "On what day was 'the boy' born?". And then just draw out the cases and count.
(Specifically, out of the 2 * 7 * 2 * 7 equiprobable cases overall for Kid 1 and Kid 2's genders and days, there are 27 cases where there's at least one Tuesday boy, and 14 cases where there's at least one Tuesday boy and also a girl. There are 3 * 7^2 cases where there's at least one boy, and 2 * 7^2 cases where there's at least one boy and also a girl.)
Many of these questions, I think, become clearer if thought of as counting questions instead of as "probability" questions (though it's all the same; the math called "probability" is just the math of various kinds of counting (from simple counting as in this case to complexly weighted continuous measurements, but still ultimately a generalized form of counting). However, despite that equivalence, the concept "probability" has developed all these other distracting connotations, such that psychologically, there can be a useful difference in perspective in switch to explicitly thinking "counting" instead. No one would long dispute that there are 27 cases with at least one Tuesday boy, etc.).
The video is wrong. The problem reads: Jones has 2 kids. What is P(he has a girl) given that he has a boy born on a Tuesday. Consider, for a moment, what information we're getting from "boy born on a Tuesday." This is no different than "boy with red hair," or "boy with 5 freckles." The fact that the BOY was born on a tuesday does not change P(day of the week girl was born). Imagine the "boy with 5 freckles" case - let 5 freckles be denoted by F5, six freckles by F6 and so on... would the appropriate calculation include enumerating P(boy F5, boy Fn) for all n? No.
The "born on Tuesday" is irrelevant. Thus you have the following scenarios: - one kid is TuesdayBoy and the other is also a boy, born at any time - one kid is TuesdayBoy and the other is a girl, born at any time
Out of these options P(Jones has a girl) is a flat out 50%. There is no need to bring in concepts of "which was born first" or enumerate all possible days of the week each child could have been born.
Ok... now all the real smartypants here can correct me :)
Of these possibilities, 27 are situations where one kid is a Tuesday boy. [Do you dispute this count?]
Of those, 14 are situations where one kid is a girl. [Do you dispute this count?]
The answer to "What proportion of cases where there is at least one Tuesday boy also have a girl?" is thus 14/27.
You have stated by fiat that certain things are irrelevant to certain other things, that certain things have probability 50%, etc, but in doing so, you have not considered the count correctly. You are likely misled by phrasing such as "the boy", when there are families with two boys in which there is no proper referent of "the boy" and no particular answer to question like "Which day was 'the boy' born?".
For each child the problem constrains to one of two possible sexes and one of seven possible days of birth.
2 * 7 = 14 possible sex/day combinations for a single child.
(2 * 7) * (2 * 7) = 196 possible sex/day combinations for a pairing of two children. To see why, you could write a program to enumerate all of them, starting with the pairing "Boy/Monday + Boy/Monday", then "Boy/Monday + Boy/Tuesday" and so on until you exhaust all possible options at "Girl/Sunday + Girl/Sunday". You'll see there are 196 options.
Now start applying the facts given to us: one of the children is born on a Tuesday (eliminate all possibilities which don't have at least one Tuesday child), and that child is a boy (eliminate all possibilities in which there is not a Tuesday child who is also a boy).
This leaves exactly 27 possible cases:
Boy/Sunday + Boy/Tuesday,
Boy/Monday + Boy/Tuesday,
Boy/Tuesday + Boy/Tuesday,
Boy/Wednesday + Boy/Tuesday,
Boy/Thursday + Boy/Tuesday,
Boy/Friday + Boy/Tuesday,
Boy/Saturday + Boy/Tuesday,
Girl/Sunday + Boy/Tuesday,
Girl/Monday + Boy/Tuesday,
Girl/Tuesday + Boy/Tuesday,
Girl/Wednesday + Boy/Tuesday,
Girl/Thursday + Boy/Tuesday,
Girl/Friday + Boy/Tuesday,
Girl/Saturday + Boy/Tuesday,
Boy/Tuesday + Boy/Sunday,
Boy/Tuesday + Boy/Monday,
Boy/Tuesday + Boy/Wednesday,
Boy/Tuesday + Boy/Thursday,
Boy/Tuesday + Boy/Friday,
Boy/Tuesday + Boy/Saturday,
Boy/Tuesday + Girl/Sunday,
Boy/Tuesday + Girl/Monday,
Boy/Tuesday + Girl/Tuesday,
Boy/Tuesday + Girl/Wednesday,
Boy/Tuesday + Girl/Thursday,
Boy/Tuesday + Girl/Friday,
Boy/Tuesday + Girl/Saturday
If you count, you'll see that of those 27, there are 13 with two boys and 14 with a boy and a girl. The probability of two boys, given that one child is a boy born on Tuesday, is thus 13/27.
If you flip 2 coins then say whatever the first coin was the the odds if you said H was HH, or HT and if you said T it would be TH, TT. However, if you flip two coins and then say if you got at least one head independently from whatever you flipped then the odds you have 3 options HT, HH, TH with equal odds.
So, the question is if the full statement was based on the data or only the truth value of the statement is based on the data.
PS: Now assuming it's truth value is based on data. if you look at all options there are 14 gender day combinations per kid and 14 * 14 = 196 gender day combinations in totoal. Only 14 of of those 196 start BT which is then split evenly 7 BTB_, 7 BTG_. However that leaves 196 - 14 other options to consider. 7 * 14 of them Start G, and 6 * 14 of them start with B not on a Tuesday, but out of those you only keep 1/14 as you need BT on the second roll. Now add them up 13B and 14G out of (13 + 14) = 27. Or 13/27 B, and 14/27G.
A much more surprising result is that most irrational are normal numbers, but we know almost no normal number (morally speaking, a normal number is an irrational number where each digit is equiprobable in any base).
One of the axioms of probability is that if you have an event (i.e. a set), then the probability of a countable union of disjoint sets is the sum of the probability of each set (event) occurring.
Assume a uniform distribution between 0 and 1. Now consider point sets of the rationals (i.e. the number 0.5 is represented by a set with just 0.5 in it). Since the distribution is uniform, each set has the same probability (i.e. the likelihood of picking a random rational).
Now consider this question: What is the probability of picking any rational between 0 and 1? Well, that's just the sum of the probabilities over all rationals (because it is a countable sum of disjoint sets). If the probability of picking any particular rational was non-zero, this sum would be infinite, which violates the laws of probability.
Thus, by convention, it's just simpler to define it to be 0.
There's no magic here. These properties were picked merely to make analysis with measure theory clean. Don't try to ascribe any real world meaning to picking a point.
[edit] strictly speaking, you would restrict yourself to a bounded interval, e.g. if you pick a random number from a uniform distribution on [0, 1], the probability that this number is rational is 0.
https://en.wikipedia.org/wiki/Almost_surely
This strange property comes from strange properties of the real numbers (and uncountably infinite sets) that give rise to things like:
https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox
Measure theory deals with resolving this:
In terms of implementation, I'm not aware of an algorithm that can randomly pick a real number on an actual computer. Perhaps a mathematician could show how to pick one on some abstract machine with infinite resources, and not constrained by finite bit representations of numbers.
Infinity Paradoxes - Numberphile - https://www.youtube.com/watch?v=dDl7g_2x74Q
... and don't get cocky once you know about it, because it's so pernicious it'll get you too if you're not careful!
