Say I ask you to draw a card and then, without looking at it, show it to me. I tell you that it is an Ace, and ask you for the probability that you drew the Ace of Spades. Is the answer 1⁄52, 1⁄4, or (as you claim about the SB problem) ambiguous?
Correct answer depends on the reward structure. Absent a reward structure, there is no such thing as a correct answer. See this post.
In your card-drawing scenario, there is only one plausible reward structure (reward given for each correct answer). In the Sleeping Beauty problem, there are two plausible reward structures. Of those two reward structures, one results in the correct answer being one-third, the other results in the correct answer being one-half.
If the context of the question includes a reward structure, then the correct solution has to be evaluated within that structure. This one does not. Artificially inserting one does not make it correct for a problem that does not include one.
The actual problem places the probability within a specific context. The competing solutions claim to evaluate that context, not a reward structure. One does so incorrectly. There are simple ways to show this.
Actually, there is no answer to the problem as stated. The reason is that the evidence I (who drew the card) have is not “the card is an Ace”, but rather “JeffJo said the card is an Ace”. Even if I believe that JeffJo never lies, this is not enough to produce a probability for the card being the Ace of Spades. I would need to also consider my prior probability that JeffJo would say this conditional on it being the Ace of Space, the Ace of Hearts, the Ace of Diamonds, or the Ace of Clubs. Perhaps I believe the JeffJo would never say the card is an Ace if it is a Space. In that case, the right answer is 0.
However, I agree that a “reward structure” is not required, unless possible rewards are somehow related to my beliefs about what JeffJo might do.
For example, I can assess my probability that the store down the street has ice cream sundaes for sale when I want one, and decide that the probability is 3⁄4. If I then change my mind and decide that I don’t want an ice cream sundae after all, that should not change my probability that one is available.
“I would need to also consider my prior probability that JeffJo would say this conditional on it being the Ace of Space, the Ace of Hearts, the Ace of Diamonds, or the Ace of Clubs. Perhaps I believe the JeffJo would never say the card is an Ace if it is a Space. In that case, the right answer is 0.”
And in the SB problem, what if the lab tech is lazy, and doesn’t want a repeat waking? So they keep re-flipping the “fair coin” until it finally lands on Heads? In that case, her answer should be 1.
The fact is that you have no reason to think that such a bias favors any one card value, or suit, or whatever, different than another.
You may think the difference between “the card is an Ace” and “JeffJo says the card is an Ace” is just a quibble. But this is actually a very common source of error.
Consider the infamous “Linda” problem, in which researchers claim that most people are irrational because they think “Linda is a bank teller” is less likely than “Linda is a bank teller and active in the feminist movement”. When you think most people are this blatantly wrong, you maybe need to consider that you might be the one who’s confused...
Yes, the fact that someone had to chooses the information is an common source of error, but that is not what you describe. I choose a single card and a single value to avoid that very issue. With very deliberate thought. Your example is a common misinterpretation of what probability means, not how to use it correctly according to Mathematics.
A better example, of what you imply, is the infamous Two Child Problem. And its variation, the Child Born on Tuesday Problem.
I have exactly two children. At least one is a boy. What are the chances that I have two boys?
I have exactly two children. At least one is a boy who was born on a Tuesday. What are the chances that I have two boys?
(BTW, both “exactly” and “at least” are necessary. If I had said “I have one” and asked about the possibility of two, it implies that any number I state carries an implicit “at least.”)
Far too many “experts” will say that the answers are 1⁄3 and 13⁄27, respectively. Of the 4 (or 196) possible combinations of the implied information categories, there are 3 (or 27) that fit the information as specified, and of those 1 (or 13) have two boys.
Paradox: How did the added information change the probability from 1⁄3 to 13/27?
The resolution of this paradox is that you have to include the choice I made of what to tell you, between what most likely is two sets of equivalent information. If I have a Tuesday Boy and a Thursday Girl, couldn’t I have used the girl’s information in either question? Since you don’t know how this choice is made, a rational belief can only be based on assuming I chose randomly.
So in 2 (or 26) of the 3 (or 27) combinations where the statement I made is true, there is another statement that is also true. And I’d only make this one in half of them. So the answers are 1/(3-2/1)=1/2 and (13-12/2)/(27-26/2)=7/14=1/2. And BTW, this is also how the Monty Hall Problem is solved correctly. That problem originated as Martin Gardner’s Three Prisoners Problem, which he introduced in the same article where he explained why 1⁄3 is not correct for #1 above.
In my card drawing problem, there is only one card rank I can report. If you choose to add information, as done with Linda the Bank Teller, you are not a rational solver.
Interesting. I hadn’t heard of the Child Born on Tuesday Problem. I think it’s actually quite relevant to Sleeping Beauty, but I won’t go into that here...
Both problems (your 1 and 2) aren’t well-defined, however. The problem is that in real life we do not magically acquire knowledge that the world is in some subset of states, with the single exception of the state of our direct sense perceptions. One could decide to assume a uniform distribution over possible ways in which the information we are supposedly given actually arrives by way of sense perceptions, but uniform distributions are rather arbitrary (and will often depend on arbitrary aspects of how the problem is formulated).
Here’s a boys/girls puzzle I came up with to illustrate the issue:
A couple you’ve just met invite you over to dinner, saying “come by around 5pm, and we can talk for a while before our three kids come home from school at 6pm”.
You arrive at the appointed time, and are invited into the house. Walking down the hall, your host points to three closed doors and says, “those are the kids’ bedrooms”. You stumble a bit when passing one of these doors, and accidentally push the door open. There you see a dresser with a jewelry box, and a bed on which a dress has been laid out. “Ah”, you think to yourself, “I see that at least one of their three kids is a girl”.
Your hosts sit you down in the kitchen, and leave you there while they go off to get goodies from the stores in the basement. While they’re away, you notice a letter from the principal of the local school tacked up on the refrigerator. “Dear Parent”, it begins, “Each year at this time, I write to all parents, such as yourself, who have a boy or boys in the school, asking you to volunteer your time to help the boys’ hockey team...” “Umm”, you think, “I see that they have at least one boy as well”.
That, of course, leaves only two possibilities: Either they have two boys and one girl, or two girls and one boy. What are the probabilities of these two possibilities?
The symmetrical summaries of what is learned are intentionally misleading (it’s supposed to be a puzzle, after all). The way in which you learned they have at least one girl is not the same as the way you learned that they have at least one boy. And that matters.
Your problem is both more, and less, well-posed than you think.
The defining feature of the “older child” version of the Two Child Problem has nothing to do with age. It is that you have ordered the children independently of gender, and identified the gender of a child in a position within that order. Age works well here, since it is easy to show why BB, BG, GB, and GG must be equiprobable by examining the event of the second birth.
But any gender-independent ordering works. It could be alphabetizing names, their seats around the dinner table (clockwise from Mother), or which bedroom each child has. You picked a specific child in an order by looking in a specific room, so the genders of the other two are independent of it and each other. So gBB, gBG, gGB, and gGG are equiprobable at that point in your acquisition of knowledge.
But your second acquisition depends on whether similar help is needed for other sports, and how many gender-specific sports there are. And why there isn’t one for girls’ sports, since we know there is a girl.
My problems are well-posed for what I intended. You didn’t “stumble upon” the information, a source with absolute knowledge told it to you, with no hint of any discrimination between genders. There is an established solution in such cases; it’s called Bertrand’s Box Paradox. That name did not, originally, refer to a problem; it referred to the following solution. It is best illustrated using a different probability than what I asked for:
I know Mr. Abbot’s two children. At least one is a boy.
I know Mrs. Baker’s two children. At least one is a girl.
I know the Curry’s two children. At least one has the gender that I have written inside this sealed envelope.
In each case, what is the probability that the family has a boy and a girl?
Clearly, the answers A1 and A2 must be the same. This is not using uniform distributions, although that is a valid justification. Probability is not about what is true in a specific instance of this disclosure of information—that’s a naive mistake. It is about what we can deduce from the information alone. It is a property of our knowledge of a world where it happens, not the world itself. Since our information is equivalent in Q1 and Q2, that means A1=A2.
But you have no significant information about genders in Q3, so A3 must be 1⁄2. And that can be used to get A1 and A2. Bertrand argued simply that if the envelope were opened, A3 had to equal A1 and A2 regardless of what it said, so you didn’t need to open it. Any change would be a paradox. But there is a more rigorous solution.
If W represents what is written in the envelope, the Law of Total Probability says:
A3 = Pr(W=”Boy”)*A1 + Pr(W=”Girl”)*A2
A3 = Pr(W=”Boy”)*A1 + Pr(W=”Girl”)*A1
A3 = [Pr(W=”Boy”) + Pr(W=”Girl”)]*A1
A3 = A1 = A2 (which all equal 1⁄2).
This solution is also used for the famous Monty Hall Problem, even if those using it do not realize it. The most common solution uses the assertion that “your original probability of 1⁄3 can’t change.” So, since the open door is revealed to not have the car, the closed door that you didn’t pick must now have a 2⁄3 probability.
The assertion is equivalent to my sealed envelope. You see the door that gets opened, which equivalent to naming one gender in Q1 and Q2. Since your answer must be the same regardless of which door that is, it is the same as when you ignore which door is opened.
Correct answer depends on the reward structure. Absent a reward structure, there is no such thing as a correct answer. See this post.
In your card-drawing scenario, there is only one plausible reward structure (reward given for each correct answer). In the Sleeping Beauty problem, there are two plausible reward structures. Of those two reward structures, one results in the correct answer being one-third, the other results in the correct answer being one-half.
If the context of the question includes a reward structure, then the correct solution has to be evaluated within that structure. This one does not. Artificially inserting one does not make it correct for a problem that does not include one.
The actual problem places the probability within a specific context. The competing solutions claim to evaluate that context, not a reward structure. One does so incorrectly. There are simple ways to show this.
Actually, there is no answer to the problem as stated. The reason is that the evidence I (who drew the card) have is not “the card is an Ace”, but rather “JeffJo said the card is an Ace”. Even if I believe that JeffJo never lies, this is not enough to produce a probability for the card being the Ace of Spades. I would need to also consider my prior probability that JeffJo would say this conditional on it being the Ace of Space, the Ace of Hearts, the Ace of Diamonds, or the Ace of Clubs. Perhaps I believe the JeffJo would never say the card is an Ace if it is a Space. In that case, the right answer is 0.
However, I agree that a “reward structure” is not required, unless possible rewards are somehow related to my beliefs about what JeffJo might do.
For example, I can assess my probability that the store down the street has ice cream sundaes for sale when I want one, and decide that the probability is 3⁄4. If I then change my mind and decide that I don’t want an ice cream sundae after all, that should not change my probability that one is available.
“I would need to also consider my prior probability that JeffJo would say this conditional on it being the Ace of Space, the Ace of Hearts, the Ace of Diamonds, or the Ace of Clubs. Perhaps I believe the JeffJo would never say the card is an Ace if it is a Space. In that case, the right answer is 0.”
And in the SB problem, what if the lab tech is lazy, and doesn’t want a repeat waking? So they keep re-flipping the “fair coin” until it finally lands on Heads? In that case, her answer should be 1.
The fact is that you have no reason to think that such a bias favors any one card value, or suit, or whatever, different than another.
You may think the difference between “the card is an Ace” and “JeffJo says the card is an Ace” is just a quibble. But this is actually a very common source of error.
Consider the infamous “Linda” problem, in which researchers claim that most people are irrational because they think “Linda is a bank teller” is less likely than “Linda is a bank teller and active in the feminist movement”. When you think most people are this blatantly wrong, you maybe need to consider that you might be the one who’s confused...
Yes, the fact that someone had to chooses the information is an common source of error, but that is not what you describe. I choose a single card and a single value to avoid that very issue. With very deliberate thought. Your example is a common misinterpretation of what probability means, not how to use it correctly according to Mathematics.
A better example, of what you imply, is the infamous Two Child Problem. And its variation, the Child Born on Tuesday Problem.
I have exactly two children. At least one is a boy. What are the chances that I have two boys?
I have exactly two children. At least one is a boy who was born on a Tuesday. What are the chances that I have two boys?
(BTW, both “exactly” and “at least” are necessary. If I had said “I have one” and asked about the possibility of two, it implies that any number I state carries an implicit “at least.”)
Far too many “experts” will say that the answers are 1⁄3 and 13⁄27, respectively. Of the 4 (or 196) possible combinations of the implied information categories, there are 3 (or 27) that fit the information as specified, and of those 1 (or 13) have two boys.
Paradox: How did the added information change the probability from 1⁄3 to 13/27?
The resolution of this paradox is that you have to include the choice I made of what to tell you, between what most likely is two sets of equivalent information. If I have a Tuesday Boy and a Thursday Girl, couldn’t I have used the girl’s information in either question? Since you don’t know how this choice is made, a rational belief can only be based on assuming I chose randomly.
So in 2 (or 26) of the 3 (or 27) combinations where the statement I made is true, there is another statement that is also true. And I’d only make this one in half of them. So the answers are 1/(3-2/1)=1/2 and (13-12/2)/(27-26/2)=7/14=1/2. And BTW, this is also how the Monty Hall Problem is solved correctly. That problem originated as Martin Gardner’s Three Prisoners Problem, which he introduced in the same article where he explained why 1⁄3 is not correct for #1 above.
In my card drawing problem, there is only one card rank I can report. If you choose to add information, as done with Linda the Bank Teller, you are not a rational solver.
Interesting. I hadn’t heard of the Child Born on Tuesday Problem. I think it’s actually quite relevant to Sleeping Beauty, but I won’t go into that here...
Both problems (your 1 and 2) aren’t well-defined, however. The problem is that in real life we do not magically acquire knowledge that the world is in some subset of states, with the single exception of the state of our direct sense perceptions. One could decide to assume a uniform distribution over possible ways in which the information we are supposedly given actually arrives by way of sense perceptions, but uniform distributions are rather arbitrary (and will often depend on arbitrary aspects of how the problem is formulated).
Here’s a boys/girls puzzle I came up with to illustrate the issue:
The symmetrical summaries of what is learned are intentionally misleading (it’s supposed to be a puzzle, after all). The way in which you learned they have at least one girl is not the same as the way you learned that they have at least one boy. And that matters.
Your problem is both more, and less, well-posed than you think.
The defining feature of the “older child” version of the Two Child Problem has nothing to do with age. It is that you have ordered the children independently of gender, and identified the gender of a child in a position within that order. Age works well here, since it is easy to show why BB, BG, GB, and GG must be equiprobable by examining the event of the second birth.
But any gender-independent ordering works. It could be alphabetizing names, their seats around the dinner table (clockwise from Mother), or which bedroom each child has. You picked a specific child in an order by looking in a specific room, so the genders of the other two are independent of it and each other. So gBB, gBG, gGB, and gGG are equiprobable at that point in your acquisition of knowledge.
But your second acquisition depends on whether similar help is needed for other sports, and how many gender-specific sports there are. And why there isn’t one for girls’ sports, since we know there is a girl.
My problems are well-posed for what I intended. You didn’t “stumble upon” the information, a source with absolute knowledge told it to you, with no hint of any discrimination between genders. There is an established solution in such cases; it’s called Bertrand’s Box Paradox. That name did not, originally, refer to a problem; it referred to the following solution. It is best illustrated using a different probability than what I asked for:
I know Mr. Abbot’s two children. At least one is a boy.
I know Mrs. Baker’s two children. At least one is a girl.
I know the Curry’s two children. At least one has the gender that I have written inside this sealed envelope.
In each case, what is the probability that the family has a boy and a girl?
Clearly, the answers A1 and A2 must be the same. This is not using uniform distributions, although that is a valid justification. Probability is not about what is true in a specific instance of this disclosure of information—that’s a naive mistake. It is about what we can deduce from the information alone. It is a property of our knowledge of a world where it happens, not the world itself. Since our information is equivalent in Q1 and Q2, that means A1=A2.
But you have no significant information about genders in Q3, so A3 must be 1⁄2. And that can be used to get A1 and A2. Bertrand argued simply that if the envelope were opened, A3 had to equal A1 and A2 regardless of what it said, so you didn’t need to open it. Any change would be a paradox. But there is a more rigorous solution.
If W represents what is written in the envelope, the Law of Total Probability says:
A3 = Pr(W=”Boy”)*A1 + Pr(W=”Girl”)*A2
A3 = Pr(W=”Boy”)*A1 + Pr(W=”Girl”)*A1
A3 = [Pr(W=”Boy”) + Pr(W=”Girl”)]*A1
A3 = A1 = A2 (which all equal 1⁄2).
This solution is also used for the famous Monty Hall Problem, even if those using it do not realize it. The most common solution uses the assertion that “your original probability of 1⁄3 can’t change.” So, since the open door is revealed to not have the car, the closed door that you didn’t pick must now have a 2⁄3 probability.
The assertion is equivalent to my sealed envelope. You see the door that gets opened, which equivalent to naming one gender in Q1 and Q2. Since your answer must be the same regardless of which door that is, it is the same as when you ignore which door is opened.
If there is no reward structure, then neither answer is meaningfully more “correct” than the other. Beliefs are for actions.