[Guest]

Alright, we've all seen the first couple of these, but the third one is kind of interesting.

1. A man has two children, the older of which is a son. What is the probability that he has two sons?

2. A man has two children, at least one of which is a son. What is the probability that he has two sons?

3. A man has two children, one of which is a son who was born on a Sunday. What is the probability that he has two sons?

[Guest]

Let's make some common notation,

() parenthesis means an ordered tuple

{} curly brackets means an unordered collection or set

| pipe means given

! means not

P[A] means probability of event A

Question 2 asks P[{M,M} | {M,F}]

I fully acknowledge and have never denied that the question does not make a reference to order. As you see here, I used {M,F} instead of (M,F) or (F,M).

My point with this, is that at the fundamental level, we cannot create {M,F} in physical reality. We can only create (M,F) or (F,M).

Gender is only one characteristic of many characteristics that these children have.

In any real situation, we may safely assume that there exists some characteristics other than gender by which these two children differ. That is what I mean when I say these children are distinct.

Let's say one child is 10 pounds and the other is 8 pounds just to pick a characteristic for this example.

Now if we assign genders in all possible ways, how many physical outcomes do we have?

Outcome 1: 10 lb male and 8 lb male

Outcome 2: 10 lb male and 8 lb female

Outcome 3: 10 lb female and 8 lb male

Outcome 4: 10 lb female and 8 lb female

BTW, I don't care about order, you can put the 8 lb baby first in the list or the 10 lb one or whatever you want.

I care about distinction, not order, order is just a convenient representation for distinction.

Your event Dej Mar, {M,F} corresponds to the union of outcomes 2 and 3.

But both outcomes cannot exist simultaneously in reality.

I know the question never mentioned anything about weight.

It doesn't matter, you can do this same exercise with any characteristic that differs between the children.

You cannot argue that such a characteristic cannot exist, and at least one such characteristic must exist.

Therefore in physical reality, your unordered set {M,F} is not realizable.

And it is from physical reality that we make the assumptions that each of these outcomes has equal probability.

The way to look at {M,F} is to say that it is an event, not an outcome, in which outcome 2 or outcome 3 takes place.

In this way, if (M,F) and (F,M) each have a probability of 0.25, then the event {M,F} has a probability of 0.5 and is twice as likely as the event {M,M} (which is the same as (M,M) )

Your way, Dej Mar, when taken to the physical realm with any differing characteristic, such as weight, age, location of center of mass of baby with respect to some origin, ..., etc

comes to a road block:

With same example as before,

Outcome A: 10 lb male and 8 lb male

Outcome B: ? lb male and ? lb female

You then assign both of these equal probability, and say prob of A is 0.5 and prob of B is 0.5.

The takeaway is B cannot be physically realized as a single outcome and physical realization is the ultimate authority regarding how we assign probabilities. The list of things explicitly mentioned in the question is not the ultimate authority.

I have no idea how to interpret your notation here.

In all cases you have a 1 and a 2 (which I think would mean order of birth) except for the ones in which order of birth (as a valid distinction) would actually matter.

I'm guessing that this cannot be the case because {M1,F1} would not make sense (exact twins? to the picosecond? really?), no, it should be {M1,F2} or {M2,F1}

Also, according to me, we go by physical outcomes, and {M1, M2} and {M2,M1} are different ways of writing the same exact physical outcome, and this outcome should be counted once.

So to use your notation (at least I think so),

{M1,M2}, {M1,F2} are the possible outcomes for question 1

{M1,M2} {M1,F2} {M2,F1} are the three possible outcomes to question 2.

By order, I meant I was using order to serve the same purpose as your 1 and 2 --- that is each distinct child is uniquely identified by their position in the gender string (MM, MF, FM, FF) in my original case or by the 1 or 2 in your case.

Very funny.

I am very confident that I am not wrong, but you may believe what you like.

I was using your definition... not mine.

The index was to indicate a unique individual, as you required - not order of birth. I was using the leftmost element as the older child. With an index for birth order, let us then identify the two male possibilities as {M1, m2} and {m1, M2} and the female-female possibilities (if there was no restriction, but two children) would be {F1, f2} and {f1, F2}. The male-female possibilites would be {M1, F2} and {F1, M2}. Given the first born is a male, we then have {M1, m2}, {m1, M2} and {M1, F2}. Thus, by your interpretation that MF and FM are not the same, there is a 2/3 probability that the 2nd born is a son. I disagree with this interpretation. As you pointed out, the result should be 1/2. Thus MF and FM ARE the same (as are {M1, m2} and {m1, M2})!!! Where one need interpet the set of elements as different is when the problem asks that they are, which this one did not. Thus in all 3 cases (unless one interprets the 3rd problem's phrase "one of which" to mean "one and only one of which", there is a 1/2 probability -- as many have agreed -- that the second child is also a male.

[Guest]

I was using your definition... not mine.

The index was to indicate a unique individual, as you required - not order of birth. I was using the leftmost element as the older child. With an index for birth order, let us then identify the two male possibilities as {M1, m2} and {m1, M2} and the female-female possibilities (if there was no restriction, but two children) would be {F1, f2} and {f1, F2}. The male-female possibilites would be {M1, F2} and {F1, M2}. Given the first born is a male, we then have {M1, m2}, {m1, M2} and {M1, F2}. Thus, by your interpretation that MF and FM are not the same, there is a 2/3 probability that the 2nd born is a son. I disagree with this interpretation. As you pointed out, the result should be 1/2. Thus MF and FM ARE the same (as are {M1, m2} and {m1, M2})!!! Where one need interpet the set of elements as different is when the problem asks that they are, which this one did not. Thus in all 3 cases (unless one interprets the 3rd problem's phrase "one of which" to mean "one and only one of which", there is a 1/2 probability -- as many have agreed -- that the second child is also a male.

Not my definition. I never had this stance, and have explicitly clarified several times. There would never be two possibilities counted separately where both children males according to my definition. I never believed that there were two outcomes in which both children were male (when considering only gender as a characteristic) so I disagree with this interpretation as well, as it is not mine.

There are different sample spaces for all three questions.

In Q1 it is impossible for a man to have an older daughter.

In Q2 is it impossible for a man to have two daughters, but possible to have an older daughter.

In Q3 it is impossible for a man to not have a son born on sunday.

All questions start with the original sample space in which all outcomes have equal probability for each combination of gender and days of week (you can consider days of week in Q1 and Q2, it will not change the result).

Eliminate the impossible outcomes above.

After elimination, each of these different sample spaces is a different size, and result in different values when taking the number of male-male outcomes (I use plural here only for when considering days of the week, so gender and day of week together make up an outcome in such cases) and dividing by the number of outcomes in the sample space.

I don't know if we will agree on this problem, so maybe we should just agree to disagree.

I think Q2 is called the Boy Or Girl paradox, you can read more about it if you want.

Edited September 24, 2010 by mmiguel1

[Guest]

I was using your definition... not mine.

The index was to indicate a unique individual, as you required - not order of birth. I was using the leftmost element as the older child. With an index for birth order, let us then identify the two male possibilities as {M1, m2} and {m1, M2} and the female-female possibilities (if there was no restriction, but two children) would be {F1, f2} and {f1, F2}. The male-female possibilites would be {M1, F2} and {F1, M2}. Given the first born is a male, we then have {M1, m2}, {m1, M2} and {M1, F2}. Thus, by your interpretation that MF and FM are not the same, there is a 2/3 probability that the 2nd born is a son. I disagree with this interpretation. As you pointed out, the result should be 1/2. Thus MF and FM ARE the same (as are {M1, m2} and {m1, M2})!!! Where one need interpet the set of elements as different is when the problem asks that they are, which this one did not. Thus in all 3 cases (unless one interprets the 3rd problem's phrase "one of which" to mean "one and only one of which", there is a 1/2 probability -- as many have agreed -- that the second child is also a male.

By the way, I solved this problem with my own notation before you even came here. So if you want my definition, it is already there is a post somewhere.

If you compare it to the fabrication that you produced and claimed was my definition, you will see that you do not have my definition, you made an incorrect interpretation of what I was saying.

You continue to make the same incorrect interpretation of what I was saying.

Also, it's a little presumptuous to assume that without reading what the other guy said, that you can just swoop in days later and enlighten him about his own understanding by giving it only a few seconds of thought.

I am not as dim as you think I am.

[Guest]

By the way, I solved this problem with my own notation before you even came here. So if you want my definition, it is already there is a post somewhere.

If you compare it to the fabrication that you produced and claimed was my definition, you will see that you do not have my definition, you made an incorrect interpretation of what I was saying.

You continue to make the same incorrect interpretation of what I was saying.

Also, it's a little presumptuous to assume that without reading what the other guy said, that you can just swoop in days later and enlighten him about his own understanding by giving it only a few seconds of thought.

I am not as dim as you think I am.

I'd rather not use your definition, but the definition given by those who have introduced combinatronics and permutations and worked extensively with such, as Euler, Gauss and many others. The answers are 1/2, 1/2, and 1/2.

[Guest]

The probability of A {a son being the other child of two children}, given B {a son being one of the children of two children} is given by the conditional probability equation P(A|B) = P(A ∩ B)/P(B).

As A and B are independent {the fact that one child is a son has no effect on the gender of the second child},

P(A ∩ B) is given to be P(A)P(B). We are given that P(B) = 1 {at least one of the two children is a son},

The probability that P(A) {the other child} is 1/2 {i.e., a son or a daughter}, therefore P(A) = 1/2,

thus P(A|B) = 1/2.

-----

Where the first born child is given first in the ordered set {a,b}, with M being a son (male) and F being a daughter (female), and x being the unknown.

Q1 asks P[{M,M} | ({M,x} U {x,M})]

Q2 asks P[{M,M} | ({M,x}]

The probability for both cases is 1/2.

Edited September 25, 2010 by Dej Mar

[bonanova]

The probability of A {a son being the other child of two children}, given B {a son being one of the children of two children} is given by the conditional probability equation P(A|B) = P(A ∩ B)/P(B).

As A and B are independent {the fact that one child is a son has no effect on the gender of the second child},

P(A ∩ B) is given to be P(A)P(B). We are given that P(B) = 1 {at least one of the two children is a son},

The probability that P(A) {the other child} is 1/2 {i.e., a son or a daughter}, therefore P(A) = 1/2,

thus P(A|B) = 1/2.

-----

Where the first born child is given first in the ordered set {a,b}, with M being a son (male) and F being a daughter (female), and x being the unknown.

Q1 asks P[{M,M} | ({M,x} U {x,M})]

Q2 asks P[{M,M} | ({M,x}]

The probability for both cases is 1/2.

Dej Mar, when you speak of one of the children and the other child, you are making the mistake of identifying one of them, specifying its gender, then finding the gender probability of the other. That probability is always 1/2.

I have two children, named Leslie and Pat.

Leslie is a boy. What's the probability that both my children are boys?

That reduces to: I have a child named Pat. What is the probability that Pat is a boy?

Read the

Edited September 25, 2010 by bonanova
Change: he To: Pat

[Guest]

Dej Mar, when you speak of one of the children and the other child, you are making the mistake of identifying one of them, specifying its gender, then finding the gender probability of the other. That probability is always 1/2.

I have two children, named Leslie and Pat.

Leslie is a boy. What's the probability that both my children are boys?

That reduces to: I have a child named Pat. What is the probability that he is a boy?

Read the

That is where it gets interesting.

I think I better understand what you meant by including the informant in the calculations after thinking a little more about the rolling a 7 puzzle.

I stopped thinking about it last night with the concept that there is a difference between knowing whether a particular child is a boy, or knowing whether at least one of the children is a boy, but not knowing which one.

I had thought of this before, days ago, but now I am a little more confused.

What is the extent of "particular" here?

In your above example you name the children Leslie and Pat. The unique characteristics serve as identifiers. What if the children had no name? Then you could easily pick a different characteristic by which to identify them.

But will any characteristic work?

Could an informant's choice be considered a characteristic of a child if the choice behaves asymmetrically with respect to the children?

If the two children have opposite gender, then the informant had to pick one of them to tell you about, wouldn't he ? (maybe not, depending on informant method)

But couldn't the informant's choice itself serve as a distinguishing and identifying characteristic?

Instead of calling child 1 Leslie and child 2 Pat, could you call child 1 (the one the informant picked), and child 2 (the one the informant didn't pick).

If we use the informant's choice as an identifying characteristic then wouldn't we always be in the case where we know the gender of a particular child? (particular here relying on the informant's choice). Then the probability would always be 1/2 wouldn't it? Is not this characteristic as much of a characteristic as being named Pat or Leslie?

Of course there is also the case that the children had the same gender when the informant looked and the informant had no choice, but to pick the gender of both, in that case are the children really distinctly identified? This is assuming the informant looks at both children and then makes the decision.

If the informant picks a child at random, the answer to the puzzle is 1/2.

Consider {MF, FM}

This event's condition occurs 1/2 the time, so to get a case where the informant says a at least one child is male (as opposed to at least one child is female), the probability for that is 1/4 (the additional 1/2 factor comes from picking male).

This 1/4 is the probability that only one child is male given that the informant chose to say at least one male, as opposed to talking about females.

If the informant sees MM, he has no choice but to pick male, and the probability for that is 1/4. This 1/4 is the probabilty that both children are male given that the informant chose to say at least one male, as opposed to talking about females.

Given the informant chose to talk about males, the probability that there are two sons is (1/4)/(1/4 + 1/4) = 1/2

If the informant decided only to look at one of the children, then even if both children had the same gender, then wouldn't they be still be uniquely identified as the one the informant picked and the one the informant didn't pick? This probability would also be 1/2 as the children are uniquely identified.

You offered, to get 1/3:

"If either child [in the inclusive sense] is a boy, I will say "at least one child is a son. Otherwise I will [have to] say "at least one child is a daughter."

I agree this works. This is a very fixed way for the informant to respond, and it is not obvious from the question that the informant is employing this method. The key thing about this is that if the solver believes that this method was employed, then it will not uniquely identify any particular child in the solver's mind.

It is surprising how the probabilities change if the informant does something different (more accurately, if the solver believes the informant does something different).

Presumably, the solver of the puzzle knows nothing of the informant's method.

Yet if the solver believes the informant did any particular method, it will change the solver's probability prediction.

Believing different things gives different probabilities, it's an old topic, yet still sometimes surprising.

In this puzzle, the solver is forced to make an assumption about the informant's method.

If he makes no assumption, then he cannot produce a probability.

For cases like these, it is a language's job to make up conventions such that if no knowledge is given, a default assumption may be assumed.

Bonanova's other post is basically asking, what is the mapping from English questions of this type (given at least one child/coin/die is ...) to informant method of transmitting information. This mapping can have logical foundations or foundations based on convention.

Very cool bonanova.

Edited September 25, 2010 by mmiguel1

[bonanova]

I agree with those statements. Let's talk about how the algorithm of the informant must be considered.

Premiss:

When we are given data, we have to consider the probabilities that lie behind them.

[1] The classic Monty Hall problem

There is a 2 to 1 favorability to switch doors when we know that Monty knows the extra door he opens will show a goat. Otherwise, we have to consider the probability that he would open a door that had a goat. If he DIDN'T know, and showed a goat, that fact increases the probability that our first choice has the car. Because if we chose the car door, it would be CERTAIN that his door would show a goat. In fact, it increases the probability that we initially chose the car to EXACTLY 1/2 [exercise left to the student], making it pointless to switch - both of the other doors have 1/2 probability.

Now let's consider what happens when the informant is known to have NO CHOICE when he gives us information. This happens when he simply responds to a question. A question that WE choose.

In Monty Hall, I choose a door. Before I open it, I point to a second door and ask Monty to open it. If it shows a car, I'm going to win. If it shows a goat, then I'm back to even odds again. The two outcomes [certainty and even odds] are weighted by the probability that I pointed to a door that has the car. That probability is 1/3. So the total probability is 1/3 x certainty + 2/3 x even odds = 1/3 + 1/3 = 2/3. That is, I will win the car 2/3 of the time if I get a peek at the second door.

And that's the same result as when Monty knows that he will show us a goat.

[2] The child gender problem

Without knowing how the informant chose to tell us about his girls, if any, the probability is unknown. We know only that it's 1/3, 1/2, or something in between. Let's remove the uncertainty.

Informant: I have two children.

Me: Is one of them a girl?

Informant: Yes.

The probability of Informant having two girls is 1/3.

There is no ambiguity here, because the informant completely describes the situation I asked about. I asked him, in effect, whether I could eliminate the BB case from the general population. He told me that I could. There is no other relevant information that he could have given. Since he had no choice, we don't have to guess his algorithm. His algorithm, as a trusted observer, is simply to give a truthful answer. His answer permits the simple calculation that p[GG] = GG/[GG + BG + GB] = 1/3.

If he had simply ventured the information that one is a girl, he might have been thinking only of the older, or the taller, or the one whose name came first in the alphabet. Then it would be the absurdly trivial case of p[GG] = p[other child = girl] = 1/2. If I don't know why he gave me the information, I know only that 1/3 <= p[GG] <= 1/2.

[3] The informed two dice totaling 7 problem

The unconstrained probability is 1/6 that fair dice total 7. If we learn from the informant that one of the dice is 6, then p[7] changes. How it changes depends on why the informant chose to tell us about sixes, rather than about fives, or threes. More to the point, since those cases are symmetric, why he chose to tell us about a number that appeared, rather than a number that was absent. Since we don't know, the probability is unknown. Let's remove the ambiguity.

Informant: I just rolled two fair dice

Me: Is at least one of the dice a 6?

Informant: Yes.

The probability now that the dice total 7 is 2/11.

Again no ambiguity. There is no other relevant information that could be given in response to the question that I chose to ask. For example, he could not tell me about fives, or threes. So we simply count the cases where fair dice show 6 [there are 11 cases] and of them, the number [two] that total 7: 1-6 and 6-1.

Does this make sense?

Yes. Consider that of the 36 possible cases, 11 show a 6 and 25 do not. So, we know the probabilities that lie behind that data we got from the Informant: p[Yes] = 11/36; p[No] = 25/36. Of those 25, there are four that total 7: 2-5, 5-2, 3-4, and 4-3. So if Informant had answered No, then the 7-total probability would have been 4/25. Note that 2/11 is slightly greater, and 4/25 slightly less, than 1/6.

So now we have the whole picture, and the weighted sum is seen to be what unconstrained result:

p[7] = 1/6

p[7] = p[7|Yes] x p[Yes] + p[7|No] x p[No]

p[7] = p[7|Yes] x 11/36 + p[7|No] x 25/36

p[7] = 2/11 x 11/36 + 4/25 x 25/36

p[7] = 2/36 + 4/36 = 6/36.

What's the takeaway here?

The takeaway is that algorithms matter.

Unconstrained, p[7] = 1/6. When Informant tells us one die is a 6, we don't know p[7] any more. Why did he choose to talk about sixes? We don't know. That leaves the answer unknown. But when WE make the choice to talk about sixes, we remove the ambiguity. We know that it will change, and we know how it will change. If he says Yes, it will become 2/11. If he says No, it will become 4/25. And the sum of these results, weighted by the known probabilities of the informant's possible responses, brings us back to the 1/6 case.

We need to know the informant's algorithm when he ventures information. What is the probability he chose [to talk about sons rather than to talk about daughters] [to talk about sixes rather than to talk about fives] and then what is the probability of the data itself [believing what he says it true.] When we chose to talk about sixes, we can calculate them - they're 11/36 for a Yes answer and 25/36 for a No answer.

With all that information, the ambiguity is gone. Without it, the answer in unknown, although we can compute bounds, so long as the information is relevant. The phase of the moon, for example, wouldn't be helpful in any of these problems.

[Guest]

I agree with those statements. Let's talk about how the algorithm of the informant must be considered.

Premiss:

When we are given data, we have to consider the probabilities behind them.

[1] The classic Monty Hall problem

There is a 2 to 1 favorability to switch doors only when we know that Monty KNEW the extra door he opens shows a goat. Otherwise, we have to consider the probability that he would open a door that had a goat. If he DIDN'T know, and showed a goat, that fact increases the probability that our first choice has the car. Because if we chose the car door, it would be CERTAIN that his door would show a goat. In fact, it increases the probability that we initially chose the car to EXACTLY 1/2 [exercise left to the student], making it pointless to switch - both of the other doors have 1/2 probability.

Now let's consider what happens when the informant is known to have NO CHOICE when he gives us information. This happens when he simply responds to a question. A question that WE choose.

In Monty Hall, I choose a door. Before I open it, I point to a second door and ask Monty to open it. If it shows a car, I'm going to win. If it shows a goat, then I'm back to even odds again. The two outcomes [certainty and even odds] are weighted by the probability that I pointed to a door that has the car. That probability is 1/3. So the total probability is 1/3 x certainty + 2/3 x even odds = 1/3 + 1/3 = 2/3. That is, I will win the car 2/3 of the time if I get a peek at the second door.

And that's the same result as when Monty knows that he will show us a goat.

[2] The child gender problem

Without knowing how the informant chose to tell us about his girls, if any, the probability is unknown. We know it's 1/3, 1/2, or something in between. Let's remove the uncertainty.

Informant: I have two children.

Me: Is one of them a girl?

Informant: Yes.

The probability of Informant having two girls is 1/3.

There is no ambiguity here, because the informant completely describes the situation I asked about. I asked him, in effect, whether I could eliminate the BB case from the general population. He told me that I could. There is no other relevant information that he could have given. So he had no choice, we don't have to guess his algorithm. His algorithm, as a trusted observer, is simply to give a truthful answer. His answer permits the simple calculation that p[GG] = GG/[GG + BG + GB] = 1/3.

If he had simply ventured the information that one is a girl, he might have been thinking only of the older, or the taller, or the one whose name came first in the alphabet. Then it would be the absurdly trivial case of p[GG] = p[other child = girl] = 1/2. If I don't know why he gave me the information, I know only that 1/3 <= p[GG] <= 1/2.

[3] The informed two dice totaling 7 problem

The unconstrained probability is 1/6 that fair dice total 7. If we learn from the informant that one of the dice is 6, then p[7] changes. How it changes depends on why the informant chose to tell us about sixes, rather than about fives, or threes. Since we're not told, the probability is unknown. Let's remove the ambiguity.

Informant: I just rolled two fair dice

Me: Is at least one of the dice a 6?

Informant: Yes.

The probability now that the dice total 7 is 2/11.

Again no ambiguity. There is no other relevant information that could be given in response to the question that I chose to ask. For example, he could not tell me about fives, or threes. So we simply count the cases where fair dice show 6 [there are 11 cases] and of them, the number [two] that total 7: 1-6 and 6-1.

Does this make sense?

Yes. Consider that of the 36 possible cases, 11 show a 6 and 25 do not. So, we know the probabilities that lie behind that data we got from the Informant: p[Yes] = 11/36; p[No] = 25/36. Of those 25, there are four that total 7: 2-5, 5-2, 3-4, and 4-3. So if Informant had answered No, then the 7-total probability would have been 4/25. Note that 2/11 is slightly greater, and 4/25 slightly less, than 1/6.

So the whole picture emerges:

p[7] = 1/6

p[7] = p[7|Yes] x p[Yes] + p[7|No] x p[No]

p[7] = p[7|Yes] x 11/36 + p[7|No] x 25/36

p[7] = 2/11 x 11/36 + 4/25 x 25/36

p[7] = 2/36 + 4/36 = 6/36.

What's the takeaway here?

The takeaway is that algorithms matter.

Unconstrained, p[7] = 1/6. When Informant tells us one die is a 6, we don't know p[7] any more. Why did he choose to talk about sixes? We don't know. That leaves the answer unknown. But when WE make the choice to talk about sixes, we remove the ambiguity. We know that it will change, and we know how it will change. If he says Yes, it will become 2/11. If he says No, it will become 4/25. And the weighted sum of these results brings us back to the 1/6 case.

We need to know the informant's algorithm when he ventures information. What is the probability he chose [to talk about sons rather than to talk about daughters] [to talk about sixes rather than to talk about fives] and then what is the probability of the data itself [believing what he says it true.] When we chose to talk about sixes, we can calculate them - they're 11/36 for a Yes answer and 25/36 for a No answer.

With all that information, the ambiguity is gone. Without it, the answer in unknown, although it becomes bounded [1/3 < p[same gender < 1/2], so long as the information is relevant. Telling us the phase of the moon, of course, doesn't limit the answer.

Monty Hall: Interesting point of view.

In your scenario, the contestant is actually picking two doors isn't he?

In either case, I don't think you recommend switching any doors, (what's the point if 1/2 and 1/2 for even odds).

Therefore, if the car is behind either of the two doors the contestant picks, he will win. Thus 2/3.

I thought that the original Monty Hall Problem was a little different than this.

In the original Monty, assuming a switch, you win if you were initially wrong, and you lose if you were initially right. You have 2/3 of a chance of picking wrong with one selection.

Your new scenario is: 2/3 of a chance of picking right with two selections.

It just seems a little different to me.

Gender Problem

I like your example.

I associate the result of 1/3 with the fact that in the transmission of information from informant to solver, no particular child was distinguished in the perspective of the solver.

Your example is a way of doing that.

I find it interesting to note, that the answer only changes if the solver believes the informant distinguished between the two children in some way when transmitting the information.

If the informant ventured the information, he may have well used any method he likes.

His methods will not change the answer though because the solver does not know what method he used.

The answer is with respect to the knowledge of the solver.

On the other hand, the informant could have used a method in which he did not distinguish the children when making his decision (like method 2 and 3 in your list), but if the solver believes that the informant distinguished between children (with symmetrical probabilities for simplicity), then the answer would be 1/2.

It's a subtle point to me, that the answer is not determined by what the informant did, but by what the solver thinks the informant did. The probabilities of a particular perspective heavily depend on what information is believed in that particular perspective.

Your example handles this problem and makes it so that the informant could not have distinguished between the children in the transmission of the information.

If the solver has no idea what the informant did, then what is most conventional thing to assume?

That is the topic in the other post.

Rolling 7

Very neat.

I was thinking about this last night.

Let's say the magician picks one number randomly to tell us (not your example in this post).

There are two cases I want to look at:

(1) Both dice are the same (both are x)

(2) The dice are different (one is x, the other is y)

The dice will be the same with prob 1/6, and diff with prob 5/6

If they are the same (lets say x), then the magician tell you at least one is x.

If they are different, the magician will randomly pick either x or y and tell you which one it is.

Assume he symmetrically picks x with prob 1/2 and y with prob 1/2.

Let's say x corresponds to die 1 (uniquely identified by this assigned number), and y corresponds to die 2.

I denote outcomes like this:

(Die 1 value, Die 2 value)

(1,5) means die 1 has value 1, die 2 has value 5, and x=1, and y=5.

Case 1: Both die are the same, magician picks x: prob for this case is 1/6

Case 2: Die are different, magician picks x: prob for this case is 5/12

Case 3: Die are different, magician picks y: prob is 5/12

Lets say the value the magician told us is v.

In case 1, v = x = y

in case 2, v = x

in case 3, v = y

Then (v,v) is the answer with 1/6 prob (according to case 1)

For case 2 and 3,

Let z represent values from 1 to 6, not including v

For case 2, (v,z) are the possible outcomes (all 5 of them).

The union of these outcomes has prob 5/12.

They are equiprobable to one another, so each particular outcome has 1/5 * 5/12 = 1/12

Analogously for case 3, each outcome has prob 1/12.

The sum of the dice, given that the magician has said at least one is v can range from 1+v to 6+v.

The sum value 2*v has a single outcome with prob 1/6, this outcome is (v,v)

All other sum values have two outcomes, each with prob 1/12.

This means that all sum values have probability 1/6. This includes rolling a 7.

All of this logic is speculation on the solver's side.

This speculation included having the magician distinguish between the dice.

When the solver believes that the information comes from the dice being distinguished, the answer is 1/6.

i.e. if one die is red, and the other die is blue, and the magician says, the red die has a value of 1, what is the prob you have rolled a 7?

Your example bonanova also prevents the solver from reaching this conclusion.

In your example, the solver assumes that the information received does not distinguish the dice. In that case the answer is 2/11.

The solver's belief in the informant's algorithm will change the probabilities that the solver will come up with.

Brilliant!

[bonanova]

In your scenario, the contestant is actually picking two doors isn't he?

Sort of. Well, yes. He's directing Monty's choice. Same as if Monty chooses randomly.

The contestant never opens two doors, tho.

If Monty certainly shows a goat, swap [third door] gives p[win] = 2/3.

But if Monty randomly opens a door, then

if a car appears, swap [car] gives p[win] = 1.

if a goat appears, swap [third door] gives p[win] = 1/2.

Given that a goat appears, the contestant can always swap with the third door.

If Monty knows where the car is, he doubles his chances from 1/3 to 2/3.

If Monty doesn't know, his original choice and the third door both go to 1/2 and it doesn't hurt.

So there's an interesting point, and I don't remember ever hearing it.

The contestant's expectation of winning is 2/3 regardless of Monty's algorithm of choice.

. . . . . If Monty knows the car's location and shows a goat, contestant is certain of 2/3.

. . . . . If he doesn't know, contestant gets a weighted average of 1 x 1/3 + 1/2 x 2/3 = 1/3 + 1/3 = 2/3.

The only thing that changes with Monty's method is p[win] for "swap with the third door."

If the strategy includes "grab the car if it appears", then p[win] is always 2/3.

[Guest]

Sort of. Well, yes. He's directing Monty's choice. Same as if Monty chooses randomly.

The contestant never opens two doors, tho.

If Monty certainly shows a goat, swap [third door] gives p[win] = 2/3.

But if Monty randomly opens a door, then

if a car appears, swap [car] gives p[win] = 1.

if a goat appears, swap [third door] gives p[win] = 1/2.

Given that a goat appears, the contestant can always swap with the third door.

If Monty knows where the car is, he doubles his chances from 1/3 to 2/3.

If Monty doesn't know, his original choice and the third door both go to 1/2 and it doesn't hurt.

So there's an interesting point, and I don't remember ever hearing it.

The contestant's expectation of winning is 2/3 regardless of Monty's algorithm of choice.

. . . . . If Monty knows the car's location and shows a goat, contestant is certain of 2/3.

. . . . . If he doesn't know, contestant gets a weighted average of 1 x 1/3 + 1/2 x 2/3 = 1/3 + 1/3 = 2/3.

The only thing that changes with Monty's method is p[win] for "swap with the third door."

If the strategy includes "grab the car if it appears", then p[win] is always 2/3.

I agree.

What but would these values coincide if the numbers of doors were different?

I feel like getting 2/3 in both cases is just coincidence here.

4 doors with Monty opening 2:

If Monty knows where the car is:

Monty will open 2 goat doors. By switching, you will win with probability 3/4.

I will indent with spaces in the following probability tree (it might be hard to see though). EDIT: The spaces are just thrown away, and the edit clock is ticking, so I can't make it look nicer... sorry

If Monty doesn't know where the car is:

Two initial cases:

Contestant's initial pick was right: P = 1/4

Monty picks 2 goat doors: P = 1

Stay and Win: P = 1

Switch and Win: P = 0

Contestant's initial pick was wrong: P = 3/4

Monty picks 2 goat doors: P = 1/3

Stay and Win: P=0

Switch and Win: P=1

Monty Picks a goat and a car: P = 2/3

Win=1

Staying and Winning: 1/4 + 3/4*2/3 = 3/4

Switching and Winning: 3/4*(1/3 + 2/3) = 3/4

I stand corrected.

What if there are 4 doors and Monty opens only one?

Monty knows where the car is:

If initial guess was correct: P=1/4

Monty opens a goat door: P=1

Switch and Win: P=0

Stay and win: P=1

If initial guess was wrong: P=3/4

Monty opens a goat door: P=1

Switch and Win: P=1/2

Stay and Win: P=0

Switching wins with probability 3/8

Staying wins with probability 1/4

Note, these do not need to add to 1.

If Monty doesn't know where the car is:

If initial guess was correct: P = 1/4

Monty opens a goat door: P=1

Stay and Win: P=1

Switch and Win: P=0

If initial guess was wrong: P=3/4

Monty opens a goat door: P=2/3

Stay and Win: P=0

Switch and Win: P=1/2

Monty opens the car door: P=1/3

Win: P=1

Staying and winning: P = 1/4 + 3/4*1/3 = 1/2

Switching and winning: P=3/4*(2/3*1/2 + 1/3) = 1/2

In this case, Monty's knowledge makes a difference. It seems you are much better off if he doesn't know where the car is (since he can pick the car himself, giving you the answer).

If he does know where the car is, switching is still better, but not by as much.

Edited September 26, 2010 by mmiguel1

[bonanova]

What but would these values coincide if the numbers of doors were different?

As Melanie Griffith said to her boyfriend in Working Girl, when he didn't like her "Maybe" after he proposed,

"Want a different answer? Ask a different girl."

Here's the puzzle I posed:

Three doors; one car, two goats. You point to a door but do not open it. Monty discloses the contents of a different door. You choose a door and win its contents. Assume Monty acts as your adversary when he can. Assume nothing else.

Find a single strategy that always gives p[car] = 2/3.

Answer:

Pick the car if Monty shows it, otherwise open the third door.

Want a different result? Pose a different puzzle.

[Guest]

0. One of which is a son.

This is quite tricky for some.

For number 3, I think the probability is...