Actually I was assuming that I did not look at the coins. (At least where I live it is generally considered not the done thing to offer a coin toss and then to look at the tossed coin before offering the “heads or tails” option to the other person. So I was naively assuming a coin toss where polite behaviour is maintained.)
But sure, if I am behaving badly, I can skew the odds.
This is more to the point in the extended set of Monty Hall problems.
The standard Monty Hall has Monty always opening a losing box and offering the option to switch. Then the probabilities are well understood. But there are three other possible behaviours (at least.)
Monty only does the box reveal randomly.
Monty takes a dislike to you and only does the reveal when you have picked the winning box.
Monty likes you and only does the reveal when you have picked a losing box.
If we are allowing bad behaviour into the problem it may simply leave us with an undefined answer.
I think part of the ambiguity is that BG is treated as distinct from GB. Unless the question calls for age, however, I don’t see why it should be.
So, for example, if the problem is framed as: A has two children. One is a boy. What is the probability that the other child is a girl? Then the answer is 1/2.
However, if the problem is framed as: A has two children. One is a boy. What is the probability that A’s youngest child is a girl? Then the answer is 1/3.
Sine this is clearly wrong, you need to wind back and look for the problem.
I was trying to point out earlier that there seems to be a confusion between two different ways of setting up the calculation. One of sets of combinations with associated probabilities, the other of counting the number of ways the participants can arise.
The first uses sets of possible combinations with probabilities.
The second counts tuples, and requires that each tuple have the same probability.
So in the first way, there are three sets, BB, BG, GG, with probability 0.25, 0.5, 0.25.
The second way has four possible tuples. BB, GB, BG, GG. Since there are four of them, the starting point is that each has a probability of 0.25.
You can proceed with either way, but the way you proceed is different. In the first you update probabilities based on new knowledge.
In the second way, you eliminate possible tuples based on new knowledge.
In both ways, the total of probabilities is always one. If it isn’t you made a mistake.
The trick is to be clear which of the two approaches you are using. If you mix them up, you will get the wrong answer.
And that is a beautifully clear way of explaining a major underlying issue. Thank you.
We’re 40+ posts in and several knowledgeable folks have come to opposite conclusions at various points. Almost certainly due to either inadvertently mix-n-matching elements of Francis’ two approaches, or from unnoticed ambiguities in the posing of the set-up allowing multiple non-crazy interpretations.
I’ve often said that the test-taker can’t be much smarter than the test-writer, lest the taker see all the mistakes, unstated assumptions, and ambiguities inherent in the writer’s sloppily conceived and sloppily written question. And therefore the smarter taker comes up with several equally plausible answers. Only one of which was the one the duller writer had thought of. That’s a sure-fire way to lose a lot of points on a test.
How is it “clearly wrong”? Before you know anything about gender and sequence, the probabilities are:
Boy followed (in age, by younger) girl
Boy, Boy
Girl, Boy
Girl, Girl
If you know nothing else but the number of children and that they follow a binary gender scheme of boy girl, then the SECOND you are informed one of the children is a boy, you know only that A has a boy. Hence you can eliminate the Girl, Girl option. But you don’t know if that boy is the youngest child or the oldest child, so you are still left with your choice of the three remaining possibilities.
So every family that has a two children, one of which is a boy has a 1/3 chance of a girl?
Why are you choosing age as the discriminating factor? Why not choose height? Order then by colour of hair, or the marks they got in their last mathematics test?
None of these matter to the person knocking on the door. The only manner in which they can distinguish between the GB BG pair is by which answers the door.
If the manner in which you distinguish between the GB BG pair is not available to the person knocking on the door, you are not using the correct method, and are confusing allocating probabilities with counting.
If the person knocking on the door cannot distinguish the BG GB pair, they must use the first method, which allocates a probability of \frac 1 2 not \frac 1 3 to the mixed pair. Once the door is opened the probabilities can be updated. You get the right answer.
My favorite version is that the new neighbor has two kids, and randomly you find out one is a boy born on a Thursday. What are the odds the other is a boy? No fun this one.
Because that’s how I wrote the scenario. I am inferring that perhaps, just perhaps, what we have here causing the confusion in the OP:
I tell you that my sister has two kids. You go to her house and knock on the door, and a girl answer the door. “Are you Dorkness’s sister’s kid?” You ask; and the girl says yes. “Is your sibling nonbinary?” you ask, and the girl says no. What are the chances that the other kid is a girl?
…is that the question was poorly worded, as often happens when mathematical word games or word games in general get retold and retold and retold. At some point, “older child” becomes “second child” becomes “other child.” Because otherwise I can’t think of any conceivable way the probability should be 1/3, as the original cite seems to insist it is. The Wikipedia page linked to later, however, contains this alternative framing:
Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
ETA: Where, to be clear, the answer is 1/2. My goal was to construct a vaguely similar scenario that would yield a 1/3 probability. So, to restate, why go by age? Because that’s how I wrote the problem, but I did so specifically because I saw another example where age was also provided, and so used that as my point of departure.
Alternatively, the original cite could simply be wrong.
It claims to be calculating conditional probability, but does not actually seem to do so. It just counts, and insists that only one of the four has been eliminated. That isn’t a calculating a conditional probability.
I agree. The first cite just seems wrong to me. I was simply trying to figure out if there was a way to tweak the question to get the answer it insists upon (1/3). And I think I have done just that with:
A has two children. One is a boy. What is the probability that A’s youngest child is a girl?
That would seem to work. Usefully there are two distinct accessible distinguishing attributes so we can sensibly reason about the counting. The earlier ones didn’t have that.
Seems to me theres a difference in whether the gender of a child is forced on the observer or observed randomly
“Mr Smith has 2 kids.
Men with 2 boys wear a special badge, always
Mr Smith does not wear the badge
Whats the chance he has 2 daughters?” 1/3
“Mr Smith has 2 kids.
Men with 2 boys wear a special badge, always
Mr Smith does not wear the badge
oh look , thats one of his kids running towards us.
Its a girl! - whata the chance she has a sister?l 1/2
Yup, that was my point exactly. I hadn’t realized you were taking it for granted that you weren’t looking at the coins, though: That does, in fact, fully establish the problem as you stated.
That depends on what precisely you mean by the statement “One is a boy”.
That also depends on how you find this out. Did you ask your neighbor “Do you have a boy born on a Thursday”, and he answered “yes”? That yields a different analysis than if you saw a boy at his house, and asked if he was the neighbor’s kid, and then asked what day he was born on.
Is ia true simple explanation that additional non-gendered information, eg day of birth, adjusts the odds because of the cases where the additional info applies to both kids?
I keep repeating - each piece of information affects the odds. How/why the information is revealed affects the odds.
For the “one child of the two is…” Gender A
If you know the order AND gender, then the resulting odds are 1/2 for the other. Because the only possible choices are those where child #X is gender A.
If you don’t know the order, you could be seeing the first child or the second. You have to consider all the cases where Child #1 is A and all the cases where child #2 is A. (GG is the same case in either circumstance)
For the coins - if you determine to expose the coins based on a predetermined choice - “I will always show the first HEADS that I see…” or a specific order “I will always show the first coin tossed.” then the possible results are different than if two coins were tossed and only one, picked at random, is shown.
The same applies to Monty Hall. If Monty Hall can arbitrarily decide to show you or not, based on his own motivation and you cannot know what his motivation is, help or hinder or humour, then all bets are off.
If he is obliged to always show you a non-winner door, then he turns a 1:3 into a 1:2.
Once he opens the door, the odds are 50-50 whether you stay or change. You have to choose to stay or change, and each of those choices has the same odds.
(“But!!”, you say, “When I picked a door it was one in three.” Not true. If Monty has to open a door, then the choice is between whichever door you pick and both the others, one of which is not a winner; so 50-50).
From what I’m seeing, the day etc. does not change the key question -M/F?
Yes, then it’s a different question. “Is the other child a boy or a girl?” has nothing to do with day, and does not affect the odds. knowing the birth order of the child you’re seeing most definitely does affect the odds.
Asking “What are the odds the other child is an older boy born on a Tuesday?” (or some such) is a completely different question which can result in another double-digit response thread… :D.
And so on.
(Similarly, at risk of scattering the focus of this thread to the wind: the birthday paradox, when you have a group of 30 people, what are the odds a pair of that 30 have the same birthday? Pretty good. Whereas, what are the odds of out of 29 other people, one has the same birthday as you? Not at all good odds. Why? Because in the second question, you’re adding a tighter condition - it’s me, not one random pair of 30 persons, that has to match up. Change the detail, change the result.)
What I don’t understand is why so many people seem to assume that birth order is a relevant factor in questions that do not call for any consideration of birth order or relative age.
Your parenthetical is actually huge. Good on ya’ for bringing it up
ISTM a lot of people facing these problems seem to get hung up on an idea like “the kids aren’t changing, so how does more info change the odds?” And the answer to that conundrum is that more info is, in effect, changing the question being asked. Different question means different answer.
Parsing the math out of the English is often (always?) the hardest part of these things.