Thanks. I will now put my hands behind my back when being smug.
However, how did you come to the conclusion that there is a 10% chance? And would this probably change if they had only one arm?
I found that higher math such as abstract algebra and topology proved to be much easier to grasp than probability.
The wonderful thing about the Monty Hall problem is that it is so simple, and yet seems paradoxical to almost all of us. The simplest explanation I have is as follows…
You pick one of three choices. This choice has a 1/3 choice of bring the prize.
Another possible choice is opened. This one does not contain the prize. The prize is in one of the choices. The remaining 2/3 probability must belong to the remaining choice.
The family with two children is a bit less neat because “one of the children is a girl” could mean “only one” or “one or more” and explaining that sort-of gives the solution. If you are told “the elder child is a girl” then you know nothing about the sex of the younger one so the odds of bring a girl are 1/2 if boys and girls are equally common (which they aren’t). The other option is harder to stick into words, so the clearest way is probably to use symbols BB, BG, GB, and GG for the possibilities, and eliminate BB because one is a girl, so the odds of the other one being a girl is 1/3.
If you didn’t get that one right you are in good company. D’Alembert was a first-rate mathematician in a country that had many fine ones, and he got the same problem (only with coin tosses) wrong.
2 Likes
This is known as ‘hand waving’ to mathematicians.
2 Likes
I think of it as the difference between:
A fair coin is tossed twice. At least one of the tosses comes up heads. What are the odds that both tosses came up heads?
and:
A fair coin is tossed and comes up heads. What are the odds it will come up heads again on the next toss?
“Clearly obvious” is the commoner’s language. Let’s just say it is axiomatic.
Answering tricky questions like these comes down to the fact that a difference in information implies a difference in probability. In the first situation, you are told “…one of which is a girl”. This information tells you that it is not the case that both children are boys, and this means that one case is ruled out. In the second situation, you are told that the oldest child is a girl. This rules out two cases – one of which is a pair of boys, and the second of which is where the oldest is a boy and the youngest a girl. Because the two different wordings give you different amounts of information, the probabilities must be different. (1/3 for the first case, and 1/2 for the second).
1 Like
My intuition says the Monty Hall question is bullshit. As I understand it, once one door is opened you should change your pick because the odds increased that it’s the one you didn’t pick. But if someone walks up at that point, their choice is 50-50. So we’re dealing with a relativistic anomaly in that the odds change between the 2 choices depending on when in the process you walked in?
To me that’s the same as saying that if you flipped heads 10 times in a row odds are the next one is going to be tails. They’re not, it’ll be 50/50 same as always.
I don’t consider it ambiguous for two reasons. First, taking the position that one means one or more in a math problem doesn’t jive with me. In your explanation, that makes 1=2 and I feel like recognizing that is the intended ‘counterintuitive’ aspect of the puzzle. Second, the word “other” loses reference to the “one of which” that is a girl.
But if you allow for both children to be girls, then I think the answer still has to be 1/2, not 1/3. In the puzzle, the word “other” is operative. Two children, one girl. What is the other, i.e. the *other child that is not included in “one girl”? This settles the “first” result, G, ruling out BB, but also BG, leaving you with an independent event with ~1/2 odds of G. Rephrased, this would be equivalent to “What are the odds that a girl’s sibling is a girl?”
A lifetime of observation of humanity?
1 Like
That’s what most people’s intuition says, but it’s wrong.
Also, you have the SAME odds as someone who walked up at that point.
When you made your original choice, you had a 1/3 chance of being correct.
If a losing door is then revealed to you, and you switch, you have a 50/50 chance of being right (just like the person walking in at this point.)
If you don’t switch, you will win approx. 1/3 of the time over many trials. If you do switch, you will win 1/2 of the time over many trials.
(You can find sites online that will let you run many trials, choosing to switch or not, and see this actually happen.)
(My apologies if your POINT was that intuition is wrong, but maybe my response will explain it to someone else in that case.)
1 Like
I like those numbers for those interpretations, but I’m still marking it zero in my book.
Because I totally asked that new family about their children. And you know what they said? These exact words: “We have two children. One of which is a girl, the other is … hold on, I have to answer this…”, and I thought they were very rude, and later when I was asked, “What is the probability the other child is also a girl?” I played my D.A.R.E. card and just said “No.”
To be fair, I have a PhD in mathematics, and that is not the most intuitive explanation of the birthday paradox. The usual explanation is to ask the exact opposite question: “How likely is it that no two people (out of N total people) have the same birthday?”
Then you line up these N people. How many different ways can we assign an unique day to the first person: 365, the second person: 364, the third person: 363, … the Nth person: 365-N+1. The number of possible days decreases by one each time because we “use up” one day at each step. Now, each of these days is taken out of a set of 365 so the probability is
q = (365/365) * (364/365) * … * ( (365-N+1)/365 )
(a product since these are independent events)
Then the probability that at least one pair has the same birthday is p = 1-q. If you like then you can either try values of N, or use some (fairly standard) approximations to that product to estimate it.
So, I tried to see if I could sit down and write out what that person was doing but instead I convinced myself that they were probably wrong. There is something broken with “then there are Θ(k) pairs each of which has probability 1/k of being equal”.
1 Like
This has the dual advantage of giving an exact answer, and of using combinatorial arguments that occur regularly (so students have to know them anyway).
As for the Θ(k) solution, I’m always leery of any solution where you have to let the number of days in the year go to infinity. My years are long enough, thank you very much.
1 Like
I don’t know if this will actually help, but one way to think about it is from information theory. Roughly, it is about what yes or no questions could you answer.
There are a few basic questions:
- Is the first child a girl?
- Is the second child a girl?
Knowing the answer to these two binary questions completely determines the kids, so we say that we have two bits of information.
Now, if we know that one child is a girl, can we answer either of those? No, but at least one of them is correct. But does knowing one of the answers determine the other? No, because they could both be answered “yes”.
So how much information was in “one child is a girl”? It is less than one bit’s worth, because you can’t ask one more yes or no question to completely determine the state. (It is kind of like you got half a bit of information.)
Now, there are different pieces of information that you could also give. Like “they have a boy and a girl”, in which case you only need to know the gender of the eldest or youngest to determine the other. So that information is worth one full bit. (It strictly contains the previous “has a girl” information as well.)
I’ll spare you the formal definition of informational entropy needed to do calculations here.
Since there are some applications of the birthday paradox to hash tables, I can see why you might want to let K be large. No hash collisions might be a bit too much to ask though.
1 Like
Yep. Language is imprecise. Most languages do not distinguish between the union of sets (cats and dogs) and the intersection (young and beautiful). Or between ‘one’ and ‘one or more’ (a plurality in patent-speak). This is why we turn to symbols and algebra when things start getting hard.
I follow your interpretation, but it was not the one that I first had. However, I did have a ‘woah, something tricky here’ feeling about that part of the problem.
The Marty Hall paradox is unusual, as it can be precisely stated in ordinary language, and yet it is still hard.
Heh. It is hard to make the siblings problem convincing because of the wording, but I think I can do it…
According to the school register, the Jones family have two children called Robin and Lindsay. Both of those names could be given to boys or to girls. I have once seen the Jones parents going to see the headmistress with a girl. What are the odds that Robin Jones is a girl?
It’s a bit contrived. I am sure Martin Gardner would have made a better job.
1 Like
Except the fact she is a girl is not 50% if children where born in = numbers then yes you could say it was 50/50 but surely the child is more likely to be a boy as 109 boys are born for every girl, if the child is born in the west then the baby boys are more likely to survive and its more likely to be a boy, but if it is in Africa then those boys are likely dead and it back to 50/50 for being a girl, or do you just treat is as 50/50 based on the fact there are only two answers, as this strikes me as something you could make as complicated or as simple as you like…
In china the chance it was boy would be even higher with the huge preference for boys.
Martin Gardner popularized this problem back in 1959. He later decided that the solution he gave was wrong, and that the problem was ambiguous. Hence my comment upthread.
1 Like
