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landmark Inner circle within a triangle 4755 Posts 
From the Drunkard's Walk by Leonard Mlodinow:
1) A mother has two children. One of them is a girl. What is the probability the other child is a girl? 2)A mother has two children. One of them is a girl named Florida. What is the probability the other child is a girl? Assume boys and girls are born with equal probability. Assume 1 out of a million girls are named Florida. Let's take answers, then I'd like to discuss some of the implicationsbecause this is blowing my mind! Jack
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TomasB Inner circle Sweden 1143 Posts 
1) What's the probability that you'd say "One is a girl" instead of "One is a boy" if there is one of each?
/Tomas 

Scott Cram Inner circle 2677 Posts 
1) I believe the probability the other one is a girl is 1 in 3 (as we don't know whether the girl in question is the older or younger child  knowing that information could reduce it to 1 in 2).
2) The name doesn't seem to change anything. I'd think it would still be 1 in 3. 

landmark Inner circle within a triangle 4755 Posts 
Not sure what Tomas is getting at . . . but Scott is only half correct!
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

TomasB Inner circle Sweden 1143 Posts 
I'm getting at the light I saw in
http://www.themagiccafe.com/forums/viewt......orum=101 Especially check out the posts by LobowolfXXX. /Tomas 

landmark Inner circle within a triangle 4755 Posts 
Well, despite my being three years too late on this, that was a fascinating thread Tomas. Thanks. I, (and I suppose Mlodinow) will need to think more about this.
Mlodinow sees these two situations as different. But now I'm going to have to rethink this in light of the thread you pointed out. Before reading Mlodinow, my thinking was the same as Scott's. After reading Mlodinow, Mlodinow confirmed my thinking for #1, but surprised me with his analysis for #2, claiming it was different than #1. This I thought was the intriguing part of the puzzle. His argument convinced me. Now that I've read that Café thread, I'm convinced that I was wrong about #1, but I'm not sure at all about #2 anymore. So, I'm going to amend my question to just #2. Those who are interested in #1 still, can check out the above link. A humbled math teacher, Jack
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

TomasB Inner circle Sweden 1143 Posts 
1) I too was convinced of the 1/3 probability before
http://www.themagiccafe.com/forums/viewt......orum=101 2) This sounds very interesting. Having two girls would increase the probability of having a girl with such a rare name. So it feels like it's a higher probability of two girls than in case 1). /Tomas 

TomasB Inner circle Sweden 1143 Posts 
2) I wonder if that's the same as saying "The child named Florida is a girl." possibly making that equivalent to the version "The oldest child is a girl." in which case the classic answer is 0.5 probability of the other child being a girl.
I'm confused. /Tomas 

landmark Inner circle within a triangle 4755 Posts 
When Tomas is confused, I worry.
When I am very confused about probability I turn to my older brother who has a terrific ability to cut through to the heart of the matter and clarify. So, here's what I now believe , based on discussion with him: First I should have, of course, worded the problem to say that the mothers have "at least" one daughter, though I assumed you knew that. 1) The question as posed above, has an answer of 1/3. Consider that given that we know there is at least one daughter, we have three kinds of families {gb, bg, gg}. Only one of those families has two daughters, hence 1 in 3. The classic answer. All the rest of the cited threads try to compare this situation with situations which are NOT analogous. For example in the case of a girl who says "I have one sibling," the probability the other child is a girl is 1/2 because out of the four possible girls, two of them have sisters. We have to distinguish between what is going on with the families as opposed to what is going on with individual siblings. For example, if the govt issued a plaque that said "Congratulations on your two girl family!", what fraction of two child families with at least one girl would have such a plaque? 1/3. But what fraction of girls who come from families with at least one girl could say they have such plaque in their home? 1/2. 2) So now, given this answer of 1/3 for question 1, does giving the girl the name of Florida make a difference? What about if the name was a much more common one? Tomas, maybe you want to explain further you reasoning for .5 for question two? Jack
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TomasB Inner circle Sweden 1143 Posts 
1) Let's get this one first. Do you really mean that "The question as posed above, has an answer of 1/3"? The question is missing the information of how often you'd say "at least one is a girl" instead of "at least one is a boy" when there is one of each.
Since that distribution is not mentioned, let's assume you flip a coin when you see mixed siblings to decide what to say. That gives the probability 0.5 of the other being a girl. If you instead say the gender you happen to lay your eyes on first it's probably 75% chance or more that you look at the female first. Not until it's 100% chance of you saying "At least one is a girl" in any mixed couple you see does it turn into 1/3 probability of the other being a girl. This is actually the strangest thing to assume since it's hard to form a reason why anyone forever and ever would only focus on the girl when there are mixed siblings. The problem _can_ be worded to force the 1/3 answer but that's not the case as the problem is stated above. One way to word it is to ask someone you know have two children: "Is at least one a girl?" and if you get the answer "Yes" it's 1/3 probability of there being two girls. /Tomas 

landmark Inner circle within a triangle 4755 Posts 
1) "A mother has two children. At least one of them is a girl. What is the probability the other child is a girl?" Sentence two is a given. I don't decide to choose to say one thing or another. I understand your point if the question read: "I met a woman with two children. I noticed that at least one of her children was a girl . . ." But that is not the question, at least as I posed it. ( It may be that in the other threads the questions were posed in such a way as to allow your interpretation. I'll have to look back.) I think common usage in the way that I have posed the question means that the mother's children belong to the set of families who have at least one girl.
Jack
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

TomasB Inner circle Sweden 1143 Posts 
1a) You know that a mother has two children and see one of them. It happens to be a girl. Then you post it as a problem saying "A mother has two children. At least one is a girl. What's the probability of the other being a girl?"
1b) You know that a mother has two children so you ask her "Is at least one of them a girl?" to which she answers "Yes." Then you post it as a problem saying "A mother has two children. At least one is a girl. What's the probability of the other being a girl?" Case a) sounds much more plausable and less convoluted to me, but since it's not stated in the question I can't know for sure. There are many other ways you can find out if at least one is a girl and it need to be part of the question for us to know the answer for sure. To answer a question like this it's easier to look at planet P which has a small but statistically correct population. There are only four twochildren families (B has two boys, G has two girls and M is mixed): B M M G 1a) It's random which of the two kids you'd spot in each family so you can say that in the first M you see a girl and in the second M you see a boy. Since we know with 100% certainty that you said "At least one is a girl." we only have one of the M and the G as the possible population (or half of each M if you like to see it that way). So with the target population of M and G we see that in half the cases the other child is a girl. 1b) You ask each family and get No, Yes, Yes, Yes. The target population this time is M, M and G. Only in 1/3 of that population is the other child a girl. In both these scenarios you have posed _exactly_ the same problem to us and we have no idea of knowing which it is, but to me a) sounds the most probable. /Tomas 

landmark Inner circle within a triangle 4755 Posts 
At this point I feel we are arguing semantics. I feel we are in the situation of Lincoln's famous, "if you call the tail of a sheep a leg, how many legs does a sheep have? Answer: Four, calling it a leg doesn't make it a leg."
Someone might translate the situation of your 1a into my words, but that doesn't make it a correct translation. They would need to say exactly what you postulated"You know that a mother has two children and see one of them. It happens to be a girl. What is the probability the other child is a girl." I maintain again that common usage of the omniscient form "A mother has two children. At least one is a girl. What's the probability of the other being a girl?" indicates neither 1a or even 1b (though the answer is equivalent), but starting with a sample space of {bb,bg, gb, gg} and given that you will only consider those outcomes which have at least one girl, what is the probability of choosing gg. Otherwise we could be in the situation 1c)You know that a mother has two children and see one of them. It happens to be a boy. Then you post it as a problem saying "A mother has two children. At least one is a girl. What's the probability of the other being a girl?" Jack
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

TomasB Inner circle Sweden 1143 Posts 
Semantics are important here, as I see it. If you pose the problem stating that you mean the omniscient form then it's clear. Otherwise we have to guess. Maybe I estimate that it's an 80% chance that you know that you need to state a problem like this clearly but doesn't. Then my answer would be 0.8*1/2+0.2*1/3 of the other one being a girl.
From your 1c) example I sense that you think that I meant a single woman. In 1a) it's the population of all families where you'd see a girl and in 1b) it's the population of all families that would answer "yes" to the question. The example I made was meerely to point out which population I meant. Since LobowolfXXX pointed this out to me I just think it's really unnecessary to state the problem ambigiously when you really don't have to leave any room for interpretation by formulating it better. /Tomas 

landmark Inner circle within a triangle 4755 Posts 
My 1c) meantbut what if I'm a liar?
Anyway, I surrender to your point about ambiguity! So, now can we go omniscient again and discuss #2?
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All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

TomasB Inner circle Sweden 1143 Posts 
2) If a girl is named a certain name with p probability she exists in a GB family with probability p, in a BG family with probability p and in a GG family with probability p(1p)+(1p)p+p^2.
The conditional probability should be to divide the last with the whole population that fullfills the requirement. p(the family is GGat least one girl has that name) = (2p(1  p) + p^2) / (2p(1  p) + p^2 + p + p) = p(2  p) / p(4  p) = (2  p) / (4  p) So when p is small it's close to 1/2 and when p is large it'd be 1/3 (like when she has the common name "girl"). Just put p = 10^6 for the exact answer. /Tomas 

landmark Inner circle within a triangle 4755 Posts 
<<So when p is small it's close to 1/2 and when p is large it'd be 1/3 (like when she has the common name "girl").>>
I love the parenthetical remark! So you throw two coins. I ask if at least one of those coins is heads. If you answer yes, the probability of HH is 1/3. You throw two coins again. I ask if at least one of those coins is a heads with a date before 1940. If you answer yes, the probability of HH dramatically rises. Yes?
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

TomasB Inner circle Sweden 1143 Posts 
Yes, to me that sounds correct.
The thing we ask for can be any property the object may have. So if you get a "Yes" to "Is at least one heads and really really shiny?" it increases the probability of the other being heads. In fact if the probability of the property you ask for is less than 1 you always end up with a higher probability of the other one being heads _if_ you get a "Yes" to your question. Quite interesting I must say. Thanks for sharing that puzzle. Now I can't stop thinking of it in other situations. Maybe it can be turned into a bar bet. /Tomas 

landmark Inner circle within a triangle 4755 Posts 
Consider this conversation:
Is there at least one head"? "Yes" "Is there at least one head that is very very shiny?" "No" "Is there at least one head before 1940?" "No" "Is there at least one head with a scratch above Lincoln's head?" "No" "Is there . . ." Can I keep asking questions until I get a "yes" to increase the probability that the other coin is a head? Seems counterintuitive if so. This is the point that got me so amazed when first reading about the rare naming.
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

LobowolfXXX Inner circle La Famiglia 1190 Posts 
Tomas, you're giving me flashbacks! How did we ever survive that dog thread!?
lol
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." 

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