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R.S. Regular user CT one day I'll have 193 Posts 
This is crazy!
(you must watch the video for the full analysis) https://www.youtube.com/watch?v=bDZieLmya_I Scenario # 1: A parent tells you that they have 2 children, and one of them is a girl. What is the probability that the other child is also a girl? Scenario # 2: A parent tells you that they have 2 children, and one of them is a girl, and her name is Julie. Now what is the probability that the other child is also a girl? Apparently, just knowing the girl's name changes the probability of there being 2 girls!! Ron
"It is error only, and not truth, that shrinks from inquiry." Thomas Paine


S2000magician Inner circle Yorba Linda, CA 3469 Posts 
I haven't looked at the video, but I presume that it depends on the fact that if they have a second girl, her name isn't Julie.
You're correct: this is an interesting situation. One that many, many intelligent people (including engineers, physicists, and mathematicians) get wrong. 

Cliffg37 Inner circle Long Beach, CA 2472 Posts 
I too find this interesting, and would also have gotten it wrong. I do take exception to one thing. He makes a big point about a boy and girl vs a girl and a boy. I don't think that's relevant to the issue, obviously, he does think it is relevant.
Magic is like Science,
Both are fun if you do it right! 

landmark Inner circle within a triangle 4682 Posts 
You asked for itit's turtles all the way down.
This has been discussed before by some really amazing folks here including Tomas B and Lobowolf XXX. There are many subthreads referred to as well. Screw your head on tight because it's about to be spinning: https://www.themagiccafe.com/forums/view......orum=101
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All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 8, 2019, Cliffg37 wrote: Think older boy, younger girl vs. older girl, younger boy. 

R.S. Regular user CT one day I'll have 193 Posts 
Quote:
On Apr 8, 2019, landmark wrote: Thanks landmark. Wow, this thread goes way back. The video I posted is helpful in the visualization, I think. Anyway, I'm still trying to wrap my brain around it. Ron
"It is error only, and not truth, that shrinks from inquiry." Thomas Paine


S2000magician Inner circle Yorba Linda, CA 3469 Posts 
At 0:24 he mentions that he was trying "to intuitively understand" what the author of the book had said.
Apart from the split infinitive, one cannot "intuitively understand" anything. Intuition means arriving at the correct conclusion without understanding why. I post a lot on a forum for the Chartered Financial Analyst (CFA) exams and have found over the last few years many, many candidates there asking for help "to understand the intuition" behind various various ideas, formulae, and what have you. It's infuriating, and it's caused primarily by prep providers who frequently refer to "the intuition behind" some idea or another when they mean the explanation behind it. I, for one, would much rather have understanding than intuition. In the book Please Understand Me, about the MyersBriggs personality types, the Portrait of an INTP includes this sentence: INTPs deal with their environment primarily through intuition, and their strongest quality, the thinking function, remains relatively hidden except in close associations. I disagree. I believe that INTPs deal with their environment primarily through understanding, but that their assessment of new situations is so quick that it appears to be intuition. I'm reminded of Holmes' first encounter with Watson, in which Holmes said, "You've been in Afghanistan, I perceive." It appears intuitive, but later in the book, A Study in Scarlet, Holmes outlines the steps in his thinking process that led to that conclusion. Understanding masquerading as intuition. 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Years ago I attended a lecture by Paul Erdős in which he posed a similar problem:
You're playing bridge. [For those of you who are unfamiliar with bridge, first, shame on you, and, second, you deal four hands of 13 cards each, face down, one to yourself, one to your partner who sits opposite you, and one each to your two opponents, one of whom sits to your left and the other of whom sits to your right.] Before you look at your cards, your righthand opponent (RHO) picks up his cards and announces, truthfully, "I have an ace." [Aces are high in bridge.] There is a probability that he has a second ace, and given enough skill, time, and motivation, you could calculate that probability. The next time you deal [in bridge the deal rotates clockwise around the table], before you look at your cards, RHO picks up his cards and announces, truthfully, "I have the ace of spades." Again, there is a probability that he has a second ace, and given enough skill, time, and motivation, you could calculate that probability. The question is: how do those two probabilities compare? Is the first greater than the second, or is the second greater than the first, or are they, in fact, equal? To simplify the understanding, Dr. Erdős proposed an easier problem. Instead of having 52 cards, you have only three: the ace of spades (SA), the ace of hearts (HA), and the king of diamonds (DK). And instead of dealing 13 cards to RHO, you deal only two. Suppose he looks at them and says, truthfully, "I have an ace." What, exactly, has he told you? The answer is . . . nothing! You already knew that he had an ace, because there's only one card that isn't an ace (DK), and he has two cards. So the probability of him having a second ace is ⅓: there are three hands he could hold (SA & HA, SA & DK, HA & DK), and only one of those has two aces. Suppose, instead, that he says, truthfully, "I have the ace of spades." Now he's told you something: he doesn't have the hand comprising exactly the HA and DK. Now there are only two possible hands he could hold (SA & HA, SA & DK), only one of which has a second ace, so the probability in this case is ˝. Going back to the original problem, as I opined earlier, it does depend on the assumption that if there are two daughters they're not both named Julie. 

tommy Eternal Order Devil’s Island 15681 Posts 
Martin Gardner published one of the earliest variants of the paradox in Scientific American. https://en.wikipedia.org/wiki/Boy_or_Girl_paradox
If there is a single truth about Magic, it is that nothing on earth so efficiently evades it.
Tommy 

yachanin Inner circle Cleveland, OH 2066 Posts 
Hi All,
There is no paradox here, only some fast talking con. Consider this: I've flipped two coins and tell you one of them has turned up "heads." What is the probability the other is a "head?" If you think it's 33%, "Go directly to Jail. Do not pass Go. Do not collect $200." The problem with the "two daughters paradox" is that the explanation of why the answer is 33%, is flawed. In that explanation, you are asked to believe that bb, bg, gb, and gg are four different possible outcomes. Given 100 pairs of siblings and assuming equally likely outcomes, there would be 25 bb, 25 bg, 25 gb, and 25 gg. Eliminate the 25 bb pairs, and you are left with 25 gg pairs divided by 75 pairs with at least one female... 33%. However, the "birth order" of the genders is not important in this probability problem and, therefore, bg and gb should not be considered as two different possible outcomes. All that talk about which was born first or second is just fast talking con to make your brain twist in your head. There are, in reality, only three possible outcomes: 1) "both male," 2) "one of each," and 3) "both female." Give 90 pairs of siblings (just to keep the math simple) and assuming equally likely outcomes, there would be 30 "both male," 30 "one of each," and 30 "both female." Eliminate the 30 "both male" pairs and you are left with 30 "both female" pairs divided by 60 pairs with at least one female... 50%. Regards, Steve
"Impossible? Your audience will think so..." TM
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Cliffg37 Inner circle Long Beach, CA 2472 Posts 
As complicated as the Original video makes this, I think Steve hit the nail on the head. As I said before I did not take birth order into account as I thought it was not relevant to the question at hand.
Magic is like Science,
Both are fun if you do it right! 

landmark Inner circle within a triangle 4682 Posts 
Quote:
There are, in reality, only three possible outcomes: 1) "both male," 2) "one of each," and 3) "both female." Give 90 pairs of siblings (just to keep the math simple) and assuming equally likely outcomes, there would be 30 "both male," 30 "one of each," and 30 "both female." Eliminate the 30 "both male" pairs and you are left with 30 "both female" pairs divided by 60 pairs with at least one female... 50%. This, I believe, is not correct. It may be true that there are only three outcomes, but they are not equally likely. Go back to the flipping two coins 100 times. Are you saying that two tails will show up as often as one of each? If so, that's not correct. In fact, one of each is twice as likely.
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 

yachanin Inner circle Cleveland, OH 2066 Posts 
Hi Landmark,
I interpret the question, as given, that the parent has two children at the time the statement was made by the parent: two boys, a boy and a girl, or two girls. The possible outcomes are fixed. Calculating the probability of what children he might have before he has any children is irrelevant. That is, the question does not ask about a future event (i.e., I will have two children and one will be female; what is the probability the other will be female?). In terms of "pennies," there are two pennies on the table in front of me as I type. One is a "tail." What is the probability the other is a "tail?" Regards, Steve
"Impossible? Your audience will think so..." TM
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S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 11, 2019, yachanin wrote: You are absolutely correct: if you assume that they're equal then they're equal. The problem lies in your assumption that they're equally likely; they're not. One of each is twice as likely as both male, and twice as likely as both female. It's that simple. 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 12, 2019, yachanin wrote: Do you know which one is a tail? 

yachanin Inner circle Cleveland, OH 2066 Posts 
Hi S2000magician,
Perhaps trying to put my explanation in the same format as that used by the gentleman in the video was a poor choice and just added some unnecessary confusion. I should have just gone with my explanation in the post above yours in the first place Regards, Steve
"Impossible? Your audience will think so..." TM
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yachanin Inner circle Cleveland, OH 2066 Posts 
Yes, I know which is the tail.
"Impossible? Your audience will think so..." TM
Thought Association Card Triangulation Word Search Detective Christmas Eve Sights  Start A Family Tradition 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 12, 2019, yachanin wrote: Howdy, Steve. The problem with your explanation isn't the format, it's the content. I'm sorry, but you're simply wrong. Try this: instead of two pennies, use a penny and a nickel. Assuming that they're fair coins, what are the (all equally likely) head/tail combinations? 

S2000magician Inner circle Yorba Linda, CA 3469 Posts 
Quote:
On Apr 12, 2019, yachanin wrote: That's the problem. 

yachanin Inner circle Cleveland, OH 2066 Posts 
It seems the problem is one of interpreting what is being asked of the problem solver. I see the question simply as "I have a second child. Is it a male or female."
Regards, Steve
"Impossible? Your audience will think so..." TM
Thought Association Card Triangulation Word Search Detective Christmas Eve Sights  Start A Family Tradition 

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