MerakSpielman
First Post
Yeah, ok, as long as it's a bright, shiny, SOLID GOLD nickle, I'm in!
(You mean an opportunity to show your true colors. Smartass.)
(You mean an opportunity to show your true colors. Smartass.)
Best...Puzzle...Ever....
- Thread starter RithTheAwakener
- Start date
Pielorinho
Iron Fist of Pelor
Hey, you told me you'd bet me any amount I want. But I won't hold you to that. Welcher.MerakSpielman said:
And I'll deny many things, but I won't deny being a smartass.
Daniel
RangerWickett
Legend
Wait a second . . . so, after reading that webpage, I see the following ridiculous situation.
I meet a lady, who is alone. I ask her how many children she has, and she says two. At this point, I know the odds of her having one boy and one girl is 50%, the odds of two boys is 25%, and the odds of two girls is 25%.
Now, the universe splits into two sub-universes, red, and yellow.
Alternate Universe 1: I ask her if she has any boys, and she replies that she does. Now I know the odds of two girls is 0%, but what are the odds of two boys vs. a boy and a girl? According to the website, it's 33% two boys, 66% girl-boy.
Alternate Universe 2: A boy runs up, and she hugs him and tells me this is her son. Now I know the odds of two girls is 0%, but according to the website, the odds are 50% two boys, 50% two girls.
In both universes, I learn that she has one boy. Why the HELL does it matter how I learn she has one boy?
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RangerWickett said:
The difference: the scenario you describe in AU2 would be twice as likely to happen with a mother of two boys, since either of the boys might have been the one to run up.
Here's a similar, slightly easier to grasp, paradox: you want to get information about average family size. You conduct two surveys, one of a random sample of parents , the other of a random sample of children. You ask the parents how many children they have, and compute the mean average. You also ask the children how many children in their family, and compute the mean average. You get a significantly higher average family size from the second sample. Why?
Because the larger the family, the more likely it is to get into your sample. A family with 6 kids is six times as likely to get sampled than a family with only one kid. But both are equally likely to make it into your survey of parents.
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MerakSpielman
First Post
Awww... Orsal, you shouln't have given us the answer. Care to go back and delete/spoiler the solution for people who haven't read it yet?
MerakSpielman said:
Oh, sorry about that. I hadn't thought of it as a solution, more as the post-puzzle discussion, and didn't think it really had more spoiler potential than the post I was replying to. But... done, on request.
Or... were you just referring to the last part, in which I explained specifically the paradox I had just described? Should I unspoiler the rest?
Sagiro
Rodent of Uncertain Parentage
Ranger, orsal's explanation is correct, and he said it better than I was saying it.
As to the riddle generally, here's my take:
-Sagiro
As to the riddle generally, here's my take:
The way the original riddle in this thread is phrased is a very simple (yet obviously counterintuitive) example of conditional probability. As stated, it presents you only with these facts:
1. There are two children
2. At least one of them is a boy.
That's it. That's all we know. While we know there is an ordering to these children, we don't know anything about that ordering. The boy could have been the first kid or the second kid, but we don't know which.
The riddle is the same as if we said: "A woman has flipped two coins. They've already landed but you're not allowed to look at them. Then you find out that at least one coin landed heads. Now, what are the odds that the other one is a tail?"
I think all mathematically-inclined people here know that if you flip two coins, there are four equally-likely outcomes:
1. HH
2. TH
3. HT
4. TT
If someone now tells you that at least one coin came up heads, the only effect that has on the problem is that you can now eliminate "TT" as a possibility. It doesn't make any of the remaining three outcomes any more or less likely. (And of specific import, it doesn't make the HH outcome any more likely.) That leaves three equally-likely possibilities:
1. HH
2. TH
3. HT
Now you're allowed to look at the coins.
If it turned out that the result was #1, then if you look at a coin that is heads, the other one is heads.
If it turned out that the result was #2, then if you look at a coin that is heads, the other one is tails.
If it turned out that the result was #3, then if you look at a coin that is heads, the other one is tails.
Thus, in two out of the three cases, the other coin is a tails, or a girl in the original riddle.
1. There are two children
2. At least one of them is a boy.
That's it. That's all we know. While we know there is an ordering to these children, we don't know anything about that ordering. The boy could have been the first kid or the second kid, but we don't know which.
The riddle is the same as if we said: "A woman has flipped two coins. They've already landed but you're not allowed to look at them. Then you find out that at least one coin landed heads. Now, what are the odds that the other one is a tail?"
I think all mathematically-inclined people here know that if you flip two coins, there are four equally-likely outcomes:
1. HH
2. TH
3. HT
4. TT
If someone now tells you that at least one coin came up heads, the only effect that has on the problem is that you can now eliminate "TT" as a possibility. It doesn't make any of the remaining three outcomes any more or less likely. (And of specific import, it doesn't make the HH outcome any more likely.) That leaves three equally-likely possibilities:
1. HH
2. TH
3. HT
Now you're allowed to look at the coins.
If it turned out that the result was #1, then if you look at a coin that is heads, the other one is heads.
If it turned out that the result was #2, then if you look at a coin that is heads, the other one is tails.
If it turned out that the result was #3, then if you look at a coin that is heads, the other one is tails.
Thus, in two out of the three cases, the other coin is a tails, or a girl in the original riddle.
-Sagiro
RangerWickett said:
MerakSpielman
First Post
I guess your post really has 2 "solutions." I was talking about the second one.
BryonD said:
But I was wrong. it isn't a 50/50 chance. There is no trick, Monty will always give the option to switch even if I pick the wrong one, and by switching I would get the right one. He will offer even if he knows he will 'lose' the game.
He always shows an empty chest. and by doing so gives me a 2/3 chance of getting it right by switching.
The only way it becomes 50/50 is if Monty doesn't know which chest the treasure is in and when he opens one it turns out to be the treasure.
RX
MerakSpielman
First Post
OK, here's the version of the treasure chest dillema I remembered, from a book on paradoxes and probability:
From Aha! Gotcha by Martin Gardner, pages 100-101
Operator: Step right up, folk, See if you can guess which shell the pea is under. Double your money if you win.
After playing the game a while Mr. Mark decided he couldn't win more than once out of three times.
Operator: Don't leave, Mac. I'll give you a break. Pick any shell. I'll turn over an empty one. Then the pea has to be under one of the other two, so your chances of winning go way up.
Poor Mr. Mark went broke fast. He didn't realize that turning on empty shell had no effect on his chances. Do you see why?
After Mr. Mark has selected a shell, at least one of the remaining two shells is certain to be empty. Since the operator know where he put the pea, his act of doing so adds no useful information for Mr. Mark to resvise his estimate of the probability that he has picked the right shell.
You can demonstrate this with an ace of spades and two red aces. Mix the cards and deal them face down in a row. Allow someone to put a finger on a card. What is the probablitly he or she has picked the ace of spades?
Clearly it is 1/3.
Suppose, now, you peek at your two cards and turn over a red ace. You can argue, like the game operator, as follows. Only two cards are face down. The ace of spades is as likely to be one of them as the other. Therefore the probability that the ace of spades has been picked seems to have gone up to 1/2. Actually it reamins 1/3. Because you can always turn over a red ace, turning it adds no new information that affects the probability.
You can puzzle your frinds with the following variation. Instead of peeking a the two unselected cards to made sure you turn a red ace, allow the person whose finger is on a card to turn over one of them. If it should be the ace of spades, the deal is declared void and the game is repeated until the reversed card is a red ace. Does this procedure raises the probability of picking the ace of spades?
Oddly enough, it raises it to 1/2. We can see why by taking a sample case.
Call the positions of the cards 1,2,3. Let's assume the person puts a finger on card 2, then turns over card 3 and it is a red ace.
There are six equally possible ways the six cards can be delt.
1. S H D
2. S D H
3. D S H
4. D H S
5. H S D
6. H D S
If the third card had been the ace of spades, the deal would have been declared void, therefore cases 4 and 6 never enter into the problem. We rule them out as possible cases. Of the remaining four (1, 2, 3, 5), card 2 (on which the finger rests) is the ace of spades in two cases. Therefore the probability card 2 is the ace of spades is indeed 2/4 = 1/2.
The result is the same regardless of which card the person chooses, and which card is exposed as the red ace. Had Mr. Mark been allowed to pick the shell to be turn over, and had it been empty, his chances of being right would have gone form 1/3 to 1/2.
end.
I highly recomend this book. The artwork is downright silly, but the reasoning is solid and understandable.
From Aha! Gotcha by Martin Gardner, pages 100-101
Operator: Step right up, folk, See if you can guess which shell the pea is under. Double your money if you win.
After playing the game a while Mr. Mark decided he couldn't win more than once out of three times.
Operator: Don't leave, Mac. I'll give you a break. Pick any shell. I'll turn over an empty one. Then the pea has to be under one of the other two, so your chances of winning go way up.
Poor Mr. Mark went broke fast. He didn't realize that turning on empty shell had no effect on his chances. Do you see why?
After Mr. Mark has selected a shell, at least one of the remaining two shells is certain to be empty. Since the operator know where he put the pea, his act of doing so adds no useful information for Mr. Mark to resvise his estimate of the probability that he has picked the right shell.
You can demonstrate this with an ace of spades and two red aces. Mix the cards and deal them face down in a row. Allow someone to put a finger on a card. What is the probablitly he or she has picked the ace of spades?
Clearly it is 1/3.
Suppose, now, you peek at your two cards and turn over a red ace. You can argue, like the game operator, as follows. Only two cards are face down. The ace of spades is as likely to be one of them as the other. Therefore the probability that the ace of spades has been picked seems to have gone up to 1/2. Actually it reamins 1/3. Because you can always turn over a red ace, turning it adds no new information that affects the probability.
You can puzzle your frinds with the following variation. Instead of peeking a the two unselected cards to made sure you turn a red ace, allow the person whose finger is on a card to turn over one of them. If it should be the ace of spades, the deal is declared void and the game is repeated until the reversed card is a red ace. Does this procedure raises the probability of picking the ace of spades?
Oddly enough, it raises it to 1/2. We can see why by taking a sample case.
Call the positions of the cards 1,2,3. Let's assume the person puts a finger on card 2, then turns over card 3 and it is a red ace.
There are six equally possible ways the six cards can be delt.
1. S H D
2. S D H
3. D S H
4. D H S
5. H S D
6. H D S
If the third card had been the ace of spades, the deal would have been declared void, therefore cases 4 and 6 never enter into the problem. We rule them out as possible cases. Of the remaining four (1, 2, 3, 5), card 2 (on which the finger rests) is the ace of spades in two cases. Therefore the probability card 2 is the ace of spades is indeed 2/4 = 1/2.
The result is the same regardless of which card the person chooses, and which card is exposed as the red ace. Had Mr. Mark been allowed to pick the shell to be turn over, and had it been empty, his chances of being right would have gone form 1/3 to 1/2.
end.
I highly recomend this book. The artwork is downright silly, but the reasoning is solid and understandable.
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