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#1 |
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FFR Player
Join Date: Oct 2005
Posts: 1
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Here's a little riddle that ought to make you think..
If a family has two children and the older child is a boy, there is a 50 percent chance the family will have two boys. So the four possibilities are: (older, younger) (b:g) (g:g) (b:b) (g:b) We know that the older is a boy, so that only leaves two outcomes, perfectly explaining why there is a 50% chance of having two boys. But say you didn't know whether the boy was older or younger, that would only eliminate the g:g possibility, leaving three others. So the odds of a b:b are 1/3. So how can the age of the boy determine the chance of the other child being a boy or a girl? |
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#2 |
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is against custom titles
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When you don't know the age of the given boy, (b:g) and (g:b) are equivalent, thereby reducing the possibility pool to two: (b:b) or (mixed). One in two chance.
--Guido http://andy.mikee385.com |
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#3 |
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FFR Player
Join Date: May 2005
Posts: 61
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Sorry, I tried to make it as clear as possible, but (b:g) represents an older boy and (g:b) is an older girl, thus making them not equivalent.
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#4 |
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FFR Player
Join Date: Apr 2004
Posts: 1,069
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If you know one of them is a boy...
...well then it's obviously 50/50 because there is a 50% chance that the other is a boy.
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-Jamie |
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#5 |
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Retired BOSS
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Guido answered the question, you just didn't bother to read it correctly.
It reminds me of the time I was explaining the Monty Hall problem with my father. You have 3 doors to pick from. You choose 1, 2, or 3. OK... then, they show you one of the other 2 doors, one that is empty. Then they ask you if you want to keep the door you picked or switch to the other door. Most people stuck with their original door, thinking it didn't matter. But, if you switch, you'll get the prize 2/3 times. Why is it not 50/50? I'll explain. I'm sure you could find this on a website explained better, but I'll do my best. Door 1 - 33% chance Door 2 - 33% chance Door 3 - 33% chance. You pick door 1. 33% chance that it has a prize. So, the combination of Doors 2&3 is a 67% chance of having the prize, or 2/3. Then, they reveal door 2 as NOT having the prize. So, you know that of the Door 2/3 grouping that has a 675 chance of having the prize, it isn't in door 2. So, door 3 has a 67% chance of having the prize and door 1 has a 33%. Therefore, its better to switch from 1 to 3 than to stay with 1. A bit of a tangent, but the main point is to explain that statistics lie. Statistics can be used to give whatever result you want. So, when you hear something like "80% of doctors recommend", you need to wonder about more questions... just as an example. "There are 3 kinds of lies. Lies, Damned Lies, and Statistics," accepted to be first said by Benjamin Disraeli.
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RIP |
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#6 |
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FFR Veteran
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Not to sound like a n00b at statistics Tassel, but if they revealed door 2 (as in your illustration) wouldn't it make it 50-50 (not 67-33) because the chances of door 1 being a prize would raise from 33% to 50%?? (making door 1: 50%, door 2: 0% obviously, and door 3: 50%) just wondering..
EDIT: duh not even on topic but yea along with Tassel it's all about how you start your problem. You could say the odds of having 2 boys are 25% and all your other satements are true, again its' just manipulating numbers |
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#7 |
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Retired BOSS
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Iggy. No, that isn't how statistics work. You chose door 1 with 33% odds, it stays 33% odds.
From google: http://math.cofc.edu/faculty/kasman/...tml?pg=explain
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RIP |
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#9 |
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Banned
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The gender of any previous children would not influence the fact that it's always 50/50 for XX/XY. I use the names of the chromosomes because of intergender people and hermaphrodites, since I like to avoid any possibility of argument ahead of time. Unless I want to argue.
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#10 |
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is against custom titles
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#11 |
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Banned
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I wasn't paying attention to the subject, what was it?
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#12 |
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FFR Player
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you all fail, for you fail to take in the chaotic facts of hermaphrodites and crazy nature experiments.
its more along the lines of a 48.5/48.5/2.9/.1 (i think that gets you to 100) |
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