Subject:
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Now the stakes are higher....
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Thu, 6 Mar 2003 18:08:59 GMT
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Viewed:
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270 times
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So now he says he's willing to put 100 dollars on the 1 in 5...
I'm swayed back to the 1 in 3...
So here's the original question--
If you have 4 marbles in a bag, 2 are red, 1 is blue, 1 is white
You grab 2 marbles from the bag at the same time
One of the marbles revealed in the hand is red.
What is the probability that the other marble in your hand is also red?
The great responses from the previous thread, y'all agreed with me on the 1
in 3, or 33 percent--but now he wants some proof (that isn't mine...)
My proof goes as follows:
r1 r2 w b
If the exposed ball is relevent, then we have all the following
possibilities in our hand:
r1 r2
r1 b
r1 w
r2 r1
r2 b
r2 w
w r1
w r2
w b
b r1
b r2
b w
All the possible combinations--so no matter which marble (column) we expose,
we eliminate 6 of those 12 possibilities for we know one is red.
Thus leaving us (if I choose to expose the 'left' marble (first column)
r1 r2
r1 b
r1 w
r2 r1
r2 b
r2 w
and out of those possibilities, that both are red, is 2 of 6, or 1 of 3
If the marble is irrelevent, then we have
r r
r b
r w
b w
as the unique combinations--but we can eliminate the last one for it doesn't
contain a red. So we're, again, left with 1 in 3
Here's his logic:
The probability is set at the grab--the chances that both are red in our
hand is 1 in 6 at the time of the grab:
r1 r2
r1 w
r1 b
r2 w
r2 b
b w
listing the unique possibilities, no matter the order
so the chance of us picking 2 red right out of the bag is 1 in 6
the fact that we revealed that 1 is red, just eliminates the last
possibility--thus leaving us with a 1 in 5 chance that the other marble is
also red.
There's a fallacy in one of these ideas and I just can't get my head around it.
Yes I know I'm probably going on just a wee bit too much on it now, but my
co-worker just won't let it go... :)
Any help would be appreciated--I'll buy lunch at the next brickfest
get-together (i plan on going this time :) )
Dave K
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Message has 1 Reply:
 | | Re: Now the stakes are higher....
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| (...) HA! Congrats on winning 100 bucks :) (...) Proof number one: (pure stats) #!/usr/bin/perl for(1..1000) { @marbles = ("r","r","b","w"); $pick = ''; for(1..2) { $n = int rand(@marbles); $pick .= $marbles[$n]; splice @marbles,$n,1; } $tot++; (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
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