Subject:
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Re: 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 18:45:31 GMT
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Viewed:
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326 times
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In lugnet.off-topic.geek, David Koudys writes:
> So now he says he's willing to put 100 dollars on the 1 in 5...
HA! Congrats on winning 100 bucks :)
> 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...)
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;
}
$tot++;
if($pick =~ /^r/) {
$rfirst++;
if($pick eq 'rr') {
$rrs++;
}
} else {
$rnots++;
}
}
print "$rfirst out of $tot had 1st as red (".($rfirst/$tot).")\n";
print "$rrs out of $rfirst were both red (".($rrs/$rfirst).")\n";
print "$rnots out of $tot were not 1st red (".($rnots/$tot).")\n";
-----------------
Output:
----------
503 out of 1000 had 1st as red (0.503)
169 out of 503 were both red (0.335984095427435)
497 out of 1000 were not 1st red (0.497)
Proof #2:
Possible choices (equal probability):
Prob. Revealed Concealed
1/12 r1 r2
1/12 r1 b
1/12 r1 w
1/12 r2 r1
1/12 r2 b
1/12 r2 w
1/12 b r1
1/12 b r2
1/12 b w
1/12 w r1
1/12 w r2
1/12 w b
Of these, only the 1st 6 choices match our sample set. IE the one revealed
in hour hand is red. SO, since we know that the 1st one is red (given), and
that each of the below are equally probable, we can revamp our probabilites:
Prob. Revealed Concealed
1/6 r1 r2
1/6 r1 b
1/6 r1 w
1/6 r2 r1
1/6 r2 b
1/6 r2 w
Now, of these, only *two* match what we're looking for. IE both red:
Prob. Revealed Concealed
1/6 r1 r2
1/6 r2 r1
The sum of the probabilities? 1/3.
Proof #3:
Look at it this way. It's the same problem as:
You pick one marble from a bag containing 2 reds, 1 blue, and one white. The
marble chosen is seen to be red. What's the probability that the *next* one
you choose (given that you don't put the one in your hand back) is red?
Same problem. Exactly. Possibilities:
Prob. Next
1/3 r
1/3 w
1/3 b
Probability? 1/3.
Proof #4:
Let's pretend order doesn't matter (it does. Ish) Possibilities for the
marbles in your hand (order irrelevant), are:
Prob. Marbles in your hand
1/6 r1 r2
1/6 r1 b
1/6 r1 w
1/6 r2 b
1/6 r2 w
1/6 b w
Now comes the part where order matters. For each of the above, there's a
chance that either the 1st marble or the 2nd marble is the one that's
revealed. And there's a 50/50 chance either way, so we can just expand each
row in the above table to *TWO* columns (and we're gonna get the same thing
we got in proof #2, in different order):
Prob. Revealed Concealed
1/12 r1 r2
1/12 r2 r1
1/12 r1 b
1/12 b r1
1/12 r1 w
1/12 w r1
1/12 r2 b
1/12 b r2
1/12 r2 w
1/12 w r2
1/12 b w
1/12 w b
And again, we can see that only 6 match our sample set:
1/12 r1 r2
1/12 r2 r1
1/12 r1 b
1/12 r1 w
1/12 r2 b
1/12 r2 w
And again we can (knowing that since this is given we can say the total
probability of this set is 1, we can re-evaluate the probabilities for each
item:
1/6 r1 r2
1/6 r2 r1
1/6 r1 b
1/6 r1 w
1/6 r2 b
1/6 r2 w
And again 2 out of 6 of these match what we're looking for:
1/6 r1 r2
1/6 r2 r1
Which adds up of course to 1/3.
Proof #5 (statistics. Again):
Have him try it. Probably the safest way being 50+ times.
DaveE
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Message has 1 Reply:
 | | Re: Now the stakes are higher....
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| In lugnet.off-topic.geek, David Eaton writes: <snip> (...) That's perfect! I owe you a lunch! Now whetehr or not my co-worker succumbs to 'outside sources' as valid enuf to change his position remains to be seen I'm trying to explain his fallacy and (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
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Message is in Reply To:
| | Now the stakes are higher....
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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 (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
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