Subject:
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Re: math question (or pattern... whatever...)
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Mon, 3 Mar 2003 22:29:56 GMT
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Viewed:
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563 times
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In lugnet.off-topic.geek, Dan Boger writes:
> On Mon, Mar 03, 2003 at 08:54:01PM +0000, David Koudys wrote:
> > r(1) r(2)
> > r(1) b
> > r(1) w
> > r(2) b
> > r(2) w
> >
> > (b w being still eliminated due to red already being in your hand) that a
> > last possible combination:
> >
> > r(2) r(1)
> >
> > fits into the mix, so 2 out of 6, which still equals 33 percent
> >
> > Any stats people out there that could shed logic on this?
>
> I'm not a stats person, but I agree with your friend. r(2) r(1) is the
> same as r(1) r(2), because order doesn't matter. If it does, then you
> end up with more combos (w r(2), etc)...
>
> --
>
> that's what I was going to say. But, I decided to run a small
> simulation:
>
> $ perl -le '
> $rr = 0;
> $count = 0;
> for (1..10000) {
> $count ++;
> @bag = qw/r r w b/;
> $hand = "";
> for (1..2) {
> $pos = int(rand($#bag));
> $hand .= $bag[$pos];
> splice @bag, $pos, 1;
> }
> $rr++ if $hand eq "rr"
> }
> print "$rr/$count = ", $rr/$count;
> '
> 3391/10000 = 0.3391
>
> so, once again, I learn I don't quite understand statistics as I think
> I do. :)
Wow! You wrote a program! Thanks Dan!
Again I'm impressed with the knowledge and skills of those that partake in
LUGNET!
Adding to my case,
Dave K
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Message is in Reply To:
 | | Re: math question (or pattern... whatever...)
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| (...) I'm not a stats person, but I agree with your friend. r(2) r(1) is the same as r(1) r(2), because order doesn't matter. If it does, then you end up with more combos (w r(2), etc)... -- that's what I was going to say. But, I decided to run a (...) (22 years ago, 3-Mar-03, to lugnet.off-topic.geek)
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