From: Emlyn O'regan (oregan.emlyn@healthsolve.com.au)
Date: Mon Oct 28 2002 - 18:31:23 MST
A question for LDC, a probability-competent individual, out of curiosity;
It took me about 20 minutes to code up and trivially test my brute-force
solution, and 40 seconds to do the run which got the final answer. How long
did the calculations below take you to do? I am guessing that this is
everyday stuff for you (remind me never to play cards for money with you),
so I figure it didn't take too long. Our answers came out close enough that
I'd call the two solutions equivalent (although I'd say 2-1 odds on a head
is a little more marginal).
Emlyn
> -----Original Message-----
> From: Lee Daniel Crocker [mailto:lee@piclab.com]
> Sent: Tuesday, 29 October 2002 11:29
> To: extropians@extropy.org
> Subject: Re: Math question
>
>
> > (Eliezer S. Yudkowsky <sentience@pobox.com>):
> > Suppose that we flip a fair coin 20 times, and that a
> biased sampling
> > procedure then randomly selects 5 coinflips from the set of
> coinflips that
> > came up heads and reports on those coinflips; if there
> aren't 5 coinflips
> > that came up heads, the biased sampling procedure reports
> all available
> > heads and enough randomly selected tails to make up the gap.
> >
> > Suppose the biased sampling procedure reports that the
> first, sixth,
> > eleventh, fifteenth, and eighteenth flips came up heads.
> Is there a
> > simple general formula for calculating the probability that
> any given
> > other coinflip came up tails? Clearly the probability is
> greater than
> > 50%, but by how much?
>
> I don't know if this qualifies as an "easy way" to figure it out,
> but here's the answer:
>
> Probability 1/2 makes the calculating the binomial distribution a
> little simpler, so the number of ways the 20 flips can result in
> exactly N heads is 20! / (N! * (20 - N!), so that's 1 way to get
> no heads, 20 ways to get 1, 190 ways to get 2, 1140 ways to get 3,
> 4845 ways to get 4, etc. Adding those 5 outcomes, which are the
> ones where your sample produces at least one tail and the probability
> of a remaining coing being a tail is therefore certainty, you get
> 6196/1048576 (the denominator here is 2^20), or a little less than
> 0.6 % as the cases that aren't interesting. That leaves 99.4% of
> the cases where your biased selector has chosen 5 heads.
>
> So, for each of these cases (N=5...N=20), calculate the probability
> that a single chosen remaining coin is a tail; this will be just
> (20 - N) / 15. For N=5, P=1 (because there were exactly five heads,
> so the probability of a reminaining coing being a tail is 1), for
> N=6, it's 14/15, etc., to N=20, where P=0 (all heads).
>
> Now write that probability (1, 14/15, 13/15....) next to the previous
> numbers you calculated in step 1 (15504, 38760, ...) Multiply them
> across, and divide by (1048576-6196). Then add the whole column.
> The number I get get is 0.66407..., so you should take the "head"
> bet if you're offered 2-to-1.
>
> Note: the hardest part of this for me wasn't figuring out how to do
> it or calculating the binomials, but having to keep scratching out
> "52" all the time...
>
> --
> Lee Daniel Crocker <lee@piclab.com> <http://www.piclab.com/lee/>
> "All inventions or works of authorship original to me, herein
> and past,
> are placed irrevocably in the public domain, and may be used
> or modified
> for any purpose, without permission, attribution, or
> notification."--LDC
>
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