[Marxism] I was wrong! (was Re: Calculating Octopus) (Was: A question to mathematicians)
shmage at pipeline.com
Thu Jul 15 06:50:56 MDT 2010
On Jul 12, 2010, at 6:33 PM, Shane Mage wrote:
>> <<On Jul 12, 2010, at 12:01 AM, Néstor Gorojovsky wrote:
>>> Can anyone calculate the probabilities that Paul the octopus has
>>> reached his results by chance?
>> 1/2^8 = . 0039
I was wrong. The calculation would be right for an experimental eight-
trial series. But there is no reason to think that was the case for
such a media-driven event. More likely than not, if Paul had gotten
one selection wrong the media would have gone away and the series
would have ended there.
Its like the start of Rosencrantz and Guildenstern are Dead--the two
courtiers pass the time riding to Elsinore by tossing a coin that
keeps coming up heads, suggesting to them and the audience that
something uncanny is happening. But they would have stopped the first
time it came up tails.
So what was the true probability? That actually would depend on how
many wrong guesses were permitted. And the range for Paul is between
1/2^8 and 1/9. How so?
Assume that the trial would end with the first wrong guess.
The chance of a right answer (R) on the first trial is obviously 1/2.
The chance of two Rs in the first two trials is given by the
possibilities RR, RW, and W:
The chance of three Rs in the first three: RRR, RRW, RW, and W;
The chance of four Rs: RRRR, RRRW, RRW, RW, W; probability=1/5
And so forth....the probability under that assumption is always 1/(n+1)
So the right answer depends on how many mulligans Paul would have been
allowed. If between one and seven the calculation gets a lot more
complicated, and I don't anymore know enough binomial theorem to
qualify as a sufficiently Modern Major General.
> Shane Mage
> "The sibyl, with raving mouth, uttering things mirthless unadorned and
> unperfumed, reaches over a thousand years with her
> voice." (Herakleitos, fr. 92)
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