On Tuesday, June 16, 2015 at 3:21:46 PM UTC-4, Thomas 'PointedEars' Lahn wrote:
> Ned Batchelder wrote:
>
> > You aren't agreeing because you are arguing about different things.
> > Thomas is talking about the relative probability of sequences of digits.
>
> There is no such thing as “relative probability”, except perhaps in popular-
> scientific material and bad translations. You might mean relative
> _frequency_, but I was not talking about that specifically.
>
> > Chris is talking about the probability of a single digit never appearing
> > in the output.
>
> I do not think that what I am talking about and what you think Chris is
> talking about are different things.
>
> > Thomas: let's say I generate streams of N digits drawn randomly from 0-9.
> > I then consider the probability of a zero *never appearing once* in my
> > stream. Let's call that P(N)
>
> In probability theory, it is called the probability P(E) of the event E that
> in n trials the probability variable X never assumes the value 0, which can
> be defined as
>
> P(E), E = {e_i | n ∈ ℕ \ {0}, i = 1, …, n} \ {X ≠ 0}, Ω = {1, 2, …, 9}
>
> where the e_i are the singular events, or outcomes, of the probabilistic
> experiment, and Ω is the sample space of the e_i.
>
> > Do you agree that as N increases, P(N) decreases?
>
> I do not agree that P(E), as defined above, decreases as n increases.
>
> See also: <http://rationalwiki.org/wiki/Gambler%27s_fallacy>
I apologize, I'm sure I've been using the mathematical terms imprecisely.
We are all intelligent people, so I still believe we disagree because we
are talking about different things.
To put us all on a footing where (I hope) we have a shared understanding,
this Python program demonstrates what I was talking about:
import random
def die_roll():
"""Roll the die once, produce a number 0-9."""
return random.randint(0, 9)
def die_rolls(n):
"""Roll the die n times, produce a list of numbers from 0-9."""
return [die_roll() for _ in xrange(n)]
def any_zeros(seq):
"""Is there any zero in `seq`?"""
return any(x == 0 for x in seq)
def probability_of_no_zero(nrolls, nseq):
"""Determine the chance of no zero in a sequence of rolls.
This is done empirically, by producing `nseq` sequences of
`nrolls` rolls of the die. Each sequence is examined to
see if it has a zero. The total number of no-zero
sequences divided `nseq` is the probability.
"""
no_zeros = 0
for _ in xrange(nseq):
seq = die_rolls(nrolls)
if not any_zeros(seq):
no_zeros += 1
return float(no_zeros)/nseq
for n in range(10, 101, 10):
# Calculate the probability of getting no zeros by trying
# it a million times.
prob = probability_of_no_zero(n, 1000000)
print "n = {:3d}, P(no zero) = {:.8f}".format(n, prob)
Running this gives:
$ pypy testrandom.py
n = 10, P(no zero) = 0.34867300
n = 20, P(no zero) = 0.12121900
n = 30, P(no zero) = 0.04267000
n = 40, P(no zero) = 0.01476600
n = 50, P(no zero) = 0.00519900
n = 60, P(no zero) = 0.00174100
n = 70, P(no zero) = 0.00061600
n = 80, P(no zero) = 0.00020600
n = 90, P(no zero) = 0.00006300
n = 100, P(no zero) = 0.00002400
As n increases, the probability of having no zeros goes down.
--Ned.
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