Testing random

[sohcahtoa82 at gmail.com]

On Tuesday, June 16, 2015 at 4:48:36 PM UTC-7, Thomas 'PointedEars' Lahn wrote:
> Ned Batchelder wrote:
> 
> > On Tuesday, June 16, 2015 at 6:01:06 PM UTC-4, Thomas 'PointedEars' Lahn
> > wrote:
> >> Your programmatic "proof", as all the other intuitive-empirical "proofs",
> >> and all the other counter-arguments posted before in this thread, is
> >> flawed. As others have pointed out at the beginning of this thread, you
> >> *cannot* measure or calculate probability or determine randomness
> >> programmatically (at least not with this program).
> > 
> > You *can* estimate probability with a program, which is what is happening
> > here.
> 
> No.  Just no.
> 
> >> I repeat: Probability is what relative
> >> frequency (which you can measure) *approaches* for *large* numbers.  100
> >> is anything but large, to begin with.
> > 
> > The number of trials in this program is not 100, it is 1 million.  You
> > seem uninterested in trying to understand.
> 
> It still would _not_ a measure or a calculation of *probability*.  So much 
> for "uninterested in trying to understand".
>  
> >> What is "large" depends on the experiment, not on the experimentator.  
> >> And with independent events, the probability for getting zero does not 
> >> increase because you have been getting non-zeros before.  It simply does 
> >> not work this way.
> > 
> > Again, if you look at the code, you'll see that we are not talking about
> > the probability of getting a zero on the next roll.  We are talking about
> > the probability of getting no zeros in an N-roll sequence.  I have no idea 
> > how you have misunderstood this for so long.
> 
> You do not understand that it boils down to the same problem.

Actually, no, they're not.  They're entirely different problems.  "What are the odds of getting 8 zeros in a row?" is a *COMPLETELY* different question from "What are the odds of getting a zero on the next roll?"


>The 
> probability of only having sons is _not_ greater than that of having
> sons and one daughter or vice-versa.  And for that it does _not_ matter
> how many children you have *because* it does _not_ matter how many
> children you had before.  The probability for a boy or a girl is *always*
> the same.  You are _not_ due for a boy if you have many girls, and not for a 
> girls if you have many boys.  But that is precisely what your flawed logic 
> is implying.

Yes, we all know what the gambler's fallacy is, but that's not what anyone is arguing.

If you pick 8 random numbers between 0 and 9 (inclusive), then the odds of getting all zeros is (1/10)^8.  Do you agree with that?

The odds of getting NO zeros is (9/10)^8.  Do you agree with that?

Note that NEITHER of these scenarios say anything about a pre-condition.  The first question is *NOT* asking "If you picked 7 random numbers between 0 and 9 and got 0 for all 7, what are the odds of getting another 0?"  The answer to that is obviously 1/10, and anybody arguing something else would certainly be committing the Gambler's fallacy.

> 
> Learn probability theory, and use a dictionary in Python when you want to 
> count random hits.

I know enough probability theory to know that you're either wrong or you keep changing the problem to something nobody else has said in order to think you're right.

> 
> -- 
> PointedEars
> 
> Twitter: @PointedEars2
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