Re: Simulating radioactive decay.

[brian whatcott <inet@INTELLISYS.NET>]

Like Ludwik, I too pondered Robert's easy correction for discrete
rather than continuous probabilities. This is not the first time
the topic has arisen, but I feel happier with the helpful dialog
this time.

At 09:52 5/14/99 -0400, Robert Cohen wrote:
> On Thu, 13 May 1999, Ludwik Kowalski wrote:
> >  ...  T=ln(2)/lambda, would produce
> > wrong results if p was used instead of lambda...

> > The curve N=f(t) can be plotted to determine the "experimental"
> > value of T. It will NOT be equal to 4.16 units. (1 unit =1 throw).

i.e. Half life is not ln(2) / (1/6) = 4.159

> After one throw, there are (1-p) left.  After n throws, there are (1-p)^n
> left.  We want to know the number of throws, n, that leave us with 1/2
> left:  (1-p)^n = 1/2
>
> Solve for n to get n = - ln(2) / ln (1-p)
> So, what is the discrepancy caused by using p rather than ln(1-p)?
> ....
> | Robert Cohen

p    -ln(1-p)   ratio   choice
.5    0.69      .72    1 in 2
.17   0.19      .91    1 in 6
.1    0.11      .95    1 in 10
.03   0.03      1.0    1 in 33
.01   0.01      1.0    1 in 100

So half life is 4.159 X 0.91 = 3.78 rolls of the die

I think Robert demonstrated a statistical rule of thumb
nicely: try and make the population/sample/choices >30

Snake eyes with two dice perhaps?
 Er...I mean one marked face up on each of two hexagonal pencils,
yes indeed!

T = ln(2) / (1/36) = 24.95 rolls
T = ln(2) / ln(1-1/36) = 24.61 rolls





brian whatcott <inet@intellisys.net>
Altus OK