Re: Simulating radioactive decay.

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

Robert A Cohen wrote:

> ... 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)

Very good. Too bad I could not do this myself yesterday.

> So, what is the discrepancy caused by using p rather
> than ln(1-p)?

That is not what I was referring to. I was talking about
using lambda instead of p. For p=1/6, according to your
formula, the half-life n= 3.801 throws. Use this to calculate
lambda = ln(2)/n=0.245. That is the paradox. We started
with p=1/6 (each pencil has six sides) and we concluded
that the probability of decay per unit time is not 1/6.

My point was that we should not identify p with lambda.
By definition, lambda is the probability of decay per a
negligibly small unit time. On the other hand, p, is the
probability that a pencil will land with its label up. It is
also a probability of decay per unit time but that unit is
not small in comparison with n. Now I am using your
notation for the half-life, n; yesterday I referred to it as T.

Nothing profound; only a good student activity. Let
them start with coins and pencils and ask them to write
a program which does the same thing by the Monte Carlo
method. The listing of such program, in True Basic, is
shown below.

input prompt" enter p= ":p         ! such 0.166 for a pencil
input prompt" enter N= ":N       ! say 100000 pencils
print "p=";p;"  N=";N
let min=N/10                             ! when to stop
DO
let count=0                                ! to count remaining pieces
for i=1 to N
let x=rnd                                   ! a random number 0 to 1
if x>p then let count=count+1  ! a surviving piece
next i
let N=count
print "N=";N
if N<min then stop
LOOP                                       ! back to DO
end

Ludwik Kowalski