Re: Geiger, not binomial ?

[John Denker <jsd@MONMOUTH.COM>]

At 10:09 AM 3/24/00 -0500, Ludwik Kowalski wrote:
> I am not comfortable with the p of a binomial distribution.
> Since the mean=p*n (using n for the number of attempts
> and r for the number of successes) then p depends on n.

Let's pin down the notation:
        p = probability of a PARTICULAR atom decaying per unit time
          = an intensive quantity
        N = total number of atoms in the sample
          = an extensive quantity
        R = mean counting rate, counts per unit time
          = an extensive quantity

So in this notation, R = pN

> Since the mean=p*n (using n for the number of attempts
> and r for the number of successes) then p depends on n.
> This conflict with a view that p is an intrinsic property of
> a setup.

Whether or not p depends on N depends on what you consider constant.

A) In the case where we are given two samples of unknown provenance, but
with the same observed R, then
  * one could have a large p and a small N, while
  * the other could have a small p and a large N.

So yes, under conditions of constant R, p depends inversely on N.  In this
case, p is not an intrinsic property of the setup.

B) In the more-common case that we have two samples of the identical
substance, then p is a property of the setup, and is constant, and R is
proportional to N.

> Is it not true that p-->r/n when n-->oo? Defined
> in that way it should not depend on n.

In case B, yes, R/N converges to an excellent estimate for p.

> Example: trying to hit a target with an arrow. My skill
> for being successful can be described by the probability
> of a hit in a single trial. If p=0.2 then I am expected to
> hit the target 2 out of 10 attempts. In reality the outcome
> will fluctuate around 2. In this case n is given and all is
> fine.

This seems to be an example of case B, because p is given.

> But in the case of a Geiger counter (where the mean was
> 2.20) both n=170 and n=340 give a satisfactory fit to
> experimental data. This implies that both p=0.0129 and
> p=0.00647 are OK. How can it be?

It's an example of case A.  You haven't presented any hint of _a priori_
knowledge of p or N.

> Isn't the objective
> probability of recording one event per unit time equal
> to 0.24 (7278/30058)?

That's certainly not p.  I don't know what to call it;  perhaps we can call
it X(1).  It's definitely not the p that you plug into the binomial formula.

You can see this easily with a spreadsheet:  choose a p on the order of
0.02 and N on the order of 100.  Use the binomial formula to calculate
X(0), X(1), X(2), et cetera.  X(1) is nowhere near p.

> Is the word "probability" used
> for two different things in this "paradox"?

The difference is in the word "probability" and/or in the word
"event".  Different events have different probabilities.

Suppose I roll two dice (one red and one green).  Rolling red=5 and green=2
is an event.  Rolling something that adds up to 7 is a different event,
with a different probability.