BicycleTree said:
Well, I plugged in 100 for M and 10 for N and got 0.131865 (in the formula (N - 1)^(M-M/N) M! / [(M/N)!(M-M/N)!*(N^M)]). Then I tried it with 200 for M and 10 for N and got 0.093636. It seems to be heading for 0 if it's going anywhere.
, you're right.
I will have to reformulate the problem once again.
Let me explain how this came up. In another thread, I am discussing the multiple worlds interpretation (MWI) of quantum mechanics. I won't go into all the esoterics of that discussion. So I'll just say that for the context of this problem with the N-sided die rolled M times, I am imagining that this results in N^M "parallel worlds," with a separate "observer" in each world. Typically in these discussions we consider preparing M identical spin-whatever particles, with spin n being an integer in [1,N] upon measurement, and the predicted probability of measuring spin = n being m_n. For this thread, let's stick with the example of an N-sided die, and define the predicted probability of getting an n on any particular throw as being the probability "measure" m_n. Typically, for a die roll, we assume that each side is equiprobable, ie m_n = 1/N. But in the more general case, we simply require that \sum_{n = 1}^{N} m_{n} = 1.
Let us suppose that an experimenter rolls the die M times and calculates the percentage of time that the n-th result shows up, p_n/M, and compares this with the predicted probability measure m_n. We will imagine that at the end of the M die rolls, we have N^M different experimenters (or "observers") who are living in N^M "parallel worlds."
One of the esoterics of this other thread is that we are entertaining the notion that each of these N^M worlds is, ontologically speaking, on an "equal footing." So here's the difficulty. If the predicted probability measure m_n is NOT equal to 1/N, then as M goes to infinity, "most" of the observers in these N^M worlds will come to the conclusion that the predicted probability measure m_n is WRONG. So my objective is to show this, but more rigorously. in other words, I want to show that m_n = 1/N is the ONLY probability measure such that "most" of the observers will conclude that the predicted probability measure, m_n, is correct.
The way that any individual observer tests the predicted probability measure m_n is to compare it with the observed quantity, p_n/M. The difference between the predicted value m_n and the observed value p_n/M is the "error" \epsilon_n = |m_{n} - p_{n}/M|. So what I am looking for is an expression for m_n such that the following Criterion is true: for any arbitrary cutoff \delta, the proportion f_n of these N^M worlds such that the error is less than the cutoff, \epsilon < \delta approaches 1 in the limit as M approaches infinity. It seems intuitive that m_n = 1/N will meet the Criterion. But what I am trying to prove is that m_n = 1/M is the ONLY expression for m_n that will meet the Criterion. Thus, my method is to define the equation lim_{M \rightarrow \infty} f_{n} \rightarrow 1 and show that this implies that m_n = 1/N.
So the issue is how to calculate f_n. Initially, I was thinking of defining \delta = 1/M, so that as M approaches infinity, the cutoff approaches zero. The reason I did this was to make the calculation of f_n easier: the denominator is the total number of worlds N^M, and the numerator is the number of worlds in which the predicted value matches exactly with the observed value, m_n = p_n/M. But as we have seen, for f_n defined this way, the limit goes to zero.
So I suppose what I should do is to define the cutoff to be arbitrary but fixed, ie not a function of M. So the numerator in the expression for f_n is the sum over all of the values of p_n that are close to m_n, ie such that |m_n - p_n/M| < \delta.
Does my statement of the problem make sense to you?
David
