In this simulation, you start with an urn containing \(k\) "Heads" balls and \(z\) "Tails" balls (with \(n = k+z\)). Balls are drawn without replacement. When the urn is empty, it is refilled with the original configuration. The last \(n\) draws are displayed below. Every \(n\)th draw is marked to indicate the end of a block.
Consider the following urn model:
Initially, the urn contains \(k\) \emph{Heads} balls and \(z\) \emph{Tails} balls, so that \(n = k+z\). Balls are drawn without replacement. When the urn is empty, it is refilled with the original \(k\) Heads and \(z\) Tails.
The key observation is that after each block of \(n\) draws, the probability of having drawn exactly \(k\) Heads and \(z\) Tails is 100\%. That is, the process enforces: \[ P(\text{exactly } k \text{ Heads in } n \text{ draws}) = 1. \]
In a classical Bernoulli experiment with a coin of real bias \(p\), the probability to get exactly \(k\) Heads in \(n\) tosses is given by \[ \binom{n}{k} \, p^k (1-p)^{n-k}. \] Generally, this expression is not equal to 1 unless \(p\) is trivial (0 or 1). In fact, setting \[ \binom{n}{k} \, p^k (1-p)^{n-k} - 1 = 0 \] yields a polynomial of degree \(n\) in \(p\). For nontrivial cases, this polynomial has complex roots.
Thus, to simulate this urn model with a single coin, we must allow the coin's probability to be complex. For example, in the case \(n=2\) and \(k=1\), the equation becomes: \[ 2\,p(1-p) - 1 = 0 \quad\Longleftrightarrow\quad p^2 - p + \frac{1}{2} = 0. \] Its solutions are \[ p = \frac{1}{2} \pm \frac{i}{2}. \]
In our simulation, drawing a ball from the urn reproduces the behavior of a coin with a complex probability, as the outcome is perfectly complementary over each block of \(n\) draws.
Author: Orges Leka
Date: February 19, 2025