\documentclass[twocolumn]{article}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{hyperref}

\begin{document}

\title{An Experimental Verification of Bayes' Theorem}
\author{BI Espen, BJ Espen}
\maketitle

\begin{abstract}
Thomas Bayes, an English Presbyterian minister, identified a method for computing conditional probabilities after new information has been added. This method is useful for determining the probabilities in “shell games”, otherwise known as the Monty Hall problem.
\end{abstract}

\section{Introduction}
Bayes’ theorem formalizes a procedure for
computing probability after new information has
been added to a problem. Rev. Bayes’ paper was
published posthumously in 1763 in the journal
Philosophical Transactions\cite{bayes}
Modern applications of Bayes’ Theorem
include subjectivist approaches to epistemology,
statistics, and inductive logic.\cite{mack} A more
commonplace application can be found in shell
games, where an object is hidden in some
fashion, and one of its possible locations is
revealed. Bayes’ Theorem predicts the change in
probability associated with that new information.

\section{Theoretical Section}
Bayes’ theorem can be simply derived from
basic statistical principles. Given two mutually
exclusive and exhaustive probabilities
\begin{equation}
	P(A)+P(A')=1 
\end{equation}
and another non-zero probability
\begin{equation}
	P(B)
\end{equation}
the conditional probability can be calculated
\begin{equation}
	P(A/B)=\frac{P(A\cap B)}{P(B)} \label{third}
\end{equation}

Bayes’ theorem is a generalized extension of this concept for a set of mutually exclusive events \begin{math}A_{1},
A_{2},\ldots A_{n}\end{math}, and another event \begin{math}B\end{math}, such that the
conditional probability \begin{math}P(B/A_{i})\end{math} is known, but
\begin{math}P(A_{i}/B)\end{math} is the desired quantity.
Equation \eqref{third}can be manipulated to give
\begin{equation}
	P(A_{i}\cap B) = P(A_{i}/ B) \cdot P(B) \label{fourth}
\end{equation}
and
\begin{equation}
	P(A_{i}\cap B) = P(B/A_{i}) \cdot P(A_{i}) 
\end{equation}
These two equations can be set equal to one
another
\begin{equation}
	P(A_{i}/ B) \cdot P(B)= P(B/A_{i}) \cdot P(A_{i})
\end{equation}
and then divided by \begin{math}P(B)\end{math} to gain the desired
quantity.
\begin{equation}
	P(A_{i}/ B)=\frac{P(B/A_{i})\cdot P(A_{i})}{P(B)} \label{eighth}
\end{equation}
However, \begin{math}P(B)\end{math} is an unknown. It can be
computed by summing
\begin{equation}
	P(B)= \sum_{i=1}^nP(A_{i}\cap B)
\end{equation}
substituting equation \ref{fourth} for the right hand side gives
\begin{equation}
	P(B)= \sum_{i=1}^nP(A_{i}/B) \cdot P(A_{i})
\end{equation}
which is then substituted into the bottom of equation \ref{eighth}
\begin{equation}
	P(A_{i}/B)=\frac{P(B/A_{i})\cdot P(A_{i})}{\sum_{i=1}^nP(B/A_{i}) \cdot P(A_{i})} \label{eleventh}
\end{equation}
\section{Experimental Section}
A simple experimental verification of
Bayes’ theorem was conducted using a human
test subject, three paper cups, and a bag of
peanut M\&M’s.
\begin{figure}
	\begin{center}
	\includegraphics{test_subject}
	\caption{Test Subject}
	\end{center}
\end{figure}

A peanut M\&M was placed under a paper
cup outside of the subject’s view. The three cups
were then revealed to the subject. The subject
was asked to choose a cup. Of the two
remaining cups, one was lifted to show the
M\&M was not beneath it. The subject was then
asked whether they would keep their first choice,
or switch their preference to the other cup. After
the choice was made, the chosen cup was lifted
and the subject was rewarded if the M\&M was
underneath the chosen cup.
\begin{figure}
	\begin{center}
	\includegraphics{exp_setup}
	\caption{Experimental Setup}
	\end{center}
\end{figure}
Data for this experiment was recorded for
fifty trials. The data consisted of whether the
subject switched their preference after an empty
cup was revealed, and whether the M\&M was
found on that trial.
The following probabilities can be assigned to
the events in this experiment:
Mi: the probability the M\&M is hidden under
cup \begin{math}i\end{math}.
\begin{math}Ri:\end{math} the probability cup \begin{math}i\end{math} is revealed.
For the sake of discussion, assume the
subject chose cup 1. What then is the probability
of finding the M\&M given the subject retains the
original choice, or chooses the other cup? The
following compound probabilities result:
\begin{align}
	P(R_{2}/M_{2})&= 0 \nonumber\\
	P(R_{2}/M_{1})&= 0.5 \nonumber\\
	P(R_{2}/M_{3})&= 1.0 \nonumber\\
	P(R_{3}/M_{2})&= 0.5 \nonumber\\
	P(R_{3}/M_{1})&= 0.5 \nonumber\\
	P(R_{3}/M_{3})&= 0.5 \nonumber
\end{align}
Using these probabilities and equation \ref{eleventh}, the
probability of success can be found in each case.
\subsection{Case 1: Stick}
Further assume that cup 2 is revealed. This
implies:
\begin{align}
		P(M_{1}/R_{2})&= \frac{P(R_{2}/M_{1})\cdot P(M_{1})}{\sum_{i=1}^n[P(R_{2}/M_{i})\cdot P(M_{i})]} \nonumber\\
	&= \frac{\tfrac{1}{2}\cdot \tfrac{1}{3}}{\tfrac{1}{2}\cdot \tfrac{1}{3}+0\cdot\tfrac{1}{3}+1\cdot\tfrac{1}{3}} \nonumber\\
	&= \frac{\tfrac{1}{2}\cdot \tfrac{1}{3}}{\tfrac{1}{3}}=\frac{1}{3} \nonumber
\end{align}
\subsection{Case 2: Switch}
\begin{align}
	P(M_{3}/P_{2})&= \frac{P(P_{2}/M_{3})\cdot P(M_{3})}{P(R_{2}} \nonumber\\
	&= \frac{1+ \tfrac{1}{3}}{\tfrac{1}{2}}=\frac{2}{3} \nonumber
\end{align}
By symmetry we can see that the
mathematics is exactly the same for any
combination of initial choice and cup revealed.
\section{Results and Discussion}
The observed probability of finding the
M\&M sticking with the original choice was
\begin{math}0.317\end{math}, which closely approximates the expected
value of \begin{math}1/3\end{math}. Whereas the probability of winning
with a switch was only \begin{math}1/3\end{math}. The test subject
indicated a preference for sticking, so switching
was only selected nine times. This provides an
explanation for the deviation from the expected
value of \begin{math}2/3\end{math}, due to an insufficient sample space.
Further experimentation to increase the sample
size was precluded due to chocolate overdose.
\section{Conclusion}
It is apparent that a larger sample space
would produce data with higher reliability.
Nevertheless, the data for sticking with the
original choice conforms to expectation due to a
larger number of trials, and can be considered
reliable. Sticking with the original choice
produces no better results than the \begin{math}1/3\end{math} chance the
M\&M was hidden beneath that cup.
A simpler exp eriment design could facilitate
future tests of Bayes’ theorem by decreasing the
time necessary. Reducing the chocolate intake
of the test subject will also increase the sample
space, and is highly recommended for future
studies.
\begin{thebibliography}{9}

	\bibitem{mack}
	  D.R. Mack,
	  \emph{IEEE Potentials}.
	  Feb 1993. 
	  pp. 41-42
	\bibitem{joyce}
	  J. Joyce,
	  ``Bayes' Theorem''
	  \emph{The Stanford Encyclopedia of Philosophy }.
	  (Winter 2003 Edition), 
	  Edward N. Zalta (ed.), forthcoming 
	  \url{http://plato.stanford.edu/archives/win2003/entries/bayes-theorem/}
	\bibitem{bayes}
	  T. Bayes,
	  \emph{Philosophical Transactions}.
	  53, 
	  (1763), 
	  pp. 370-418
	\bibitem{lamport94}
	  R. J. Tallarida,
	  \emph{Pocket Book of Mathematical Formulas and Integrals}.
	  3rd edition, 
	  CRC Press, Boca Raton, 1999.

\end{thebibliography}


\end{document}