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‎src/QuantitativePrimer.pdf

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‎src/QuantitativePrimer.tex

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% C: Increment with minor error corrections and additions
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% When B is incrememted, C is reset to zero
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% When A is incrememted, B and C is reset to zero
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\newcommand{\docversion}{V1.1.5}
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\newcommand{\docversion}{V1.2.0}
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\begin{document}
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%\doublespacing
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\input{answers/drop20cards.tex}
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\clearpage
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\subsection{Face-to-face, 1 hour}
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\begin{question}{bayescoins}
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\index{questions!Bayes' law and coin flips}
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You have a bag with 1000 coins in it.
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One of them is a double headed coin, the other 999 are fair coins.
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I pick one coin from the bag at random, and flip it ten times.
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It comes up heads all ten times.
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What is the probability that I have selected the double headed coin?
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\end{question}
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\clearpage
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\input{answers/bayescoins.tex}
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\section{Old friends}
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Two questions stood out:
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I got the \emph{Air Force One} question three times in 2014, and I got the \emph{Stick Breaking} question more than three times during 2017.\footnote{I never got \emph{Air Force One} in 2017, nor did I get \emph{Stick Breaking} in 2014. Questions seem to go in and out of fashion and there is evidence of cross-pollination.}
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to foldernames?
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\item What programming languages does the team use? R, Python, C++, Julia, Matlab?
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For data analysis? For scripting? For production?
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\item Is the team Bayesians, frequentists, or whatever gets the job done?
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\item Is the team made up of Bayesians, frequentists, or whatever gets the job done?
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\item What are the backgrounds of the other team members? What would someone with my background contribute?
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\item Are modellers expected to write the code to implement their models, or is this handled by developers?
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\item How are datasets accessed? SQL (Postgres, MySQL, MSSQL, sqlite), Hadoop, csvs, Excel, shared folders, or do people throw around USB sticks?

‎src/answers/bayescoins.tex

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\begin{answer}{bayescoins}
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This is another question that tests your knowledge of Bayes' law.
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Let's define some notation to use.
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Define $H$ as the event that a coin comes up head, and $T$ that a coin comes up tails, and let $10H$ denote getting ten heads from ten coin flips.
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Let $C_{F}$ be the event where we select the fair coin from the bag, and $C_{R}$ the event that we select the rigged coin.
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This is one of the simplest questions about Bayes' law as there is not much to unpack.
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You want to know the probability of the rigged coin being selected, given you saw ten heads.
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By rote application of Bayes' law:
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\begin{align}
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\label{eq:1000coins:bayeslaw1}
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P( C_{R} \vert 10H)
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&=
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\frac{
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P( 10H \vert C_{R} )
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P( C_{R} )
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}{
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P( 10H \vert C_{R} )
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P( C_{R} )
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+
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P( 10H \vert C_{F} )
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P( C_{F} )
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}
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\text{.}
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\end{align}
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Consider
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$P( 10H \vert C_{R} )$, the probability of getting ten heads in a row with the rigged coin.
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Since this will happen with certainty
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$P( 10H \vert C_{R} ) = 1$.
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For the fair coin each flip is independent, so you have
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$$P( 10H \vert C_{F} )=P( H \vert C_{F} )^{10}= ({1}/{2})^{10} = {1}/{1024}.$$
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Since you picked te coin out of a bag of 1000 coins, the probability that you selected the rigged coin is
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$P(C_{R}) = 1/1000$ and the probability that the coin you selected is fair is
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$P(C_{F}) = 999/1000$.
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You can substitute these into \eqref{eq:1000coins:bayeslaw1} to get
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\begin{align*}
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P( C_{R} \vert 10H)
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&=
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\frac{
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(1)
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\left( \frac{1}{1000} \right)
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}{
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(1)
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\left( \frac{1}{1000} \right)
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+
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\left(\frac{1}{1024}\right)
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\left(\frac{999}{1000}\right)
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}
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\\
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&=
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\frac{
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1
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}{
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1
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+
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\left(\frac{999}{1024}\right)
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}
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\\
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&=
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\frac{
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1
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}{
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\left(\frac{2023}{1024}\right)
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}
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\\
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&=
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\frac{1024}{2023}
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\text{,}
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\end{align*}
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which is slightly more than $1/2$.
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This question is so well known that your interviewer likely won't even let you finish it.
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Once they see that you are on the right track they will move on to the next question.
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My interviewer didn't care about the answer, but he wanted me to describe \eqref{eq:1000coins:bayeslaw1} in detail.
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Since Bayes' law is just the application of conditional probability, you can derive it from first principles:
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\begin{align}
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\label{eq:1000coins:bayesexplain}
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P( C_{R} \vert 10H)
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&=
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\frac{
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P( 10H , C_{R} )
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}{
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P( 10H )
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}
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\end{align}
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and even a frequentist will agree.
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The nominator is the joint probability of ten heads and the rigged coin being selected, and it is easier to split this into another conditional probability:
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\begin{align*}
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P( 10H , C_{R} )
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=
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P( 10H \vert C_{R} ) p( C_{R} )
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\text{.}
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\end{align*}
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Technically, we can also say
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\begin{align*}
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P( 10H , C_{R} )
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=
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P( C_{R} \vert 10H ) p( 10H )
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\text{,}
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\end{align*}
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but this is not helpful, as it contains the quantity we are trying to solve for, $P( C_{R} \vert 10H )$, and will lead to circular reasoning.
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You can expand denominator in
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\eqref{eq:1000coins:bayesexplain}
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using the law of total probability to consider all the possible ways you can see ten heads.
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Since you only have two types of coins---either a fair coin or a rigged one---there are only two ways ten heads can happen:
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\begin{align*}
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P( 10H ) =
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P( 10H \vert C_{R} )p(C_{R})
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+
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P( 10H \vert C_{F} )p(C_{F})
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\text{.}
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\end{align*}
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If the interviewer wants you to explain even further, you can note that this is derived form the marginal probability
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\begin{align*}
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P( 10H ) =
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P( 10H , C_{R} )
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+
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P( 10H , C_{F} )
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\text{,}
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\end{align*}
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by applying the law of conditional probability to each of the terms.
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Putting all this together yields \eqref{eq:1000coins:bayeslaw1}.
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My interviewer used this Bayes' law question to test my handle on probability concepts.
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It is easy to confuse these basic concepts, which is another reason for doing proper interview preparation.
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Some interviewers ask this question and demand an ``intuitive'' solution that doesn't rely on algebraic manipulation.
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In this case, you could opt for a visual explanation of Bayes' Law as discussed in answer \ref{a:bayeslawdisease}.
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If that's not what your interviewer wants, use the following argument.
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You know the probability of choosing the rigged coin is $1/1000$.
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You also know the probability of getting ten heads from a fair coin is $1/2^{10} = 1/1024$.
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These two events are about equally likely, meaning the probability that we have the double headed coin is about a half.
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If you only had nine heads in a row, the fair coin would give a probability of $1/512 = 2/1024$.
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That means the outcome of nine heads is about twice as likely with the fair coin as the probability of selecting the rigged coin from the bag.
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So the odds of ${fair}{:}{rigged}$ are $2{:}1$, leading you to assign a probability of about $1/3$ to the rigged coin being selected.
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%TODO: could link this up to section on gambling mathematics
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\end{answer}
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