Moved question to book

Next, will add links to it

Author: dwcoder

src/QuantitativePrimer.tex

@@ -783,6 +783,22 @@ \subsection{Phone interview, 1 hour}
 \input{answers/drop20cards.tex}
 
 \clearpage
+
+\subsection{Face-to-face, 1 hour}
+
+\begin{question}{bayescoins}
+You have a bag with 1000 coins in it.
+One of them is a double headed coin, the other 999 are fair coins.
+I pick one coin from the bag at random, and flip it ten times.
+It comes up heads all ten times.
+What is the probability that I have selected the double headed coin?
+\end{question}
+
+
+\clearpage
+\input{answers/bayescoins.tex}
+
+
 \section{Old friends}
 Two questions stood out:
 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.}

src/answers/bayescoins.tex

@@ -0,0 +1,137 @@
+\begin{answer}{bayescoins}
+This is another question about Bayes' law.
+Let's make some notation to use.
+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.
+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.
+This is one of the simplest questions about Bayes' law as there is not much to unpack.
+You want to know the probability of the rigged coin being selected, given you saw ten heads.
+By rote application of Bayes' law:
+\begin{align}
+\label{eq:1000coins:bayeslaw1}
+ P( C_{R} \vert 10H)
+ &=
+ \frac{
+    P( 10H \vert C_{R} )
+    P( C_{R} )
+ }{
+    P( 10H \vert C_{R} )
+    P( C_{R} )
+    +
+    P( 10H \vert C_{F} )
+    P( C_{F} )
+ }
+ \text{.}
+\end{align}
+Consider
+$P( 10H \vert C_{R} )$, the probability of getting ten heads in a row with the rigged coin.
+Since this will happen with certainty
+$P( 10H \vert C_{R} ) = 1$.
+For the fair coin each flip is independent, so you have
+$P( 10H \vert C_{F} )=P( 10 \vert C_{F} )^{10}= ({1}/{2})^{10} = {1}/{1024}$.
+Since you picked te coin out of a bag of 1000 coins, the probability that you selected the rigged coin is
+$P(C_{R}) = 1/1000$ and the probability that the coin you selected is fair is
+$P(C_{F}) = 999/1000$.
+You can substitute
+\begin{align*}
+ P( C_{R} \vert 10H)
+ &=
+ \frac{
+    (1)
+   \left( \frac{1}{1000} \right)
+ }{
+    (1)
+   \left( \frac{1}{1000} \right)
+    +
+    \left(\frac{1}{1024}\right)
+    \left(\frac{999}{1000}\right)
+ }
+ \\
+ &=
+ \frac{
+    1
+ }{
+    1
+    +
+    \left(\frac{999}{1024}\right)
+ }
+ \\
+ &=
+ \frac{
+    1
+ }{
+    \left(\frac{2023}{1024}\right)
+ }
+ \\
+ &=
+    \frac{1024}{2023}
+ \text{,}
+\end{align*}
+which is slightly more than $1/2$.
+
+This question is so well known that your interviewer likely won't even let you finish it.
+Once they see you can answer it they will move on to the next question.
+My interviewer didn't care about the answer, but he wanted me to describe \eqref{eq:1000coins:bayeslaw1} in detail.
+Since Bayes' law is just the application of conditional probability, you can derive it from first principles:
+\begin{align}
+\label{eq:1000coins:bayesexplain}
+ P( C_{R} \vert 10H)
+ &=
+ \frac{
+    P( 10H , C_{R} )
+ }{
+    P( 10H )
+ }
+\end{align}
+and even a frequentist will agree with you here.
+The nominator is the joint probability of ten heads and the rigged coin, and it is easier to split this into another conditional probability:
+\begin{align*}
+    P( 10H , C_{R} )
+    =
+    P( 10H \vert C_{R} ) p( C_{R} )
+    \text{.}
+\end{align*}
+Technically, we can also say
+\begin{align*}
+    P( 10H , C_{R} )
+    =
+    P( C_{R} \vert 10H  ) p( 10H  )
+    \text{,}
+\end{align*}
+but this is not helpful, as it contains the probability we are trying to determine and will lead to circular reasoning.
+
+You can expand denominator in
+\eqref{eq:1000coins:bayesexplain}
+using the law of total probability to consider all the possible ways you can see ten heads.
+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:
+\begin{align*}
+    P( 10H ) =
+    P( 10H \vert C_{R} )p(C_{R})
+    +
+    P( 10H \vert C_{F} )p(C_{F})
+    \text{.}
+\end{align*}
+If the interviewer wants you to explain even further, you can note that this is derived form the marginal probability
+\begin{align*}
+    P( 10H ) =
+    P( 10H , C_{R} )
+    +
+    P( 10H , C_{F} )
+    \text{,}
+\end{align*}
+by applying the law of conditional probability to each of the terms.
+Combining the nominator and the denominator yields \eqref{eq:1000coins:bayeslaw1}.
+
+My interviewer used this Bayes' law question to test my handle on probability concepts.
+It is easy to confuse these basic concepts, which is another reason for doing proper interview preparation.
+Some interviewers ask this question and demand an ``intuitive'' solution that doesn't rely on algebraic manipulation.
+In this case, you could opt for a visual explanation of Bayes' Law, like that of answer \ref{a:bayeslawdisease}.
+If that's not what your interviewer wants, use the following argument.
+You know the probability of choosing the rigged coin is $1/1000$.
+You also know the probability of getting ten heads from a fair coin is $1/2^{10} = 1/1024$.
+These two events are about equally likely, meaning the probability that we have the double headed coin is about a half.
+If you only had nine heads in a row, the fair coin would give a probability of $1/512 = 2/1024$.
+That means the outcome of nine heads is about twice as likely with the fair coin as the probability of selecting the rigged coin.
+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.
+%TODO: could link this up to section on gambling mathematics
+\end{answer}
+