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| \begin{answer}{bayescoins} | ||
| This is another question that tests your knowledge of Bayes' law. | ||
| Let's define 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 | ||
| $$P( 10H \vert C_{F} )=P( H \vert C_{F} )^{10}= ({1}/{2})^{10} = {1}/{1024}.$$ | ||
| 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 these into \eqref{eq:1000coins:bayeslaw1} to get | ||
| \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 that you are on the right track 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. | ||
| 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: | ||
| \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 quantity we are trying to solve for, $P( C_{R} \vert 10H )$, 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. | ||
| Putting all this together 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 as discussed in 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 from the bag. | ||
| 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} | ||
|
|