Solving Monty Hall with the List monad (or Probabilistic Programming)

Continuing from my Bayes Theorem-based blog post on solving the Monty-Hall problem, I will now show a rather elegant way to model the solution that exploits the humble List monad. To remind the reader, the Monty Hall problem involves a contestant initially choosing one of three doors, hoping to find a car behind it. Monty Hall then opens a different door, revealing a goat, and asks the contestant if they want to switch their initial choice. The goal of the solution is to guide the contestants to make this switching choice such that they have a higher probability of winning. This post was inspired by a less-than 100 lines implementation of a probabilistic programming monad in Haskell

We will begin by representing discrete distributions as a Haskell datatype:

type Prob = Double -- probability of an outcome

newtype Dist a = Dist { unpackDist :: [(a, Prob)] }


This is a nicer reformulation of distributions that the List monad could naturally model. For example, in an experiment with 5 tosses (modelled as data Toss = Head | Tail), that leads to 4 Heads and 1 Tail, our Dist type will model this as Dist [(Head, 0.8), (Tail, 0.2)]. However, we could have modelled this as a plain list as [Head, Head, Head, Head, Tail]. It is simply that the Dist type is nicer to work with (but it is a wrapper on the plain List).

Next, to compute probabilities accurately, it is easier for our representation if we normalise the distribution. We can do that using:

normP :: [(a, Prob)] -> [(a, Prob)]
normP xs = map (\(x, y) -> (x, y / (sumP xs))) xs
where
sumP :: [(a, Prob)] -> Prob
sumP = sum . map snd


Given the above, we can now define the uniform distribution:

uniform :: [a] -> Dist a
uniform = Dist . normP . map (, 1.0)

Tail | 0.2000


The Show instance of Dist ensures that it sums up the probabilities of the same outcome, giving the above. We don’t show the Show instance here, that and other useful probabilistic combinators are presented in this blog post.

The Monad instance of Dist is the most important construct for modelling Monty Hall. We will use the Monad instance to model conditional discrete distribution. Given two discrete random variables X and Y, we can compute their joint distribution (given that they are not mutually independent) as follows:

$Pr({X = x} \cap {Y = y}) = Pr(X = x | Y = y) . Pr (Y = y)$

We can convert this idea to the Monad instance quite naturally:

instance Monad Dist where
(Dist xs) >>= f = Dist $do (x, p) <- xs (y, p') <- unpackDist (f x) return (y, p * p')  The p * p' is multiplying the probabilities of the two outcomes. The join function of the Monad instance can allow us to consider an alternate view of the above. Considering a distribution of distributions (please don’t try to visualise), we want to flatten it into a single distribution as follows: join :: Dist (Dist a) -> Dist a join (Dist dist) = Dist$ do -- dist  :: [(Dist a, Prob)]
(Dist dista, prob) <- dist -- dista :: [(a, Prob)]
(a, prob2)         <- dista
return (a, prob * prob2)


Monty Hall

Armed with this small set of combinators above, we now model the Monty Hall problem. We start by defining the outcome type:

data Outcome = Win | Loss deriving (Show, Eq, Ord)


And now, let’s intuitively specify the switching strategy : If our first choice is the winning door, followed by which Monty Hall opens a door that contains a goat, if we switch, then we will certainly lose. That is because we already selected the winning door, and the switch will cost us the victory. However, if our first choice is the losing door, followed by Monty Hall opening the door containing a goat, if we switch, then we certainly win. Because we chose the losing door, Monty Hall opened the other losing door and the remaining door certainly assures our victory.

In the specification above we used the language of certain victory or certain loss, so we need to model certainty. And that is modelled as:

certainly :: a -> Dist a
certainly a = Dist [(a, 1.0)]


So, one outcome that has a probability of 1.0. And, with that, we can easily translate the English specification above, to the Haskell snippet below:

switching :: Dist Outcome
switching = do
firstChoice <- uniform [Win,Loss,Loss]
if (firstChoice == Win)
then {- switching will -} certainly Loss
else {- switching will -} certainly Win


The comment {- switching will -} is added such that the specification almost translates to literal English. Now, we can observe the distributions:

> switching
Win | 0.6667
Loss | 0.3333


This shows us that switching our choice will allow us to win 66% of the time while losing 33% of the time. The first choice has the following distributions:

> uniform [Win, Loss, Loss]
Win | 0.3333
Loss | 0.3333
Loss | 0.3333


Our computation proceeds by certainly losing if we switch after winning the first round, while certainly winning if we switch after losing the first round. Hence the calculation proceeds as:

 Win ~> certainly Loss| 0.3333 * 1.0
Loss ~> certainly Win | 0.3333 * 1.0
Loss ~> certainly Win | 0.3333 * 1.0

==>

Loss | 0.3333
Win  | 0.3333
Win  | 0.3333

==>

Win | 0.6667
Loss | 0.3333


Hence, the contestant should switch!

Source code for the above is available at Abhiroop/bayes

Written on July 25, 2024