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Monty Hall in Common Lisp

2018-10-13

I've been doing a bit of reading this weekend including a chapter on the Monte Carlo method out of Data Analysis with Open Source Tools and making it about half-way into Practical Common Lisp. It occurred to me it might be fun to write a little program combining the two.

The Monty Hall Problem

The Monty Hall problem is a well-known, albeit counter-intuitive, puzzle demonstrating a bit of probability theory. The premise is:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

The trick is, the first time you choose, your chance of being correct is 13 (or, put otherwise, 23 wrong). The claim is, switching your choice gives you a 23 chance of winning.

Monte Carlo Method

Broadly speaking, Monte Carlo methods are those which adhere to the following pattern "in order to solve problems that might be deterministic in principle":

  1. Define a domain of possible inputs
  2. Generate inputs randomly from a probability distribution over the domain
  3. Perform a deterministic computation on the inputs
  4. Aggregate the results

There is, of course, a well-known explanation for the Monty Hall problem, but I'll be honest, seeing a proof for it is pretty dense. Performing a simulation is a nice way to prove it to yourself (as apparently was the case with Paul Erdős).

Why Common Lisp?

Having finished up with Touretzky's Common Lisp: A Gentle Introduction to Symbolic Computation, I'm really enamored with the idea of "symbolic" computation. While Janert's Data Analysis performs an analysis in Python, I've been won over to the idea of mental friction imposed by forcing what should be a symbolic "thing" into Python's data model and treating it as a string. It's also just fun.

A Simulation

A first step is to write a helper function for performing a random choice from a list of options (the function nth takes an index-postion from a sequence, you can imagine Python's list[index] syntax):

(defun random-choice (options)
  (nth (random (length options)) options))

From there, I've defined a make-choice function, it locally defines the "doors" and performs a random selection as the first choice, returning the first choice and the list of remaining doors:

(defun make-choice ()
  (let* ((doors '(prize goat goat))
         (first-choice (random-choice doors)))
    (if (equal first-choice 'prize)
        (setf doors (list 'prize  (random-choice (cdr doors)))
        (setf doors (list 'prize first-choice)))
    (list first-choice doors)))

I've modeled each of the three choices as a function, operating on two parameters the first selection and the remaining choices:

(defun remain (first-choice choices)
  first-choice)

(defun blindly-choose (first-choice choices)
  (random-choice choices))

(defun switch (first-choice choices)
  (first (remove first-choice choices)))

blindly-choose is the strategy of randomly picking between the first pick and the remaining. switch achieves it's namesake by removing the first-round-pick from the list of remaining, and returns the "first" (and only) option left. With those stategy functions, we can define a test, which performs some number of trials repeatedly using a given strategy:

(defun test-strategy (strategy num-trials)
  (loop for trial from 1 to num-trials collect
       (apply strategy (make-choice))))

The output of test-strategy is a list of the results for a given number of "games" played. If we wanted to try 10 games of choosing blindly:

* (test-strategy 'blindly-choose 10)
(GOAT GOAT PRIZE PRIZE GOAT PRIZE GOAT GOAT PRIZE GOAT)

If we're following the steps outlined above, we've got a domain of inputs, random input for the first choice and the three strategies outline our deterministic computation. All that's left is to aggregate the results:

(defun aggregate (test-output)
  (let ((wins (count 'prize test-output))
        (losses (count-if-not #'(lambda (x) (equal x 'prize)) test-output)))
    (list wins 'wins losses 'losses)))

I rather like this as a high-level operational view of what we're doing, the resulting call ends up looking like:

* (aggregate (test-strategy 'blindly-choose 100))
(49 WINS 51 LOSSES)

In order to do them all at once over a larger number of trials I can also write a short function to summarize each strategy:

(defun summarize (&optional (trials 100))
  (let ((results (mapcar #'(lambda (s) (cons s (aggregate (test-strategy s trials))))
                         '(remain blindly-choose switch))))
    (format t "~{~{~14a ~a ~a ~a ~a~}~%~}" results)))

* (summarize 1000)
REMAIN         322 WINS 678 LOSSES
BLINDLY-CHOOSE 500 WINS 500 LOSSES
SWITCH         662 WINS 338 LOSSES

Just like that, a computer simulation of one thousand takes on the Monty Hall problem utilizing the three strategies available. Remaining with the original choice results in winning 13 of the time, blindly picking between the two remaining in the second round gives you even odds, and switching gives you a 23 chance of winning.

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