code guessing, round #96 (completed)

function randomWithCondition(cond, ...)
	local number

	repeat
		number = math.random(...)
	until cond(number)

	return number
end

function simulateOneIsCar(doSwapOrNot)
	local doorWithCar = math.random(3)
	local initialDoorChoice = math.random(3)

	local revealedDoor = randomWithCondition(function(n)
		return n ~= doorWithCar
		   and n ~= initialDoorChoice
	end, 3)

	local unrevealedDoor = randomWithCondition(function(n)
		return n ~= revealedDoor
		   and n ~= initialDoorChoice
	end, 3)

	if doSwapOrNot then
		return unrevealedDoor == doorWithCar
	else
		return initialDoorChoice == doorWithCar
	end
end

function simulate(doSwapOrNot)
	local total = 0
	local goats = 0
	local cars = 0

	local prob
	local timeoutInit = 100000
	local timeout = timeoutInit

	for n = 1, math.huge do
		total = total + 1

		if simulateOneIsCar(doSwapOrNot) then
			cars = cars + 1
		else
			goats = goats + 1
		end

		local newProb = cars / total * 100

		if prob and math.floor(newProb * 1000 + .5)
		         == math.floor(prob    * 1000 + .5) then
			timeout = timeout - 1
		else
			timeout = timeoutInit
		end

		prob = newProb

		if timeout <= 0 then
			break
		end
	end

	return prob
end

function assign(key, value)
	_G[key] = value
	return value
end

print ([[
The Morty Call problem is a famous statistics conundrum in which the
problem resolver is presented with the following situation:

- 3 doors, 2 with "goat" behind them, 1 with "car"
- the presented-to must pick one door and obtains the "*" behind it
- for unspecified reasons, it is desirable to obtain uncaring cold
  machine ("car") rather than friend ("goat")
- after initial pick, presenter will reveal the "*" behind another door,
  where "*" is guaranteed to be "goat"
- the presented-to may now change its choice with the newly-obtained
  information available

The problem statement is: Which choice leads to more desirable outcomes
(that is, obtainage of "car")?

- changing the choice to the other door that was not revealed by the
  presenter, making use of the new information
- double-down on the initial pick.

It is postulated that "changing the choice to the other door" is the
correct choice to make in accordance with problem statement and
preamble. We consulted our probaballistics law team to find out if
this postulate is true or not. Here are their findings:

- "CHANGE CHOICE": "car" is ]]..("%.1f%%"):format(assign('pOfChange', simulate(true)))..[[ likely
- "KEEP INITIAL": "car" is ]]..("%.1f%%"):format(assign('pOfDont', simulate(false)))..[[ likely

Therefore, analysing our results, the conclusion is that:

- "]]..(pOfChange > pOfDont and "CHANGE CHOICE" or "KEEP INITIAL")..[["

shall be the correct solution to the Morty Call problem.
]])