Geeky is Awesome: The Monty Hall problem

This is an interesting mathematical problem in probability which has an intuitive answer. The only problem is that the intuitive answer is wrong, which is what I love about it. It's called the Monty Hall problem and it goes like this:

Imagine you're in a game show and presented with three doors.

Behind one of the doors is a prize. You are to guess behind which door it is. At this point we should all agree that the chance of guessing the right door would be 1/3. You decide to pick door number 1.

You think that you have sealed you fate, but in an surprising turn of events, the game show host says that in order to help you, since at least one of the two doors left must be wrong, the host is going to tell you that door number 3 is surely wrong.

So now you know that the prize is either behind the door you chose or the other remaining door, door number 2. The host now asks you if you'd rather stick with your first choice or choose the other door.

Did the host actually help you? Is there some advantage to knowing which one of the three doors is surely wrong? Intuitively you'd say that your chances of winning by sticking with your first choice or choosing door number 2 are both 1/2. It's either behind one or the other right? So your chances of winning are equal right? Right guys? Guys?

Most people will not be convinced that this intuition is false. So here's a program I've written in Python 3 which simulates this situation 100 times and it will count the number of times you would have won if you stayed with the first choice and the number of time you would have won if you swapped your choice.

import random

stay_win = 0
swap_win = 0
doors = {"Door 1", "Door 2", "Door 3"}
for _ in range(100): #For 100 times do the following...
    #Select the correct door
    correct = random.choice(list(doors))
    #The contestant makes their choice
    first_choice = random.choice(list(doors))

    #The door which can be eliminated cannot be the correct door or the contestant's choice
    can_eliminate = doors - { correct, first_choice }
    #Select the eliminated door
    eliminated = random.choice(list(can_eliminate))

    #The contestant has only one door left to choose if they swap their door
    choice_left = doors - { first_choice, eliminated }
    #Make this unchosen, non-eliminated door the second choice for swapping the door
    second_choice = choice_left.pop()

    if first_choice == correct:
        stay_win += 1
    if second_choice == correct:
        swap_win += 1

    print("Correct:", correct, " | ", "First choice:", first_choice, " | ", "Eliminated:", eliminated, " | ", "Second choice:", second_choice, " | ", "Won if stay:", first_choice == correct)
print()
print("stay:", stay_win, "swap:", swap_win)

In order to let you scrutinize the result, all the individual situations will be outputted as well:

stay: 29 swap: 71

What? That can't be right! You win by swapping almost twice as often as when you stay with your first choice. Let's try it again!

stay: 39 swap: 61

Um...

stay: 37 swap: 63

Hmm... Why would you win twice as often when you swap than when you stay? The only way to know this is by seeing what needs to happen in order to win by sticking with your first choice and to win by swapping.

Here is what the different possibilities are when the correct door is door number 1:

And here is what the different possibilities are when the correct door is door number 2:

And finally here is what the different possibilities are when the correct door is door number 3:

Did you get it? No? Well, look carefully at what needs to happen in order to win using the first choice. You need to guess the correct door on first try. What are the chances of you doing that? It's 1/3 of course. Could you ever win by choosing a wrong door on first try and stick to it? No.

Well then what do you need to do in order to win by swapping doors? Clearly, you will only lose if you choose the correct door on first try, but if it's the wrong door you've chosen on first try then you'll win. What are the chances of you choosing the wrong door on first try? 2/3, twice as many as choosing the correct one.

Clearly you're twice as likely to win by swapping than by staying. Obviously this doesn't mean that you'll always win, it just means that you're more likely to win. This illustrates a very important point which is that intuition isn't perfect. Sometimes it's just not enough to use intuition in order to predict what will happen, you will need to get down and dirty and actually do boring work in order to make a correct judgement. Never overestimate yourself.

Geeky is Awesome

Friday, April 12, 2013

The Monty Hall problem

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