 | The Monty Hall Problem |
This classic probability problem — "The Monty Hall problem" — is described on countless pages across the web, each offering its own interpretation. So why cover the same ground again?
Well, the aim of this article is to help those of us who might feel less-than-persuaded by the mathematical proofs offered in most descriptions. I work through several explanations, many of which you may have seen elsewhere, but the last of which is original, and I hope of use to some readers.
While I long ago 'intellectually accepted' the counterintuitive answer to the problem, I remained unsettled by a nagging sense that I hadn't fully understood it. Why did the mathematical proof not enable me to 'feel' the answer at a more visceral level? And why was I so easily 'tricked' into giving the wrong answer?
Years later, rethinking the subtleties of the problem (time of writing: 2008), I tabulated the data (see Table 4 at the very end of the article) and had a mini 'eureka' moment. The key, for me at least, was seeing all the permutations represented visually. I suddenly 'internalised' not only why the answer was what is was, but why the wrong answer was so tempting. This article, then, is an attempt to elucidate this notoriously difficult-to-grasp concept in a way that might satisfy your gut as well as your brain.
To that end, I would judge the use of equations and technical nomenclature to be ineffectual, and as such, this article is entirely devoid of those intimidating little symbols: 
and 
.
I am indebted to my good friend, and former professor of mine, Roger Marsden, for having introduced me to this brain-teaser back in 1999.
This problem is a delight; its conclusion, at first blanch, nothing short of incredible. It beautifully highlights one of the most common 'school-boy errors' in the conceptual minefield of Probability Theory, while remaining very simple in its presentation. But along with the intrigue of the apparent paradox comes another important lesson with far-reaching psychological implications: to what extent should we trust our intuition? How often does putative 'common-sense' lead us down the wrong path?
This puzzle has been presented in myriad guises, perhaps the first of which being a 1959 publication by mathematics and science writer, Martin Gardner. His framing of the problem concerned three condemned prisoners, one of whom was to be pardoned at random. Arguably the most famous version, however, goes by the name of "The Monty Hall problem/trap/paradox". It was highly popularised by an article written for Parade magazine in 1990 by Marilyn Vos Savant (who, incidentally, is cited in The Guinness Book of World Records under "Highest IQ"). This interpretation involves our guessing behind which door a prize is to be found.
In this article, I present the problem in an admittedly less colourful way, focusing on boring old playing cards, but my choice is designed to help highlight an additional point I make in the conclusion.
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| Fig 1 Three overturned playing cards (©) |
I have 3 playing cards (as in Fig 1) turned face-down on a table.
Under one of the three cards, I've placed a feather (as in Fig 2).
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| Fig 2 A feather under one of the cards (©a) |
You have to guess which of the 3 cards is covering the feather.
What's the probability — all things being equal — that you'll guess the correct card?
(Note that there's no 'trickery' involved here. The card covering the feather is just as flat as the other two. There's no way to tell where the feather is.)
1 in 3. (Otherwise put: 0.3 or 33.333˙% or ⅓.)
So far, so good.
We're agreed that your choice of card has a 1 in 3 chance of being correct. After all, there are 3 cards and only 1 feather, with an equal chance of the feather being under any one of the cards. Probability theory confirms this. No catches. This all seems like common sense, right? Utterly intuitive.
But let's make things a little more interesting.
I'm not going to reveal whether or not you've guessed the correct card.
What I am going to do is to tell you something about where the feather is not. I'll point to a card that you did not choose, and tell you that it is not covering the feather.
Let's be clear about this: I won't comment on your chosen card. The feather may be there, or it may not. I will merely point to 1 of the 2 cards you didn't choose, and tell you categorically that the feather ain't there.
(And I'm not allowed to lie, in case you were wondering!)
Let's take a 'for-instance'...
We'll refer to the 3 cards as #1, #2 & #3, from left to right.
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| Fig 3 You guess #2. I then tell you that the feather is not under #1. (©b) |
Let's start the game from the top, and imagine that you guess card #2 (that's your hand pointing in Fig 3).
Now that you've chosen, I reveal that the feather's not under card #1 (hence the red cross in Fig 3).
This means, without a shadow of a doubt, that the feather is in one of two places:
Agreed?
Knowing that it's not under card #1, tell me: What's the probability that your guess was correct?
Got it? Simple, right?
I'm predicting that you reckon there's a 50/50 chance (1 in 2). The same chance as flipping a coin. Either the feather's under the card you chose (card #2), or it's under card #3. You're thinking: "There's an even chance that the feather's under either one of them."
If you were indeed thinking that you had a 1 in 2 chance of being correct (as almost everyone does), then you'd be wrong.
Shock, horror!
Here's the truth — the correct answer according to probability theory, and the way the Universe really works:
Bear with me now. I appreciate that this may well fly in the face of everything you think you understand about calculating odds. I 'get' that I'm asking you to accept something that appears to clash with common sense. But I can assure you that mathematicians are unanimous.
You might want to go back and re-think the problem, simple as it is, before reading on for the explanation.
Before explaining, however, there's a juicy extension to the problem that will mess with your head even more, while at the same time providing the key to fully understanding it...
Let's start from the top, repeating the game in exactly the same way. For the sake of simplicity, we'll also use the same hypothetical choices that were presented in Fig 3. I.e.:
You chose #2. I then tell you that the feather is not under #1.
This is all exactly the same as before: We know it's not under #1, so it must be under either #2 or #3. But here's where we add a new step...
Before revealing the feather's location, I'll give you an opportunity to either stick with your original choice, or to change your mind. In this case, that would mean either sticking with #2 or changing your mind and going for #3.
Again, I'm predicting that you reckon there's a 50/50 chance (1 in 2) regardless of whether you stick with your original choice or change your mind.
If you were unsettled, incredulous, or perhaps enraged by the answer to the simple version of the problem (I've heard stories of friendships ending over it!), then you'd better sit down for this one (does anyone surf the internet standing up?)...
Please don't shoot me. I didn't invent the laws of probability. I am but a messenger.
If you have the time, run through the steps for either the simple or the extended versions of the problem in real life with a friend. Perform the guessing game many times, and make a note of how often the guesser picks the correct card, and you will see proof before your very eyes.
If the odds were truly 50/50 we would expect the guesser to pick the correct card approximately 50% of the time. So, after 100 repetitions, it's likely that the number of correct guesses would fall somewhere in the peak of the bell-curve. I.e. It's very likely that you would have been correct 45 — 55 times, with all the other guesses having been wrong.
In reality, however, you will find that running the game 100 times results in approximately 33 correct guesses. You will be wrong around two thirds of the time.
If you run the game 100 times, always changing your mind when given the opportunity, you will indeed win approximately 67 times.
100 runs, always sticking with your original choice, results in approximately 33 wins.
Luckily, thanks to the genius mathematical minds of Gerolamo Cardano, Pierre-Simon marquis de Laplace, Andrey Nikolaevich Kolmogorov, and many others, we are equipped with the tools to demonstrate the veracity of such claims without resorting to tedious practical repetition. But my intent here is to help the truth shine through without resorting to an equation.
I want to start by elucidating why it seems so intuitive that the chance be 50%, and to help us understand what's going wrong with our thinking.
For now, I'll restrict the discussion to the simple version of the problem (where no opportunity was given for you to change your mind).
Table 1 generalises the problem. It shows every possible outcome when we're asked to pick between 3 things that all have an equal chance of being correct. Here, we will imagine that Option #2 is correct, and we'll mark the outcomes accordingly:
| Table 1 | ||
|---|---|---|
| If Option #2 is the correct answer... | ||
| Your Guess | ⇒ | Result |
| Option #1 | ⇒ | Wrong |
| Option #2 | ⇒ | Correct |
| Option #3 | ⇒ | Wrong |
This, of course, is precisely the scenario we had in the 'Easy Question' of the simple version of our problem. (Before we introduced the element of my revealing a card that was not covering the feather). You may be wondering what value there is in putting something as simple as this into a table. Well, I'll remind you that there's something going wrong with our reasoning; something fundamental that we're screwing up when we calculate probabilities. So, in order to be absolutely clear, I've started with this simple table upon which we will build later.
Blindingly obvious as all this is, work with me for a second... Notice the word "Correct" appears once, and the word "Wrong" appears twice. There are 3 possible outcomes, and 2 of them result in "Wrong", while 1 of them results in "Correct". It follows that two-thirds of the time we would be wrong, and one third of the time we would be correct.
Table 1, as established, is examining the outcomes on the presumption that Option #2 is correct. Naturally, we could re-draw the table to show the outcomes if either of the other two options were correct. But the table would always show 1 instance of "Correct", and 2 instances of "Wrong".
Let's reintroduce the step of my pointing out a card that does not cover the feather. This is where we enter the intellectual quagmire. Refer back to Fig 3, which represents such a situation. With this in mind, look over Table 1 again, and ask yourself whether or not the table is still accurately representing every permutation.
You see, even though we've been told that the feather is not under card #1, the outcomes in the table (2 instances of "Wrong" and 1 instance of "Correct") still apply. The critical point is this:
At the moment of our making a choice, the probability that we were guessing correctly was 1 in 3. Anything we learn after the fact is irrelevant. The choice was made before we learned the additional information, so the additional information has no bearing on the odds.
If you're already persuaded, you can skip the next bit and scroll down to the section called "Extended Problem Explanation".
Otherwise, read on...
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| Coin toss (©c) |
The ineluctable truth — that receiving additional information after the fact (after we make our choice) doesn't alter the probability — becomes obvious when we take an even simpler example.
I've tossed a coin. Guess whether it's gonna come up heads or tails.
Here, there are only two possible outcomes (forgetting the highly improbable, though certainly possible, outcome of a coin landing on its edge).
It'll either be heads or tails. 50/50 chance of getting it right.
The fact that there are only 2 outcomes, rather than 3, makes a big difference to how readily we can 'see' the truth...
With me so far?
What I'm getting at is this: If my knowing the outcome doesn't change the odds, why should your knowing the outcome make any difference?
You've learned the result, so you know definitively that your guess was wrong, yet the chance that you might have been right remains 1 in 2.
Contrary-wise, if you had guessed heads, and the answer was indeed heads, there was still an equal chance of being right or wrong, even after you learned whether you were right or wrong.
Allow me to rephrase all that:
There's a 50% chance of predicting the outcome of a coin toss. If you guess heads, but the result is tails, you then know for sure that your guess was wrong. But that does not mean that there was a 100% chance that you would be wrong. It is certain now that you were wrong, but it was not known then. So, knowledge learned after a guess doesn't affect the original probability.
While that whole coin toss thing may have seemed self-evident, it serves as a useful reference when returning to our cards.
Increasing the number of possible outcomes from 2 (such as the sides of a coin) to 3 (such as our cards), we find ourselves in an identical situation: "The feather isn't under card #1" is knowledge learned 'after-the-fact', i.e. after we've made our guess. It therefore cannot affect the original probability, just as with the coin.
The difference between the coin and the cards — and the consequent source of all our confusion — lies in how we relate certainty and causality:
Any additional information about the coin results in knowing the answer with certainty. Given that there are only two possible outcomes, learning what it is not tells you what it is. There are only two possible facts that could be disclosed, and both of them — "It is not tails" or "It is heads" — reveal the answer.
When there are more than two possible outcomes, however, additional information such as "the feather isn't under card #1" is not sufficient to know the answer with certainty. It's the lack of certainty that misleads us. The fact that, despite the new knowledge, the answer is still not known with certainty suggests to us that it is reasonable to re-evaluate the probability. We deduce, erroneously, that the fact that more light has been shed on the scenario must mean that the odds have changed.
To truly equate the card game with a coin toss, imagine we are told: "The feather is not under card #1, nor is it under card #3".
With cards #1 and #3 'eliminated', we have an exact analogy to the coin toss. Every outcome has either been revealed, or can be deduced by a process of elimination. #1 & #3 are out, so it must be #2. We have effectively been told the answer, just as with the coin toss when we were told which side it was not.
Think it through once more: you had a 1 in 3 chance of guessing the correct card. I've now told you that you did in fact guess correctly. But the question remains: "What was the chance that you originally guessed correctly?"
The question is not: "Are we now certain that you are right?". The question is concerned with the odds you had (past tense!). Whether or not you now know that you are correct is irrelevant. The chance was 1 in 3 at the moment that you were guessing. Nothing — no additional after-the-fact information — can ever change that. It doesn't matter if the information provides certainty or not.
We could say that in the absence of certainty, we have a tendency to imply 'backward causation'.
If we change the order of the events, thus:
...then the odds are indeed 1 in 2 of guessing correctly.
Card #1 never had a look-in. It truly was eliminated from the calculation. With the steps in this new order, it is tantamount to saying:
Put that way, the 3rd card is an unnecessary complication, extrinsic to the problem.
Reversing steps 2 & 3, however (as we had in the original game), brings that 3rd card into play: It has not been eliminated before you make your choice. Hence, the odds of guessing correctly were 1 in 3.
If this hasn't helped, and you're still thinking "Fine, but regardless of when I learned that is wasn't under card #1, the fact that we eventually ruled it out means that there was really only a choice between 2 cards, not 3!", then take a look at Table 2.
| Table 2 | ||
|---|---|---|
| If #2 is the correct answer, and a card is ruled out AFTER you've chosen, then: | ||
| If you picked card #1 | and then a card is ruled out ⇒ | you would be wrong |
| If you picked card #2 | and then a card is ruled out ⇒ | you would be correct |
| If you picked card #3 | and then a card is ruled out ⇒ | you would be wrong |
Count how many possible outcomes are wrong. Count how many are right. The table is showing every possible outcome, and we can see that there are more possible wrong answers than there are correct answers. In order for the probability to be 50%, there would have to be the same number of right answers as there were wrong answers. As it is, however, there are 2 wrongs and 1 right. So, there's a ⅔ chance of being wrong, and a ⅓ chance of being right.
To remind you, the extension to the problem was the additional step of "You can stick with your original choice or change your mind".
The counterintuitive, though accurate, answers were:
We've done the ground-work for this already, so all that remains is to show every possible outcome (technically called the 'sample space') by building upon our simple tables.
| Table 3 | |||
|---|---|---|---|
| If Option #2 is the correct answer... | |||
| Original Guess | What will do you after a card is eliminated? | ⇒ | Result |
| #1 | Change | ⇒ | Correct |
| #2 | Change | ⇒ | Wrong |
| #3 | Change | ⇒ | Correct |
| #1 | Stick | ⇒ | Wrong |
| #2 | Stick | ⇒ | Correct |
| #3 | Stick | ⇒ | Wrong |
If you probe deeply enough, you will find information in this table that answers two questions:
There are 6 possible permutations to the problem, all of which are listed in Table 3.
Of the 6:
3 permutations result in 'Correct', &
3 permutations result in 'Wrong'
Given that there are 3 permutations for both 'Correct' and 'Wrong', we are tricked into supposing that there is an even chance — i.e. a 50/50 chance (a probability of ½) — that we've guessed the correct card, no matter what we decide to do.
BUT...
If we add a little extra information to the table (see Table 4 below), we see that:
We are correct 2 out of 3 times ( = ⅔ probability), if we change, while
We are correct 1 out of 3 times ( = ⅓ probability), if we stick.
| Table 4 | |||||
|---|---|---|---|---|---|
| If Option #2 is the correct answer... | |||||
| Original Guess | What will you do after a card is eliminated? | ⇒ | Result | Tally | |
| #1 | Change | ⇒ | Correct |  |
2 Correct 1 Wrong |
| #2 | Change | ⇒ | Wrong | ||
| #3 | Change | ⇒ | Correct | ||
| #1 | Stick | ⇒ | Wrong | |
1 Correct 2 Wrong |
| #2 | Stick | ⇒ | Correct | ||
| #3 | Stick | ⇒ | Wrong | ||
This game has been used by many a con-artist. It's power to deceive us is impressive, but there's more to it than merely fooling our intuition...
Human beings, in general, have a tendency to 'stick to their guns'. When given the option, a very large majority of people elect not to change their mind, preferring to stay with the card they originally chose. Perhaps this is attributable to superstition, to a belief in fate, or simply to our desire to affirm our resolve; to avoid appearing weak-willed. But whatever the reason, it long ago came to the attention of street magicians that this game can be used to rake in a lot of money.
The poor unsuspecting pedestrian, lured into this cunning trap, is invited to place a bet on whether or not they will guess the correct card. It seems so utterly obvious that their chance of wining is ½, and so the stakes seem pretty good to them. Of course, if they only knew that changing their mind and picking the other card would result in the excellent odds of ⅔, they would be far more likely to win the bet. Ironically, owing to a well-documented propensity of our psychology, the chances are that they'll stick with their original choice. And so the con-artist wins the money two thirds of the time.
What genius, to earn a profitable living by taking advantage of a common misunderstanding about the workings of Probability Theory, combined with a little pop-psychology. At least now, you won't get caught out.
Matheson Bayley © 2008
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