I started writing this entry ages ago, back when the subject was vaguely topical again amongst a few of the ScienceBloggers, but because I still suck at blogging, it’s taken this long to get around to finishing it. But I wanted to put something coherent together about this, because although I’ve made several off-the-cuff attempts to explain it during various conversations before, I’ve never yet organised my own thoughts on the Monty Hall problem into an actual blog post. So, here goes. Fretter ye not: there will be as little actual maths here as possible, although there is a lot of head-bendy number-wrangling to get to grips with in the problem itself, and it is interesting whatever any of you say.
What’s the dealio?
Imagine that you’ve used all your wit and cunning to make it through to the final round of a game show, eliminating the competition mercilessly with your infallible knowledge of pointless trivia and/or ability to press a button really fast. The rewards for your victory so far are numerous and many, but pale in comparison to the grand prize… A very, very shiny thing. Oh, such glorious dreams that might finally be realised this night.
And the path to true delightenment is so very simple. There are three doors. Behind one of them is your shiny thing, garnished with an elegant silk ribbon and all ready to go. Behind each of the other two is a small tuna sandwich, as a rather sad and pathetic consolation prize. You’re not that hungry, but you do like shiny things, so it is vital that your final reserves of tactical brilliance are brought into play now.
Your first task in this
deadly game is to pick one of the doors. This door will not be opened, but held provisionally as your first, tentative choice. You’re not committed to anything yet.
Next, I, your charismatic and photogenic host, will show you where one of the sandwiches is to be found, by opening up one of the other two doors. Your initially chosen door will not be opened, and the grand prize will not yet be revealed.
Then, you simply get to open one of the two doors that remain closed; whatever lies behind is yours to keep, whatever remains hidden behind the third door is lost to you forever.
Your choice, of course, is between resolutely and firmly sticking to your convictions and opening the door you chose in the first place, or flip-flopping indecisively and picking the other one. So, which is it to be?
Well, if you let my ever-so-subtly inflammatory language sway you, then bad luck. If you stick with your original choice, you’ve got a 1-in-3 chance of winning. You can double those odds, to a much more friendly 2-in-3 (I know I said there wouldn’t be much maths, but I really hope you’re not struggling already) if you switch your choice, and open the other door.
Curse you, logic!
So, what’s up with that? After that first sandwich is out in the open, it looks like a fairly simple case of 2 doors, 1
cup shiny thing1. Your prize is either behind one door or the other, so it’s just a 50/50 decision, right? It shouldn’t make any difference which of the two doors you pick, because you’ve got a 50% chance of picking the one with the prize in each case, right? So it doesn’t make any difference whether you switch or not… right?
No. Not right. The opposite of right. (There should be a word for that.) Hence the controversy. There’s some great stuff on the wikipedia page about the thousands of people who wrote to Marilyn vos Savant to tell her she was wrong, “including several hundred mathematics professors”, and a great deal of debate still rages on, as evidenced by the discussion page for that article. Some of this active disagreement is to do with important semantic details of how the game works, but a lot of it comes from people thinking they’ve managed to outsmart hordes of mathematicians by saying “but it’s obvious!!”
So although you can embugger about with the precise wording and turn it into a different problem with a different solution, the way I’ve phrased it above is the best-known form, as well as probably the most counter-intuitive and perhaps the most interesting. So, let’s see if I can persuade you that you’re wrong in time for tea, when I will proceed to use the Banach–Tarski paradox to prove that my slice of cake is actually equivalent in size to yours, even though it appears to have twice the volume.
… or door number three?
The semantics are important here, and it’s worth thinking about where that “1-in-3” number even comes from. What is it describing? Obviously it relates to the probability of a shiny thing being behind a door, but it’s not strictly true to simply say “The probability of the prize being behind this door is 1-in-3”. Whichever door you happen to be pointing at, the odds of the prize being behind it, actually, are either 100%, or 0%.
Your prize isn’t hovering behind each door in some misty field of quantum indeterminacy like a half-dead cat in a box, and opening the doors doesn’t collapse any wave-functions to change the probabilities. If you open door number 2 and find a sandwich, the odds of that door hiding the prize don’t then suddenly become zero; you never had a hope with that door, let alone a 1-in-3 hope.
Understanding this might help you get around some of the more awkward sloppy thinking that gets in the way of understanding the paradox. What the 1-in-3 odds refer to is the chance of a randomly chosen door turning out to hide the prize. If you’re about to close your eyes and point, you’ll have a 1-in-3 chance of choosing the right one, because if you repeated the experiment numerous times, you’d get it right one time out of every three.
Similarly, when there’s two boxes left, picking one randomly would give you a 1-in-2 chance of winning. But you’re not doing that. You can do better than just make a random guess, because you have more information available to you, so you can improve your odds. I’ve cut a lot of wordy rambling from the first draft of this article at this point, because I think it’d be more helpful to dive in with a semi-practical example.
In a world of pure imagination
Close your eyes and picture a scene. Wait, don’t go drifting off into your own fantasy-land yet, I’ll describe it to you first. It’s okay, this isn’t an excuse for me to divert your attention while I creep up on you and throw a massive spider onto your face and laugh and laugh. It just helps to have a visual for this, and waving an actual deck of cards at you isn’t really an option from where I’m sitting. (After all, not only am I far away, but from my perspective as I type this, you’re reading it in the future. Spooky.)
Anyway, close your eyes if it helps, and imagine I’m waving a deck of cards in your face. We’re going to play a variation on the classic Monty Hall game, I tell you. (Don’t worry about how I got into your home with these cards in the first place. It’s all fine. Just relax. There is no massive spider.) In this variation, there are cards instead of doors, and fifty-two of them instead of three. The ace of spades represents the prize, the others are less rewarding than a Jo Caulfield stand-up show. (Not witty, hardly scathing, but I haven’t mentioned to anyone lately that Jo Caulfield isn’t funny, so it had to be done.)
Much like in the original, you start off by picking one of the cards. You pull it across the table (yes, there’s a table now, don’t strain your imagination) toward you, keeping it face down and unseen, while I hold the other fifty-one cards. I then look through my cards, and throw fifty of them face-up on the table, all of which are not the ace of spades.
So, do you stick with the one you first chose, or would you rather change your mind and have the one I’m left with from my batch of fifty-one?
Hopefully it’s obvious why I hope this answer is obvious. After you made your initial pick, and before I showed you those fifty cards there, it would have been pretty clear that I was shuffling through looking for the ace of spades, so that I could hold that one back and throw the rest of them down. It’s possible that you happened to pick the exact right card on your first try – your odds were 1 in 52, which isn’t really that remote at all – in which case, when I looked through my 51 cards, I would’ve seen that I could’ve chosen any batch of 50 to show you, and kept any one of them behind, and you’d have been better off sticking.
But it’s far more likely that you grabbed something other than the ace of spades the first time, leaving me with that ace along with fifty other cards, so I had no choice but to show you those fifty. If someone else was brought in to the game at this point, and just saw two cards face-down on a table and tried to pick the ace, they could do no better than to guess at random with 50/50 odds. But you know more than they do about these two cards: you know which one of them you picked at random, and which one I (probably) selected carefully. So you know that sticking with your first choice means that you’re relying on 1-in-52 odds, which you can deliberately choose to avoid, giving yourself a 51-in-52 chance of coming out on top.
Probability is weird. Our brains aren’t wired to be able to naturally handle complex and unnatural likelihood analysis. This is meant to be confusing. Even if you get that several reds in a row doesn’t mean that a roulette wheel is “due” a black on the next spin, this stuff is hard. Don’t sweat it.
1 One of the less successful spin-offs, currently with only 103 views on PornoTube.
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