Supposing you are a participant on a game-show where the host presents you with three doors to choose from, behind one of which there is some desirable prize, such as a million dollars which you will win if only you have chosen the door behind which it lay. You are given a chance to choose one of the three doors as an opening gambit – then before the door is opened to you, your host does you the kindness of revealing that behind one of the other two doors, there was nothing (or else at least there was not the prize – popularly people say there was a goat instead; imagine whatever you like). Now, the contestant is always asked whether they would like to rethink their selection, and so you are, as expected, asked by the host to reconsider your selection. Most contestants think that the odds of them having the correct door is actually 50%, since there are only two doors, and behind one of them there is a million dollars. Suppose you stay with your door, and you will most likely loose – you may not immediately realize it but your chance of winning with the door you originally chose was exactly 1/3, and it would have been 2/3rds if you had decided to change your selection from your door to the other door.
Draw it out if you must:
[_1_] [_2_] [_3_]
Supposing that you chose door 1, and the host shows you that door 2 has nothing desirable behind it. Notice that the host has just shown you one of the two remaining doors which he knows there is nothing behind. You might be tempted to think that, since it is true that there is one million dollars behind one of two remaining doors, either door you choose is equally likely to be the correct door, and there is a probability of 50% for either. However, that’s not really the case – two thirds of the time the door you start with will not have the prize, and the remaining door will.
Do it trial and error – or methodologically. Let’s imagine that you decide to never change your choice first choice. First, you choose door 1, and it is behind door 1, and you are shown door 2, and thus you choose door 1 – you are correct. Second, you choose door 1, and it is behind door 2, and you are shown door 3, and you choose door 1 – you loose. Third, you choose door 1, it is behind door 3, you are shown door 2, and you choose door 1 – you loose. Those are all the possibilities, and you lost 2/3 times by sticking with the same original choice. In other words, in two out of three logically possible worlds, sticking with your choice yields the result that you do not get the prize.
Suppose you do just the opposite. Suppose you say you will always switch instead. First, you choose door 1, and it is behind door 1, you are shown door 2, and thus you choose door 3 – you loose. Second, you choose door 1, it is behind door 2, you are shown door 3, you choose door 2 – you win. Third, you choose door 1, it is behind door 3, you are shown door 2, you choose door 3 – you win. You just won 2/3 times by methodologically changing your selection of doors.
If this isn’t clear why the previous paragraph covers all the logically possible scenarios (somebody might think ‘why not start picking door 2 at first instead of door 1’ or something like that -without realizing that the situation is just going to be logically equivalent), you can write it out the longer way substituting all instances of ‘door 1’ in the paragraph above for ‘door 2’, ‘door 2’ for ‘door 3’ and ‘door 3’ for ‘door 1’; then afterwards run through them again once you substitute from the same above paragraph ‘door 1’ for ‘door 3’, ‘door 2’ for ‘door 1’ and ‘door 3’ for ‘door 2’ – you will have literally imagined all the logically possible worlds one at a time (though anyone who realizes that no matter which door you choose the formula is going to work out exactly the same way, yielding 2/3, can just move on immediately). Either way, you will come out with 6/9 times, or 2/3 times, that you win if you decide to switch doors, and you will loose that same ratio of times if you do not switch.
Curiously, most people think this is just completely counter-intuitive. It overturns all our intuitions about probability, because we tend to think that if presented with two doors, and you know behind one of them there is a million dollars, then either one you choose you will retain a probability of 50% chance of being correct. However, in the case of the Monty Hall problem, this is demonstrated to be untrue.
Is this really untrue? No. If you were presented with two doors and simply told that behind one of them lay a prize, you would intuitively think that you had a fifty-fifty chance of choosing the correct door, and you would be correct. Your intuition is not at fault – rather, your assessment of your situation is at fault – you aren’t really being given a choice between two doors, but rather between one door and “one of the other two doors”.
The problem is precisely that when shown one of the doors, we immediately think that the game has changed – we think that instead of being offered 3 doors, we are now just being presented with two doors, and this is the whole mistake. The game hasn’t changed – the chances one of the two doors we didn’t choose was the right one was 2/3, and since the host knows which one it is, the host chose to reveal the door that it wasn’t – however, the one door which you didn’t originally choose retains that whole 2/3 chance of being the correct one, because the one door isn’t one door, it is just ‘one of the two doors’ which you didn’t choose. The problem is not with our intuitions about probability (those are correct) but with our assessment of the situation in which we find ourselves.
In other words, the hosts action isn’t a game-changer at all, and that’s the whole trick of it. All you know is that the door you chose has one third a chance of being true, while it is 2/3 likely that “one of the other two doors” has the prize. When the host shows you that he knows which door it isn’t behind, it still remains true that it is 2/3 likely that the prize is behind “one of the other two doors”. In essence your choice is either door 1, or else “one of the other two doors”.
Therefore, the Monty Hall problem is not a good example of our intuitions about probability being incorrect, but a good example of a language-trick; what has changed, we are tempted to think, is the grammar of the game, but the grammar of the game hasn’t changed at all. Once one understands the philosophy of language, one can plausibly discern more easily that the host’s action is not a speech-act with respect to the game of choosing whether it is behind your door, or else one of the other two. The fact that the host has narrowed down for you which of the other two it isn’t, doesn’t change the fact that choosing the other door represents choosing “one of the other two doors.” I think this becomes clearer if the host were to, instead of showing us one of the doors it isn’t (since there is bound to be one), the host were simply to say “ok, how about this, I will give you the choice between the prize being behind the door you chose, or else the prize being behind one of the two doors you didn’t choose – which say you?” without him showing you anything, you would immediately see that, of course, it is 66.6% likely that “the prize is behind one of the two doors you didn’t choose”.