Goats and cars revisited
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Last week's piece about goats, cars and the laws of probability has attracted a considerable postbag. The paradox began by putting you in the position of a game show contestant who has to pick one of three doors. All you know is that one of the doors has this week's star prize of a car behind it; the other two offer only a mangy goat. When you have picked one door, the host (who knows which door conceals the star prize) opens one of the doors you did not choose and shows a goat behind it. He then offers to let you change your mind. Most people then stick to their original choice. In fact, as explained last week, you will double your winning chance if you change your mind.

Antonio Carvalho writes from Henley on Thames: "Let's freeze the game show at the point when the host asks the contestant if he would like to change his mind. To avoid clutter, let's remove the door already shown to have a goat behind it. As far as the contestant is concerned, the circumstances remain unchanged, ie the door initially chosen by him has only one-third of a chance of having the star prize behind it. Now let's bring in a passer- by from the street, show him the remaining two doors, and give him a go at the star prize. The odds for the passer-by cannot but be 50-50! But this is impossible, because the odds must be the same for the passer-by and the initial contestant. Unless probability is a subjective concept, in which case a whole chapter of maths goes down the plughole. Can you help? (I'd like to get some sleep!)"

That, I think, sums up the psychological aspect of the paradox rather better than I did last week. Ian Bellamy identifies the paradox as one cited by Frederick Mosteller of Harvard 30 years ago. Mosteller's explanation, he says, is to draw up a table of probabilities of the various possible outcomes when the contestant picks door A from three doors A, B and C. Four things can then happen:

1: A and B are goats; host opens B;

2: A and C are goats; host opens C;

3: B and C are goats; host opens B;

4: B and C are goats; host opens C.

Since the probability that the prize is behind any particular door is one in three, the probability of 1 and 2 are each one in three, while 3 and 4 each have only one in six chances of happening. If the contestant sticks to his original choice, he wins in cases 3 and 4, but loses in 1 and 2. The total probability of a win is therefore 1/6 + 1/6 = 1/3. If he changes his mind, he wins in cases 1 and 2, and loses in 3 and 4: overall chance of a win 1/3 + 1/3 = 2/3 . Mr Bellamy concludes: "The Paradox, Mosteller sternly insists, is apparent only to those who fail to employ the proper sample space, which has to include the actions of the game show host."

But I am sure that will not totally cure Mr Carvalho's sleeplessness. For the real paradox lies not in the calculation of the probabilities, but in the nagging feeling that the host's action in opening one door after you have made your choice is completely irrelevant.

Jim Bragg, however, has given us a very simple example that should cure any lingering example:

"I had difficulty intuitively (not logically) understanding why you should always opt to change your door, until someone explained it to me thus: The game show host offers a million doors to you. After you choose one, the host closes all doors except yours and one other. The odds on the other door having the star prize now seem a lot better, don't they?"

And that explains everything. A chap coming in from outside would still see only two apparently equal doors. You, however, would know that one door (your original choice) was picked at random from a million, the other was selected from 999,999 as the only one that might have a goat behind it. That's a very special door. Change your mind!