Here's something I think I understand, and something arising from it that I don't understand. So somebody out there may be able to put me straight.
What I think I understand
A new version of an old story. You're a participant in a game show and have three doors in front of you. Behind one door is a complete set of Wisden Cricketers' Almanack and behind each of the other two doors is a signed photograph of Mary Chapin Carpenter. Being a rational person, you naturally want the Wisdens. You choose a door - let's call it door A - and after you've done so, the game show host, before opening door A to let you see what you would have won, opens another door - say, C - to reveal one of the two unwanted signed photographs. He or she now offers you the following option: you can stick with door A and get whatever is behind it; or you can alter your initial choice and go for door B. Should you stick with A or go for B, or does it make no difference which you decide to do?
A standard intuition here is that, whether you stay with door A or switch to door B, you now have a fifty/fifty chance of getting the Wisdens. One of the photos is 'taken out'; one door has the Wisdens, the other door the second photo. So you might as well leave your original choice unchanged, although it will do no harm if, on a whim, you do decide to switch. This is wrong, however. You should switch. If you do, you have a 2 in 3 chance of getting the Wisdens; if you stay with your original choice you have only a 1 in 3 chance of getting them. It's known as the Monty Hall problem, and I now display the breakdown of probabilities with the aid of the following table. It gives you a complete picture of how the Wisdens and the photos of Mary Chapin Carpenter might be distributed behind the three doors, A, B and C.
(1) A Ws B Ph C Ph
(2) A Ph B Ws C Ph
(3) A Ph B Ph C Ws
These three cases exhaust the possible ways in which the prizes can be arranged vis-à-vis the three doors. Now, if your choice is door A, then... In case 1, the game show host can open either of the other two doors to reveal a photo of Ms Carpenter, and by altering your choice to the third of the doors, you will lose the Wisdens and get the other of the two photos. In case 2, the host must open door C to reveal the other photo, and by shifting from door A to door B, you'll get the Wisdens. And in case 3, the host will reveal a photo behind door B, and you jump from door A to door C, so getting the Wisdens again. That's two successes out of three. You can run this any way you want, with your initial choice being door B or door C rather than door A, and you'll get same result: two successes and one Mary Chapin Carpenter - failure! - from three attempts.
The thing I don't understand
OK, so what puzzles me is this. Suppose that one evening there are two participants playing at the same time: you and me, rather than you alone. We both want the Wisdens. Neither of us wants a photo. Suppose, further, that things are so arranged that a door can be opened for you to see what's behind it, without my being able to, and that a door can be opened for me to see what's behind it, without your being able to. So we make our initial choices, and let's imagine - just to keep things simple - that we don't go for the same door. Let's say that I go for door A and you go for door B. Look again now at cases 1 and 2 above. The host can show both of us door C. And, if what I've said is right, it's in my interests to switch from door A to door B and it's in your interests to switch from door B to door A. If I stay with door A, I have only a 1 in 3 chance of getting the Wisdens, but if you go for door A, you have a 2 in 3 chance of getting the Wisdens. The same disparity, though in reverse, vis-à-vis door B. Does this mean that probability is agent-relative? Even if it does, I can't escape the air of paradox in the proposition that it's better for you to choose the door I'm abandoning, and better for me to choose the door you're abandoning - that we both improve our chances thereby. Any offers?