And the scores on the doors are...

A new twist on an old problem

Professor Geras at Normblog proposes a novel twist on the Monte Hall problem:

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?

Norm correctly states that while intuition leads us to assume there is no difference, actually switching doors doubles our probability of winning.

He goes on to suggest a modification:

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?

Well, here’s mine:

The original problem

Norm picks a door; any door. As the Wisdens is behind one of the 3 doors, the odds of Norm winning are 1/3. Accordingly, the probability of it being behind the other two is 2/3.

If Norm sticks with his door, his probability of winning the Wisdens is 1/3.

Now the host opens one of the doors, revealing the photo. [It is important to note he can always do this: regardless of which door Norm picks, at least one of the remaining two has a photo behind it.] If Norm swaps, this is effectively the same as being allowed to look in the other two, as he opens one and the host opens one. So his probability of winning is 2/3 if he swaps, as opposed to 1/3 if he does not.

Swapping doubles his chances.

The Norm Variation

In this scenario, both Norm picks a door, and I pick a door. Now, the host needs to open a door - but he cannot always do this.

Suppose both Norm and I both choose a door in front of photo: the door we have not picked has the Wisdens behind it, and the host cannot open this door to reveal a photo.

Ringing the changes

One way to adapt the problem is to generalise to the n-door case. In this instance, there are (n-1) photos, and 1 Wisdens. Suppose I pick a door at random, my chance of winning the Wisdens is 1/n. The host opens another door, displaying a photo, and offers me the chance to swap.

If I do not swap, my chance of winning remains at 1/n

If I do swap, my chance of winning can be calculated by noting that the probability of the Wisdens being behind one of the remaining n-2 doors is (1 - 1/n), and so the probability of the Wisdens being behind any given one of them is (n-1) / [(n)(n-2)]. Swapping leads to an improvement of the chances of winning by a factor of (n-1)/(n-2). [1]

Clearly, swapping is the correct strategy regardless of the number of doors, but the factor by which you improve your chances decreases with increasing number of doors, tending to 1 as the number of doors tends to infinity.

[1] Quick reality check: for n=3, these results imply not swapping leads to a Pr(Wisdens)=1/3, swapping to Pr(Wisdens)=2/3, an improvement factor of 2, as expected above.

[Edited fur spoling]

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