What are the odds?
Now for something completely different. My former poli sci TA at Yale, Brett Marston, has raised one of my all-time favourite math/probability problems on his fine blog. Okay, now that you're riveted, here's the "Monty Hall Dilemma." (I'm taking the description from this website.)
Here is the situation. Finalists in a tv game show are invited up onto the stage, where there are three closed doors. The host explains that behind one of the doors is the star prize - a car. Behind each of the other two doors is just a goat. Obviously the contestant wants to win the car, but does not know which door conceals the car. The host invites the contestant to choose one of the three doors. Let us suppose that our contestant chooses door number 3. Now, the host does not initially open the door chosen by the contestant. Instead he opens one of the other doors - let us say it is door number 1. The door that the host opens will always reveal a goat. Remember the host knows what is behind every door! The contestant is now asked if they want to stick with their original choice, or if they want to change their mind, and choose the other remaining door that has not yet been opened. In this case number 2. The studio audience shout suggestions. What is the best strategy for the contestant? Does it make any difference whether they change their mind or stick with the original choice?
The answer (highlight to read): Yes, it does make a difference. In fact, switching improves your chances of getting the car from 1/3 to 2/3. Read the linked page's explanation. And if that doesn't help, read my explanation of it on Brett's site, linked above. The answer is pretty counter-intuitive. If you still don't get it, comment here and I will get back to you. For me, it took a while for it to sink in. And apparently there is still some debate among mathematicians about the answer to this problem. How cool is that? (Answer: Ice cold.)
posted by T.R. Cooper @ 11/25/2004 09:48:00 AM
3 Comments:
Thanks for this Tim, an intriguing problem that I have not come across, reminding me of the algebra you used to send from Saudi Arabia.
I have not yet come around to full agreement. I understand the 1,000,000 point well enough, but don't think that really applies to the neat symmetry of the three doors. It is the certainty of the goat appearing regardless of the original choice that so galls the intuition.
So answer me this: if, after the goat is revealed, we cast the situation back to a status quo. The odds are 50/50 on the two doors. Why are the odds worse if I then "choose" to stick, instead of "choose" to switch? In both cases I effectively get 2 of the 3 doors.
If I were more computer literate, I would simulate this over 1,000,000 times - the program would be simple enough. Would the switcher win more often? I still struggle to see why, on the narrow basis of the 3-way game.
Something to discuss more coherently in person, I assume, and I am also coming at this fresh without reading the long list of explanations, so perhaps I am missing an essential component? If so, enlighten.
Cancel that - it just sunk in over here while typing up some Conflict of Laws notes. I can now get on with it clearly!
The trick really is in such simple deception, in making you think the puzzle involves the making of a NEW choice. Recast the problem as the choice between your original selection (1 of 3) on one side and the two unopened doors (2 of 3) on the other. Why should I get worried when Monty reveals one of my two options to be a dud in a zero-sum game?
So switch, I guess. Although personally I would stick with the Red Snapper. Very tasty.
That's a great story. Waiting for more. » » »
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