The Monty Hall Problem, Part Two
I'm afraid I'm very pressed for time today, so only brief blogging.
The answer to Variation One of the Monty Hall Problem, given at the end of yesterday's post, is that it now makes no difference whether or not you switch. Really.
The answer to Variation Two is that your best strategy is to stick with your initial choice until there are only two doors remaining. No other strategy will give you as high a probability of winning.
I'll go into some of the mathematical aspects of these problems in later blog entries, though some of my commenters to the previous post have already done an excellent job of it (note particularly Micahel Ralston's excellent comment explaining the difference between the situations where Monty Hall does, or does not know where the car is.)
Richard Wein points out that the Times version of the problem, quoted in yesterday's entry, is rather vague, since it is not stated explicitly whether Monty Hall knows where the car is. He's right about this, though I think it's stronlgy implied in the Times version that Monty Hall does know where the car is and therefore always reveals a goat.
If this is the point Somerby had in mind in saying that in the Times version it makes no difference whether or not you switch, then I will be very impressed. I don't believe that to be the case, however. Somerby's reply was so snide and harsh towards the Times reviewer that I think he was treating it as obvious that after Monty Hall opens one of the doors there is a 50-50 chance of winning. I don't think he was making subtle points about what Monty Hall does and does not know.
As of this writing Somerby has not revisited this problem.
Anyway, there's lots more to say about this problem, and I'll try to get around to it in the near future. Let me thank all the commenters to the previous post for their thought-provoking replies.
In the meantime, you can have a look at Wikipedia's entry on this subject. It contains several other variations in addition to the ones I mentioned.