03 January 2009

The Problem with the Monty Hall Problem

If you're not familiar, the Monty Hall problem is a frequently discussed (in probability) and often confusing problem (for almost everyone).

Jeff Atwood recently posed a question to his readers that involves similar reasoning. The poor phrasing of Jeff's question and the response in the comments got me thinking about old Monty again.

I think the problem is not just the wording of the boy/girl part of Jeff's question, but when you get your information. The 2/3 answer assumes that you are making a prediction. This is the same problem with the Monty Hall problem.

Exploring the Problem

Say I was going to flip a coin twice. If I flipped it once and it was heads, the second flip is not more likely a tail since the events are independent.

However, if I said I was going to flip a coin twice and I want you to bet on the result before I flip any coins. If a coin is heads, you bet what the other coin will be, and if one is tails you bet what the other will be. You should be the opposite (for heads bet tails).

If a coin is going to be heads, it's more likely that the other coin is tails. (Possibilities: HT TH HH)

Or, to make this more like Jeff's original problem, you can only bet assuming one coin is heads. If both are tails, then we throw out that round (and this action is part of what makes the result not 50%).

This explanation also demonstrates why programming a simulator and examining the results is misleading. The simulator is working on predictive probability. If you wanted your simulator to work without predictive probability, then you'd have to program it at the point the decision is made. You should assume the initial condition (that you chose the left door and the middle door was shown to have a goat). Then decide whether to switch or stay.

I'll leave the programming as an exercise for the reader : )

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