# Monty Hall of Fame

Every month I receive dozens of letters about Fuzzy Math, many of them claiming I have made a mathematical error. I appreciate the interest in the articles, and I must say I’m pleasantly surprised to learn that so many folks feel strongly about math! In this blog I will attempt to address some of the comments, questions, and flat-out disses I have received.

For starters, here is a typical response about a column on the famous Monty Hall scenario:

The reasoning behind “Fuzzy Math: The Monty Hall Scenario” (July 2006) sounds plausible enough, but is in fact fallacious. When a door is picked the odds are 1 in 3 that it’s the one with the prize behind it, the same as the odds for each of the other doors. After Monty eliminates one of them, the odds become fifty-fifty not just for the door left that you didn’t choose, but also for the door you chose. This can be clearly borne out by running a simple computer simulation a few thousand times. Sticking with your original choice is correct 50% of the time, and switching doesn’t increase your success rate one iota. –M. S.

The Monty Hall scenario is a classic problem in probability, and the error that M.S. makes is very common. One might say that it is the “intuitive” answer, and many smart people have mistakenly thought it to be correct. The key point is that switching is tantamount to betting that you initially chose an empty door, the odds of which are always 2/3, even after one of the (empty) doors is eliminated.

One helpful way of looking at it is to use a “decision-tree” analysis. Let’s label the doors with the numbers 1, 2, and 3; with the prize behind door #2. We now analyze what happens after each of the three possible initial choices. If you initially pick door #1 (empty), then door #3 (empty) is eliminated, and switching lands you on door #2 (you win!). If you initially pick door #2 (prize), door #1 or #3 is eliminated (both empty), and either way switching lands you on an empty door (you lose!). Lastly, if you pick door #3 (empty), then door #1 (empty) is eliminated and switching lands you on door #2 (you win!). Conclusion: switching wins in TWO OUT OF THREE trials.

Another way of thinking about this problem is to consider limiting cases. (This is a useful technique is a great many math and physics problems!) What if instead of 3 doors there were 1000 doors (it’s a big room, ok?) and the prize is hidden behind just one of them. You make your initial selection, the host then eliminates 998 empty doors and offers you the chance to keep your initial choice or swap for the one remaining door. Are the odds fifty-fifty that you picked the prize out of 1000 possible locations on your first try? Or is it more likely that you were wrong initially and the host, constrained to open only empty doors, revealed the most probable location of the prize by process of elimination? Think about it.

Of course you don’t have to take my word for it. One can do an experiment to find out for sure. It is a simple matter to write a computer program to execute millions of trials and tally the results. Such programs can be found online; here is one with the corresponding results for 100,000 trials. As you can see, in 100,000 games, swapping won 66,676 times. Here is another, more interactive version. Finally, for those who are sill unconvinced, I suggest playing the game on this website.

It’s a cool, deceptive little problem. Thanks to all those who wrote in, and I look forward to your future comments.