Thursday 26 January, 2006

Excel solution: the Monty Hall problem

Filed under: All things Excel

It seems that as was the case when Marilyn vos Savant published the correct answer in the New York Times back in 1990, my entry is causing some (seemingly) intelligent people to blunder. (I have to be somewhat polite here, as one of the blunderers is my hosting provider.)

The rationale in the post below is pretty straightforward, as highlighted by the friend who originally sought my help:

"Perfect ... even I understand that!"

Nonetheless, it seems some people need some further evidence. What better way to help than via the powers of Excel. Here's a spreadsheet that takes you through the logic behind both the no switching option and the switching option. Each row in each of the sheets represents a unique experiment, with each element that is up to chance being driven by an independent random number.

In the no switching option, the only things that are random are the prize-winning door and the door originally chosen by the contestant. In the switching option, both of these elements are still random, as is the door that Monty opens, although this is sometimes forced.

The spreadsheet contains 50 rows of experiments, although you can copy a row down as far as your computer's memory will allow to see where your odds settle. Using all 65,536 rows (65,532 experiments of each type), I recorded a 33.15% win rate for games in which I stuck, and a 66.72% rate for games in which I switched.


Posted by dan at 12:07pm | Permalink | Comments (3) | Trackbacks (0)
Comments

Not wishing to upset your hoster but:

Odds the car is behind your initial choice of door is 1/3.
Therefore the odds that the car is behind one of the other two doors is 2/3. The fact that you are shown that one of these other doors is the wrong one doesn't change those odds. Therefore the odds that the car is behind the door you didn't choose[i.e. the one you are swapping to] is still 2/3.

QED

Posted by AJ, 9:35pm, Friday 27 January 2006

I agree... my spreadsheat had similar results. But I am trying to think outside the box and not assume switching or not switching is already decided upon. Only once that percentage is found out (the percentage of people that think they should switch) will we know the real percentages. But I agree that the optimul choice is to switch.
cheers

Posted by josh finnie 9:17am, Saturday 28 January 2006

I'm not sure I understand. The percentage of people that would switch vs. stick under the circumstances depends on the audience. (There is a third choice that some people will take: that to reselect at random, which in the end will result in a stick or switch.)

If you allow for this, then the percentage of people who end up winning the game will always be somewhere between 33.3% and 66.7%. It will tend towards the former if there are a lot of people who don't know the right choice playing, and the latter if there's lots of savvy mathematicians...

Posted by Dan 10:00am, Saturday 28 January 2006
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