Coding Horror: Monty Hall, Monty Fall, Monty Crawl

Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by olavk on Monday June 22, 2009 @01:55PM (#669148)

A reframing which makes intuitive sense (to me):

There are two bucket containing a number of balls. A third of the balls contain iPhones inside, two-thirds contain only navel-fluff. The balls are mixed at random. You have to chose a single ball from one of the buckets.

Obviously the chance of any ball you pick to contain an iPhone is 1/3 regardless of which bucket you choose from - even if the buckets does not contain the same number of balls.

But when you decide on one bucket to pick from, the host takes the other bucket, and removes half of the balls. However, he doesn't remove balls at random - he primarily removes balls containing fluff.

After this you are allowed to reconsider your choice of bucket. I think it is pretty obvious now that there is a better chance of finding an iPhone if you chose from the other bucket, where the host have removed some of the fluff-balls.

The critical point in the Monty Hall problem is that the host eliminates a bad choice, but he always eliminates it from the set of doors you didn't chose. However, this is a subtle detail in how the store is told. If for example the host randomly choose between opening any door containing a goat - including potentially the one the protagonist chose if he happens to choose a door with a goat behind - then it doesn't improve the odds to choose a different door.

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Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by jcl on Monday June 22, 2009 @08:57PM (#669666)

In fact, it is immensely important that the problem states that Monty Hall always reveals a losing door. If the problem merely states a particular series of events without specifying the rules -- as both Jeff's problem statement and Marilyn's problem statement do -- you can't know the probabilities.

Suppose, for example, that Monty has the policy of only revealing a losing door if the player has chosen the winning door; if, on the other hand, the player initially chooses a losing door, then Monty reveals no door and makes no offer to switch, sticking the player with the loss. In that case, if you encounter the same sequence of events in Jeff's problem description, then switching always has zero chance of winning. With the opposite assumption -- that Monty only reveals a losing door if the player has initially chosen a losing door -- switching always has a 100% chance of winning.

Both Jeff and Marilyn nevertheless press forward with the unstated assumption that Monty has no choice in the matter. It turns out that this assumption is not only faulty, but it's also unrealistic: on the actual game show, Monty was under no such obligation.

I'm a little surprised that Jeff, who apparently has read up a little on the topic, didn't bother to grab the much better problem specification from the Wikipedia article.

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Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by olavk on Tuesday June 23, 2009 @04:33AM (#670010)

Good point. The paradox is not the math per se, but rather that the rules the host follows are unintuitive not to mention unrealistic. When these rules furthermore is not stated explicitly, it is not surprising that even seasoned mathematicians gets tricked.

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Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by cruise02 on Monday June 22, 2009 @02:10PM (#669182)

Everyone agrees that your odds get better. Most people intuitively guess that the odds go from 1/3 to 1/2. It's getting them to understand how you got to 2/3 that's the problem.

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Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by charltones on Monday June 22, 2009 @06:16PM (#669514)

I found the easiest way was this: You pick door A, the host opens door B or C and offers to let you switch to door C or B. He is effectively letting you choose between either A alone (by sticking) or B & C together (by switching) - i.e. by switching you open two doors instead of one, hence twice the probability.

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Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by cruise02 on Monday June 22, 2009 @06:25PM (#669523)

I've found that the most enlightening explanation is to increase the number of doors. What if the game is played with 100 doors, and the host opens 98 of them after you've chosen yours? At the start you only have a 1% chance of winning. Most people see right away that the host can't be increasing the odds to just 50% by opening so many doors. They don't always see the logic immediately, but at least they begin to question their intuition. Getting people to admit that they might be wrong is often the first step toward getting them to accept the correct answer. (That last sentence is a general observation that applies to a much wider range of problems.)

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