A Riddle. Let the Procrastination begin.

Paradox Play

It’s amazing what I’ll do instead of homework. I was researching paradoxes for English, and I came across this deligthful riddle, a-la an old time game show.  So in keeping with the random nature of my column, here it is. Originally part of Joseph Bertrand’s book Calcul des probabilities (its cool because it’s French), published in 1889, and most commonly known because of its part on the TV show “Let’s make a deal,” it’s the Bertrand Box problem! Because the host of “Let’s make a deal” was named Monty Hall, it’s also called the Monty Hall problem. We will use it in the version seen on the show.

The statement of this famous problem in Parade Magazine is as follows:

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, donkey. You pick a door, say No.1, and the host, who knows what’s behind the doors, opens another door, say No.3, which has a donkey. He then says to you, “Do you want to pick door No.2?” Is it to your advantage to switch your choice? (Whitaker 1990)

Answer in the comments! (Googling is cheating.)

`Spoiler Alert! Click here only when you ready for answers: [spoiler]

Many people argue that “the two unopened doors are the same so they each will contain the car with probability 1/2, and hence there is no point in switching.” As we will now show, this naive reasoning is incorrect. To compute the answer, we will suppose that the host always chooses to show you a donkey (When the problem and the solution appeared in Parade, approximately 10,000 readers, including nearly 1,000 with Ph.D.s, wrote to the magazine claiming the published solution was wrong. Some of the controversy was because the Parade version of the problem is technically ambiguous since it leaves certain aspects of the host’s behavior unstated, for example whether the host must open a door and must make the offer to switch).

Assuming that you have picked door No.1, there are 3 cases:

A Math Look

 

 

Notice that although it took a number of steps to compute this answer, it is “obvious”. When we picked one of the three doors initially we had probability 1/3 of picking the car, and since the host can always open a door with a donkey the new information does not change our chance of winning.

The Picture Look

Basically, switching is to your advantage, resulting in you winning the car 2/3^rds of the time.

http://www.math.cornell.edu/~mec/2008-2009/TianyiZheng/Conditional.html   [/spoiler]`

Thanks to Parade Magazine and Cornell University for the Summary and Answers.