Answer to "Ask Marilyn"

* If you haven't read the post from 7-26-2005 and you'd like to try a little puzzle, take a look at the "Ask Marilyn" post before reading on.

So this puzzle was evidently designed to lead the intuition away from the cold, hard facts of statistics. After she explained what I'm about to explain, I'm told Marilyn had angry college math teachers and statisticians writing in telling her it was time to admit she was wrong. However, I stand by Marilyn.

To properly analyze this puzzle, we must first realize it is all about probability and statistics. It is just a little more complicated than asking, "If I flip a coin, what's the chance of getting heads?" but to understand the puzzle we have to remember the difference between an independent event and a dependent one.

Two independent events are, in fact, independent. I might ask, "If I flip a coin and roll a die, what's the chance of getting a heads and a six?" The chance is 1/12, which is equal to 1/2 chance in flipping heads, and 1/6 chance in rolling six. These probablities are multiplied as they are independent events.

So going back to our goat & Porsche problem, the chance that the Porsche is behind any particular door is 1/3.

~ Door A~Door B~Door C

1. Porsche ~ goat ~ goat

2. goat ~ Porsche ~ goat

3. goat ~ goat ~ Porsche

Each of the three rows are equally likely, given what the game show host has told us. At random, the contestant picks a door. This is another independent event which, when multiplied by the first three events, yields nine equally likely combinations. To illustrate this, I've taken the three rows above which represent the three distributions possible of the Porsche, and expanded each row times three to show the chance of choosing any particular door. The selected door is bolded.

~ Door A~Door B~Door C
1. Porsche ~ goat ~ goat

1. Porsche ~ goat ~ goat

1. Porsche ~ goat ~ goat

2. goat ~ Porsche ~ goat

2. goat ~ Porsche ~ goat

2. goat ~ Porsche ~ goat

3. goat ~ goat ~ Porsche

3. goat ~ goat ~ Porsche

3. goat ~ goat ~ Porsche

All should agree that each of the nine rows above is equally likely. If you were to count up the chance right now of picking Porsche, you'd see it's 3/9 which equals 1/3.

Now here's where it gets tricky. For simplicity's sake, let's assume that door A is chosen. What I illustrate next is door A selected. This is the first row of each group of three from the nine equal possibilities above.

~ Door A~Door B~Door C
1. Porsche ~ goat ~ goat
2. goat ~ Porsche ~ goat
3. goat ~ goat ~ Porsche

Notice, you still have a 1/3 chance of having selected the Porsche. What the game show host does next is NOT an independent event. He opens a door to reveal a goat, but he does so knowing which door you have picked (he doesn't open that one) and he knows where the Porsche is, so he doesn't open that door either. So in row 1 above, the game show host might open either door B or door C. In row 2, he will open door C. In row 3, he will open door B. Remember that row 1 is as equally likely as either 2 or 3, but in row 1, the host has a choice of doors to open. In my next illustration, we know there's a goat, so I'll illustrate only the closed doors.

1. Porsche ~ ~ ~ ~ ~ goat (1/6)

or Porsche ~ goat ~ ~ ~ (1/6)

2. goat ~~ Porsche ~ ~ ~ ~ (1/3)

3. goat ~ ~ ~ ~ ~ ~ Porsche (1/3)

The way I have it illustrated almost makes it look like a 50/50 chance. Two doors closed, one Porsche, one goat. But remember each numbered line is equally likely. Row one has two equally likely possibilities (the host had a choice of opening door B or door C). We can clearly see that as things stand, there is a 2/3 chance that the goat is behind door A. Therefore, it would be to the contestant's advantage to change his selection to the closed door. There's a 2/3 chance the Porsche will be there.