Figure 7-4 and explanation of the Monty Hall problem contain a logical flaw. Figure 7-4 includes both goats as possibilities, even though Monty, by revealing the goat behind door number 2, has already shown you that you did NOT pick one of the two goats!

It is no wonder "hundreds of math professors" wrote in. It is so simple, it's hard for me to conceive of how you could get it wrong!

In case you didn't get it, here is a more detailed explanation:

When the author states that "The most common (and thoroughly incorrect) answer is that it makes no difference, as each remaining door has a 50% chance of leading to the car.", he is tossing out the correct answer!

Here's why:

(C is the car, G are goats, "-" are revealed goats or invalid scenarios (based on the scenario having a car where Monty revealed a goat))

Door: 123
Scenario X GGC
Scenario Y GCG
Scenario Z CGG

You pick door number 1.
Monty reveals door number 2.

That takes scenario Y out of play, since the second door has a goat (not a car, as happens in scenario Y).
It also takes column B out of the equation, since (if you are logical and sane) you would not pick a door already revealed to be a loser.

So, here is your matrix after Monty reveals a goat behind door number 2:

Door: 123
Scenario X G-C
Scenario Y ---
Scenario Z C-G

Now, your chances are 1/2 and it doesn't matter which of the two doors you pick (unless you bribed Carol Merrill and know the correct door).

NOTE: If you have Monty reveal door a goat behind door number 3, instead, it takes out scenario X, not scenario Y.
This time, you would delete Door 3 from each scenario, instead of Door 2, for the same reason (it is a known goat).

Door: 123
Scenario X ---
Scenario Y GC-
Scenario Z CG-

For brevity's sake, I won't go through the remaining scenarios, since they all end up the same way.

It is easy to see that number 3 below figure 7-4, which states: "Switching wins two out of three times." is false, since according to the problem, Monty has already revealed one of the goats and it is, therefore, not the same problem anymore.

It is true that when you first picked the door, your chance was 1 in 3.

But when one of the losers is revealed, you have two doors left: a winner and a loser. The author's scenario does not acknowledge that revealing a goat changes the probability.

Take another example. You have a normal deck of 52 playing cards. The four aces are the only winners. You pick a card. "Monty" reveals 47 cards, none of which are aces, and your card is still face down. You now have a 4 out of 5 chance of winning, even though you started with a 4 in 52 (1/13) chance, because "Monty" has taken away the other losers. Of course, you still don't know which of the five cards is not an ace, so it won't change your odds to switch from the card you initially chose.

no chart provided on "missing CD" cite