Warning: Off-topic. This is an update of this post on "Deal or No Deal." Math and other non-zoning-related stuff below the jump.

Here's the setup: Suppose the DOND contestant opens 22 of the 24 briefcases without hitting $1 million. There are two briefcases left -- the one she chose at the start of the game and the one left after the others have been opened one by one. The contestant can either open the case she originally chose (and keep whatever's there), or she can switch to the "leftover" case.

I claim that the contestant does not improve her odds by switching. (The opposing view is that there is a 23/24 chance that the $1 million is in the other briefcase.)

There is a succinct proof using Bayes' theorem that I'm right.

Let A and X be events in a sample space. Let P(A) be the probability of A, P(A l X) be the probability of A given the occurrence of X (i.e., the conditional probability), and P(~A) the probability of A *not *occurring. Then Bayes' theorem states:

(There is a good discussion of Bayes' theorem here.)

Let's apply Bayes' theorem to "Deal or No Deal." At the beginning of the game, the contestant chooses one briefcase as "her" briefcase. Call the one she chooses A. ~A is the set of the remaining 23 briefcases. There is a 1/24 chance that A contains $1 million.

The game begins and the contestant opens one briefcase at a time. Call the first briefcase she opens "X." Let's calculate the probability that A contains $1 million after she's opened X (assuming, of course, X does not contain the $1 million).

We want to know the probability that A contains $1 million given that X does not. I should write P(A contains $1 million l X does not contain $1 million) but that's cumbersome, so I'll abuse notation slightly and write P(A l ~X).

Then, by Bayes' theorem:

P(A l ~X) = P(~X l A)*P(A)/[P(~X l A)*P(A) + P(~X l ~A)*P(~A)].

(Note: I substituted ~X for X throughout the formula, but that notational change doesn't make any difference.)

Calculating individual terms:

P(A) = 1/24.

P(~X l A) = 1 (since it is impossible to "hit" the $1 million if it was in the briefcase the contestant initially chose).

P(~A) = 23/24.

P(~X l ~A) = 22/23 (because, if the $1 million is in one of the 23 briefcases the contestant did not initially choose, there is a 22/23 chance it won't be hit on the first try).

Then P(A l ~X) = 1*(1/24)/[1*(1/24) + (22/23)*(23/24)] = 1/23.

If the "opposing" view were right, P(A l ~X) should equal P(A) = 1/24; opening X shouldn't improve the odds that the contestant initially chose the briefcase with $1 million. In any event, simply update the probabilities and play again with the remaining 23 briefcases. Again, let "X" be the first briefcase opened (and, as before, assume it does not contain $1 million). After this round, P(A l ~X) =1/22. Continue. By induction, when just two briefcases remain, the probability that the briefcase the contestant initially chose contains $1 million is 1/2.

The contestant does not improve her odds by switching.

Incidentally, the same approach shows why the contestant on "Let's Make a Deal" *does *improve her odds by switching.

The contestant faces three doors, and chooses the one (door A) she hopes is hiding the prize. Monty Hall opens one of the two other doors -- invariably revealing a "goat" (i.e., bad prize) -- and then asks the contestant whether she wants to switch.

The contestant should always switch, because there is a 2/3 chance the prize is behind the door Monty did *not *open.

Proof: Again, we want to know P(A l ~X). The numerator in Bayes' theorem -- P(~X l A)*P(A) -- is equal to 1/3.

The denominator is 1/3 + P(~X l ~A)*P(~A). P(~A) = 2/3. But here's the trick: P(~X l ~A) = 1. That is, *even if prize is behind one of the two doors the contestant did not choose, Monty Hall will always open the door without the prize. *The probability that the prize won't be behind X is 1.

Do the arithmetic and you get P(A l ~X) =1/3, which is exactly what the contestant started with. This means there is a 2/3 chance the prize is behind the other door.

The Monty Hall "paradox" isn't really a paradox. People intuitively update conditional probabilities when they get more information (even if they don't do it right). The Monty Hall format tricks us into updating the conditional probabilities despite the absence of new information.

This trick doesn't apply to "Deal or No Deal." When the first briefcase is opened it is opened *at random*; this gives the contestant more information about all of the remaining briefcases, including the one she chose before the game began.