Strangely, it depends on * WHO * is doing the eliminating and with what knowledge!

In other words, are they eliminated in such a way that the GP must be in play at the end, or in such a way the game might have been aborted prematurely?

A) If the contestant chooses, or the cases are chosen randomly (i.e. the GP was at risk at every stage), then the probability is the same for each case, at each stage, right to the end. It doesn’t matter either way if the contestant switches. This is the way Deal or No Deal is played.

B) If the game-show host, or some knowledgeable party removes cases from play (knowing they do not contain the GP), then it is better to switch. Incidentally, this version is known as “The Monty Hall Problem”, after the host of the 70’s game show Let’s Make a Deal (In which a contestant would be offered three doors, one of which conceals a fantastic prize; the contestant chooses one door, and then Monty Hall would eliminate one of the remaining doors that doesn’t have a prize; the contestant is then given a chance to switch to the last, unopened door— an opportunity which should always be taken!)

This seems paradoxical, doesn’t it? The knowledge and intention of the person removing cases from play seems to change the probabilities.

But this really does make sense.

—-

In (A) the probabilities remain equal, in effect, because no action has been taken that changes the relative likelihood of any outcome. Suppose we have, at a given stage, N equally likely possibilities, and one is removed at random, if the game does continue (which it might not) then there now (N-1) possibilities— all of which are still equally likely, etc.

In (B) the actions change the relative probabilities. This is a little harder to explain, but in a nutshell, *the host sweetens the deal:* your original choice is just as likely to hold the prize, but the other choices have become more likely to be winners, since a losing choice has been removed. Let’s count out the possibilities:

Suppose we have three briefcases are a, b, c, and the prize is in case a. We will list them in the order of

“case chosen by the contestant, case eliminated, case remaining”

The game would have ended if case a had been eliminated, so this leaves only

a b c (contestant should keep)

a c b (contestant should keep)

b c a (contestant should switch)

c b a (contestant should switch)

In (A) each of these is equally likely, since each of the choices was made completely at random. Any of the six sequences

a b c

a c b

b a (stop)

b c a

c a (stop)

c b a

was equally likely (Since there is 1/3rd chance the contestant will pick a,b or c; then there are two equally likely possible ways for one of the remaining case to be eliminated; the final case, if there is one, is determined)

Now 1/3rd of the time the game ends prematurely, but if the game finishes, there is 1/2 probability that the contestant should switch– it’s 50-50 either way.

In (B) though, the choices are not equally likely.

There is still a 1/3rd chance that the contestant chooses a, 1/3rd b, 1/3rd c.

If the contestant chooses a, it is equally likely that the host opens case b or c;

On the other hand, if the contestant chooses b, the host will certainly open c; if the contestant chooses c, the host will certainly open b.

so this gives

a b c 1/6th of the time

a c b 1/6th of the time

b c a 1/3rd

c b a 1/3rd

2/3rds of the time, the contestant is better off switching.