Whether or not that's how it works in any ordinary game Deal or No Deal is irrelevant.
The question he asked is one scenario of Deal or No Deal where all but four cases have been revealed to not have the million. In this scenario, it fits the Monty Hall problem. Yes, you are correct that Deal or No Deal itself does not guarantee you will keep the million in play, but in this one instance he creates in his question, it did play out to have the million left in a remaining suitcase. That is what matters here. In this one single scenario he created in his question, the Monty Hall problem exists, only because he has already guaranteed (to use your own word) 22 cases to be without the million.

All I can say is that there is a flaw in your logic here. The odds just don't work that way. You seem to have decided that a situation looks like the Monty Hall situation, therefore it must be the same, and that just isn't correct.

I mishandled the html in a previous response, let me ask this again-

Let's say the contestant chose case #1, and has already opened cases 5 through 27.

Please tell me why you believe that case #2, #3, and #4 have better odds of containing the million than case #1.

Now let's suppose that the contestant chose initially chose case #4, and left #1-3 in play. Would case #4 magically be an inferior choice now, while #1 somehow has its chances improved?

The reality is that all four cases are the ones that the contestant has chosen to leave in play at this point in the game, and all four have equal chances to hold the million. The fact that one of them has been sitting in front of the contestant (instead of left on the stage) is irrelevant to the odds.

In Monty Haul: Use the red twos and the Ace of Spades, deal one card to the player. Look at the remaining two cards and discard a red two. Then record if the player would win by switching. They will win by switching anytime they were dealt a red 2 (2/3rds of the time).

In Millionaire (lets go with 27 total cases), use all the red cards and the Ace of Spades. deal one card to the player. Now throw out all but the bottom 3 cards in the remaining deck. Is one of them the Ace of Spades? If not then the results of this experiment are discarded. This is one of the 23 of 27 scenarios where the Ace of Spades is already gone by this point. In the scenarios we do record about 1/4th of them will result in the player winning with his original choice. If the Ace was the top card in the original deck, he wins. If it was one of the bottom 3 cards it is a scenario we will count. If the Ace was anywhere else, we're not looking at this scenario. The Ace is on top in 1 of the 4 scenarios that matter to us, and we will win with our original choice in 1/4th of those scenarios.

First, before we get to conditional probabilities, here are the initial straight probabilities:

In Monty Haul - the random variable at the start has three possible outcomes - car, goat 1, and goat 2.

In Millionaire - the random variable at the start has 28 possibilities - the million, other#1, other#2, other#3 and then other#4 etc...

So there is a tree with 3 branches in the first case, and a tree with 28 branches in the second case.

Now for the conditionals - In Monte Haul we are told (GIVEN) that we are on a branch where Monte will reveal a goat. Well, he reveals a goat ON ALL THREE BRANCHES. Saying that we will see a goat, eliminates none of the possible future worlds. Had we chosen differently, it would not have been otherwise, we still would have seen a goat.

But...in the other game saying that we get down to 4 cases tells us something different. It tells us that we are not in a world where we picked other cases #4-#28. Those branches are eliminated, we know we don’t go down those paths (or didn't go down those paths), so now we are looking at a tree with only 4 branches that we have not eliminated, and the chance that we are on the winning branch is 1 in 4.

So, at the start there are 28 possible worlds. 24 of those possible worlds end up with the million being revealed before we get to the final 4. 3 of those worlds have the million in some remaining case, other than ours at the final 4, and 1 world has us winning. At the start 1 of the 28 possible worlds has us winning. By 4 cases, however, we know we are not in the other 24 worlds, and one of the 4 possible worlds is a winner for us.

Again, in Monty Haul, there were three possible worlds at the start. After he reveals the goat, we still could be in any of the three possible worlds. Our chance of winning with our original choice is still 1 in 3.

Thank you for demonstrating your credibility by stating you are a statistician, that makes this discussion easier for both of us.
Take note that the original post stated: "after the 22 other cases have been opened and none of them have the $1m in them"
This correlates to the idea you stated: "ALWAYS reveals a goat," because all 22 cases had the proverbial goat in them. Thus, Monty Hall's problem remains, just with 27 doors and the fact that you revealed the goats, not the host.
And with you being a statistician, you will already know that the Monty Hall analysis is correct that switching being the proper choice.

In DoND, you do not always open 22 cases and reveal 22 losers.

If that was guaranteed to happen (by the host interfering and removing losing cases for you), then the Monty Hall situation applies and you would be increasing your odds by switching cases.

But it's not.

Interestingly, the Monty Hall problem scenario DOES apply here, because we are under the assumption that upon reaching the four cases left, one of those remaining four is the million. Whether or not the "game master" was the one who removed them is irrelevant, all that matters is that the condition above exists..

The Monty Hall scenario does not apply here. It is extremely relevant as to how this situation was reached.

In the Monty Hall scenario, the gameshow host always interferes with the game's odds, and always does so by removing a losing prize. The contestant will always find himself in a situation where the winning prize is still in play.

In DoND it's completely different. There is never any interference from outside. The contestant will very rarely find himself in a late-game situation where the million dollars is still in play - and if he does, it's because he's been very lucky already.

In Monty Hall, the winning prize is ALWAYS there, and the contestant will ALWAYS be in a situation where he can win it.

In DoND, the winning prize will often be eliminated early, and it's rare that the contestant will be in a situation where he can win it.

They're entirely different scenarios that you cannot use the same math for. One is a single-decision, rigged-gameplay show... the other is a step-by-step show with many decisions and a lot of luck involved. They aren't the same at all.

I do agree that the instantaneous probability does change, clearly, because at that instant in time there is a 25% chance that the case has a million.
I do not agree that this instantaneous probability takes precedence or re-evaluates the original probability, which is that picking the first case was a 97% fail rate. Regardless of how many cases are left, that 97% remains, under the presumption that the game is still in continuation (e.g. if the million is revealed prior to the four cases being left, the game of finding the million is already over). This is true because a moment in time existed when the case had a 97% fail rate.

Let's say the contestant chose case #1, and has already opened cases 5 through 27.

Please tell me why you believe that case #2, #3, and #4 have better odds of containing the million than case #1.

Now let's suppose that the contestant chose case #4, and left #1-3 in play. Would case #4 magically be an inferior choice now, while #1 somehow has its chances improved?

The reality is that all four cases are the ones that the contestant has chosen to leave in play at this point in the game, and all four have equal chances to hold the million. The fact that one of them has been sitting in front of the contestant is irrelevant to the odds.

Well, first I'll reassert my above argument. Also I suppose I should add my profession is a statistician, and I actually went to school and studied things like this.

But now I'll add the description of the Monty Haul game from the classical stat point of view.

In that game we start with a 1/3 probability of having picked the car. 2/3 probability of picking the goat. That probability is fixed and can not change, just as in the millionaire game. However, the conditional probability in both game can evolve. IN MH we are told that Monty ALWAYS reveals a goat. No matter what you picked in the first place he will always show you a goat. So what is the conditional probability that you picked the car, given that Monty showed you a goat? Still 1/3rd. Because 100 percent of the scenarios are still possible. (1/3) / 1.00 = 1/3. We should switch to the remaining door.

Let's return to the millionaire game. Let's suppose that we pick case #1 and start revealing 28,27, etc, until we get to 4 cases left, 1,2,3 and 4. Now let's think about our probability. GIVEN that cases 5-28 are no good, what is the conditional probability of picking correctly? 1 in 4. (1/28) / (4/28) = (1/4). That is: There is a 1/28 initial probability of picking correctly. But that many cases will be opened successfully in only 4 of 28 scenarios, and that is what we are conditioning on.

But now I'll add the description of the Monty Haul game from the classical stat point of view.

In that game we start with a 1/3 probability of having picked the car. 2/3 probability of picking the goat. That probability is fixed and can not change, just as in the millionaire game. However, the conditional probability is both game can evolve. IN MH we are told that Monty ALWAYS reveals a goat. No matter what you picked in the first place he will always show you a goat. So what is the conditional probability that you picked he car, given that Monty showed you a goat? Still 1/3rd. Because 100 percent of the scenarios are still possible. (1/3) / 1.00 = 1/3. We should switch to the remaining door.

Let's return to the millionaire game. Let's suppose that we pick case #1 and start revealing 28,27, etc, until we get to 4 cases left, 1,2,3 and 4. Now let's think about our probability. GIVEN that cases 5-28 are no good, what is the conditional probability of picking correctly? 1 in 4. (1/28) / (4/28) = (1/4). That is: There is a 1/28 initial probability on picking correctly. But that many cases will be opened successfully in only 4 of 28 scenarios, and that is what we are conditioning on.

Returning to DA