The orthodox solution to the Monty Hall riddle is incorrect.

For the purposes of this debate, the Monty Hall riddle is the following:

Suppose you are walking down the street and someone calls your attention to a set of three closed doors. This person--we'll call them The Host hereafter--offers you a large sum of money if you can guess which door the money is hidden behind. Two doors are empty, and one contains the prize.

You choose a door.

Before the Host opens the door to show you whether you won or not, they instead reveal what is behind one of the doors you did NOT choose. Behind that door is nothing of value. There are now only two closed doors, one of which is the door you originally chose.

The Host now offers you the chance to switch doors.

Should you agree to switch or should you stick with your original choice?

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The orthodox "answer" to this riddle is to always switch.

I believe this answer is incorrect.

If you disagree, accept this debate challenge.

Please keep in mind that I have defined the Monty Hall riddle above for the purpose of this debate. Read it carefully before you accept this challenge. Some tellings of the riddle vary slightly on some of the details.

First, I would like to thank my opponent for starting this debate and extending the challenge to anyone. With that said, let us proceed.

Let us assume that you ignore what the host does and choose the door you originally believe to be the correct door. When you chose, the 3 doors were equally likely, so you win the car with probability 1/3. Now lets back up for a moment and assume that you didn't ignore the host, and decided to switch; you choose a door, wait for the host to expose the door with nothing behind it and then switch to the other remaining door. You win if and only your initial door had nothing of value behind it (think it through). How likely is it that your initial choice had a nothing of value behind it? Two thirds, of course. Keep in mind that in the beginning, the probability of getting the right door was 1/3. This probability would stick with the door the candidate chose, so when one of the other doors is removed from the game, the remaining probability is two thirds in order to have all probabilities sum up to 1

If you don't buy that, lets expand this hypothetical situation a bit. Suppose there are a hundred doors instead of just three. Only one has something of value behind it. You get to pick a door, then the host opens 98 loser doors, then he allows you to switch to the other remaining closed door. Would you? Of course, it's almost sure that the prize is behind the other door, and very unlikely that it is behind your original choice.

Still not convinced? Here's another try. Suppose the host had three boxes . . . one of them contains a hundred dollar bill, the other two contain nothing. You have to pick, but before you do, the host agrees to help you out: You can point to two boxes, and he will gladly combine their contents into a new box. Now, after that's done, would you go for the new box, or for the old untouched one? Of course for the new one, it is twice as likely to have the bill. Now to make it more like the original problem, suppose that the host doesn't combine the contents into a new box, but instead throws away an empty box from among the two you pointed out to him. That's effectively the same thing, and has the same result: the other box is twice as likely to have the bill as the untouched one.

Still not convinced? Lets get some empirical evidence then:

http://www.remote.org...

http://www.grand-illusions.com...

In both of these simulated programs, you'll notice that the probability for victory will remain at 66-67% most of the time

Here is a mathematical proof that confirms my claims:

http://www.remote.org....

Conclusion: Changing your door is the best solution.

First, let me say that my opponent did a fine job explaining the orthodox solution to the problem. It is a solution that I am well aware of and which is correct GIVEN A CERTAIN ASSUMPTION, which I explain below.

From the definition of the problem--and this is the way the problem is often, though not always, defined--it still can not be determined whether or not switching doors gives you a better chance to win the prize.

Here's why:

In the problem, we are not told that the Host will ALWAYS offer the player a chance to switch doors. The player does not know how many times this game has been played before with someone else as the player. For all the player knows, this is the first time this particular host is offering this game to any player. So, we (the player) do not know what strategy the Host is using.

There are several different possible strategies for the host. A few of them are listed below but there are more.

1) Always open one empty door and offer the player a chance to switch doors.
2) Always open one empty door but offer the player a chance to switch ONLY if the player chose the correct door to begin with.
3) If the player chose an empty door to begin with, simply reveal the contents of the door. Otherwise, open one of the empty doors and offer a chance to switch doors.
4) Always open one empty door but offer the player a chance to switch ONLY if the player chose an empty door to begin with.

Each of these strategies by the host implies a different optimal strategy for the player.

My opponent, as most other people do, assumed that the host was playing strategy #1 above. As I have stated, my opponent is correct that IF the host is playing strategy #1, then the player should ALWAYS switch, as that gives the player a 2/3 chance to win (not 1/2 as many people assume).

However, if the host is playing strategy #3, then things are reversed. The host will only offer the chance to switch if you have chosen correctly originally. Therefore, switching now gives you a ZERO percent chance of winning and sticking with your original choice gives you a 100% chance.

The host can also use a complex mixture of strategies 2 and 3 where, for example:

If the player chose the wrong door to begin with, offer a chance to switch 25% of the time, and 75% of the time just show them they chose a losing door.

If the player chose the correct door to begin with, offer a chance to switch 100% of the time.

Depending on the host's strategy, the host can manipulate your success ratios any way he wants to (but never make it less than 1/3 without cheating).

The point is that without knowing the host's strategy, there is no way to know what strategy the player ought to play (always switch, or never switch).

Furthermore, we need to look at the motives of the host. Does he want you to win? If he is a host of a game show and is trying to increase ratings, then it is probably best if they allow the player to win every once in awhile. But if he is just some sort of street con-artist type, then he probably doesn't want you to win. This will also influence the type of strategy the host takes. Since the host's motive is often not made explicit in the riddle, it becomes even more difficult to determine the player's best strategy.

In summary, I agree that IF the player knows that the host will ALWAYS open one empty door and offer a chance to switch, then my opponent's solution (the orthodox solution) is absolutely correct. However, from the way I defined the problem, there is no way to know which strategy the host is playing and therefore there is no way to know which strategy the player ought to play to get the best result.

Incidentally, there was another parameter to the problem which I meant to state but forgot, but which is also quite important. I didn't state whether or not the host knew ahead of time which door contained the prize. This changes the probabilities as well. However, since I meant to make that parameter explicit and since my opponent assumed that parameter to be true (otherwise his answer is even more incorrect), lets just assume that it was part of the original problem.

First, my opponent insinuates that there is no flaw in the answer I am supporting for this problem unless we take into account the possibility that the Host is using a particular strategy used to undermine the contestant's confidence. However, the problem with this line of thought is easy to spot. In the riddle (and like all riddles), the answer is based off of the information we are aware of. The host' use of multiple strategies is not part of the Monty Hall riddle my opponent typed in round 1.

For instance, when playing poker, the initial phase in the game requires that you receive a hand of cards. If you're given the first card, you'll note that there is a 1/52 chance of you getting an Ace of Hearts. Then again, if the card shuffler has fixed the deck to get you a hand purely made up red and black digits, there is a 0/0 chance that you'd end up with the Ace of Hearts. That said, can such an assumption be warranted? Possibly, but without prior knowledge of the card dealer? Not really.

The same applies to this riddle as we're given no information that can allow us to take into account these other variables which my opponent brings up. All we know is that the host opens one door and tells the contestant that he/she has the option of switching. All that stuff about being a con artist (in which case, the best idea would be have all three of the doors contain nothing of value) or trying to increase the ratings cannot be applied without having been insinuated in the riddle. There is nothing in the riddle to suggest that the host would go about any method other than to always open one empty door and offer the player a chance to switch doors. Using my opponent's approach, one can conclude that most (if not all) riddles are unsolvable

With that said, you can dismiss strategies 2-4 (since they aren't implied by the riddle). With those dismissed, the con has already won this case as my opponent has conceded to there being no fault in the reasoning used to conclude that switching is the best option.

On top of that, my opponent does not address my empirical evidence used to conclude that switching is the best option.

IMPORTANT NOTE FOR THE VOETERS: I realize that many here may be tempted to discount my round 2 response (and vote against me) since it was not as LONG as my opponent's. I suggest that if you don't wish to read the arguments, you shouldn't vote on them.

With that said, I await my opponent's final rebuttal.

"In the riddle (and like all riddles), the answer is based off of the information we are aware of. The host' use of multiple strategies is not part of the Monty Hall riddle my opponent typed in round 1."

The above is the crux of my opponent's objection to my previous argument.

In short, my opponent is saying that my argument relies on an assumption that is not implied in the original riddle.

However, it is in fact my OPPONENT who has assumed something not originally in the riddle.

My opponent assumed that the Host would ALWAYS open an empty door and offer a chance to switch. The original riddle neither states or implies this to be the case. All it says is that you were given a choice between 3 doors. You chose one. The host opened an empty unchosen door and gave you a chance to switch.

Why would you believe the host was OBLIGATED to give you this choice? It is not stated in the riddle. Yes, as I have said, SOMETIMES it is stated in the riddle when others tell it, but often it goes unsaid as it did here.

Again, nowhere in the riddle does it state or imply that the host was OBLIGATED to give you a choice.

And if the host is NOT obligated, then you can not know what strategy the host is using. And, if you do not know the host's strategy then, as I showed previously, it is impossible to know what the player's best strategy is (to switch doors or not to switch doors).

I have already agreed that my opponent's analysis of what happens when the host IS obligated to offer a switch is spot-on accurate, so it is not true that I failed to address my opponent's analysis. My point is that his analysis is incomplete, since it only takes into account ONE of the many possible strategies that the Host may be using. And since we have no way of telling from the original riddle which strategy is being used, we therefore have no way of knowing whether we should apply my opponent's analysis or not.

To conclude, my opponent has accused me of assuming something which was not stated or implied in the original riddle. However, it is in fact my OPPONENT who has done that. My opponent incorrectly assumed that the Host was OBLIGATED to show an empty door and offer the player a chance to switch. This is clearly NOT stated nor implied in the original riddle. Therefore, my opponent has failed to successfully argue against my contention that we have no way of knowing what the player ought to do without at least knowing what the Host's motives are (do they want us to win or lose), or knowing what strategy the host is using.

My opponent states that I tell you that his argument relies on an assumption that is not implied in the original riddle. This is absolutely correct, as he is making the assumption that the host may have some kind of deceptive agenda. Furthermore, he is the one to insinuate that the host having an agenda here is a definite possibility, This is a good example of argument ad ignorantiam as my opponent is essentially demanding that I prove that the host doesn't have some kind of deceptive agenda. Keep in mind that it is the job of my opponent to prove this possibility since he was the one to bring it up.

However, I've already addressed this in round 2 with my poker game analogy (which my opponent has apparently dropped in round 3). To reiterate, in calculating the possibility of receiving an ace of hearts from the card dealer, you do not conclude that it is impossible to determine the answer since the card dealer could have rigged the deck to purposely give you a hand with no aces. Problem solvers do not consider this possibility unless there is evidence to support it. In this problem, there is nothing stated about the host' attitude/behavior that could lead us to accept strategies numbers two, three, and four.

What the player knows is that the host gives him the option of changing doors after he has revealed a door with nothing of value behind it. Thus, in the eyes of the player, the percentages are determined as I mentioned in my analysis. Thus, it would be of the player's best interest to switch.
As for obligation, again, it is the job of my opponent to show why this is a variable in this debate. He has not done this, thus being the reason as to why dismissing this argument is well warranted.

Of course, to pacify my opponent, there is actually strong evidence against the host needing to be deceptive as well as the host not being obligated to switch. Going back to how my opponent attempted to take this from a realistic approach in round 2, it would really make little sense for the host (whom the contestant has randomly met on the street corner) to rely on deception considering that he wouldn't gain anything out of it. After all, in my opponent's wording of the Monty Hall riddle, the contestant doesn't exchange anything of value with the host nor does the host have any ratings to gain since this doesn't occur on television. Thus, deception would be utterly pointless since there would be nothing for the Host to gain. Therefore, going with the idea that the host is obligated to switch is quite logical. Since my opponent has already conceded to my analysis being flawless, switching is hence the best solution even according to his wording of the riddle. Therefore, the pro wins today's debate.

Voters: My opponent has failed to show you the evidence needed to take his extra variables into account, dropped my poker game analogy mentioned in round 2, and has conceded to my original analysis of switching being the best solution. I've presented evidence for all of my claims, thus I maintain the stance that the orthodox solution for the Monty Hall riddle is correct.

With that said, I thank my opponent for debating with me and hope to debate with him in the future. I also thank the audience for reading.

Good night. :D