The orthodox solution to the Monty Hall riddle is incorrect.

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.