I’ve got a couple variations on the Monty Hall problem that are interesting. The first, which I came up with, seems to be easy to solve, but I wanted to run it by y’all. The second, which was pointed out to me by a guy on another message board, is not so easy to solve, and may be a genuine paradox.
So here’s the first scenario. You pick a door, Monty opens another, and gives you the opportunity to switch. You switch, and Monty then offers you the chance to switch back. The hasty reader might see shades of the envelope paradox here, but it’s not so bad–you’re actually choosing the other two doors when you switch the first time, and picking two doors naturally increases your chances of winning. Am I right here?
The second scenario is a little more troubling. Suppose Alice and Bob are playing simultaneously. Alice chooses door A, Bob chooses door B, and Monty opens door C. Now should either of the players switch? By the analysis of the simple version, they’re both better off switching, but then each of them would be picking a door that somebody else was better off switching from. By the logic of the previous paragraph, they’re each picking two doors–however, that would suggest that each of them should win with probability 2/3 when switching, and they can’t both do that if they picked different doors. Is there a resolution?
First off, one of the most important aspects of the original question was knowing what was going on in Monty’s head. In the original question, he’s showing you a door he knows is not the big winner. It is also assumed he knows going in that he’s going to show you a door and offer you the chance to switch. Got that? It’s important. If, in the original question, Monty might take a different path depending on what door you pick, the whole reasoning process goes to hell. Before any door picking, Monty knew he was going to show you a bum door and allow you the chance to switch.
A) Assuming Monty was compelled to make you the offer to switch, and was set on showing you a bum door, you switch once and stick with it. If he also is compelled to offer you another chance to switch (like, the producers said, no matter what happens, show them a lousy door, and offer the chance to switch twice), you switch once and stick with it. If Monty has any latitude, sometimes offering the second switch and sometimes not, it could be him messing with your head. There’s no good mathematical answer. You have to know what Monty’s thinking.
B) Again, you’ve changed something crucial. If Alice chooses A, and Bob chooses B, what does Monty do if the good prize is behind door C? In your example, Monty shows you door C and it’s a bum prize - he can’t always do that. Given that Monty won’t ever show the good door, you know he must have some latitude - Alice chooses A, Bob B, and the good prize is in C must lead to some other behavior on Monty’s part. Therefore, your example is a conditional probability. Another way of saying it is, “If Alice chooses A and Bob chooses B, and you know the good prize is in one of the doors they’ve chosen, should they switch?” In that case, it doesn’t matter - it’s a 50/50 proposition.
Will Monty always open whichever door was not choosen by either Alice or Bob, no matter what is behind it?
Your first scenario analysis looks fine.
Your second scenario is completely different. I’m assuming that in the scenario, Monty opens an empty door, so we’ve ruled out the possiblity of the prize being behind door C. In the normal Monty Hall problem, it’s more likely that your original pick is wrong, so when Monty opens the empty door, chances are that the third door has the prize.
In this case, however, Monty is forced to open the third door. What if the third door has the prize behind it? Does Monty still open it?
The way I’m interpreting this, when Monty opens the (empty) door, the only information you’ve gained is that one of you picked the correct door.
The trick in the orginal problem was that he opens one of two doors. You don’t know if he was forced to open that door (because it’s the only empty door left), or if, in fact, he could have opened either door (since your original guess was correct and the other two are empty). The former is twice as likely as the latter, so you benefit more often from switching.
In this version, I don’t see any such “trick”. You pick a door, the other contestant picks a second door, the third door is shown to be empty. All you know now is that the prize is behind one of the two unopened doors–with 50% probability each.
The first thing to know about the MHP is that people rarely state it correctly. Take your post for example and your 2nd scenario. Which door does Monty (or Monte) open? In the original problem, if you state it carefully, you will state that Monty deliberately opens a door he knows does not contain the big prize. With two doors available to him, he clearly has such an option. In your 2nd scen., he no longer has such an option in some cases. (Door C has the big prize.)
Please restate your problem completely and accurately. You will then note that the situation has changed completely from the original problem and thence the answer is different.
Remember, state the problem completely and accurately. Don’t be a Marilyn.
For the second scenario, let’s assume that we’re only interested in the case were either Alice or Bob picked the correct door.
Again, it’s a conditional probability.
What are the odds that you’ve picked the correct door as Alice, given that your or Bob picked the correct door?
Can I be a prick and complicate things by pointing out that Alice and Bob could choose the same door?
Let’s assume that doesn’t happen either.
The Alice&Bob situation can’t come up if neither of them pick the winning door, so it’s a pretty artificial situation. In reality, they could both choose prizeless doors, which would force Monty Hall to reveal the prize.
That being said, Alice should switch and Bob should not. Alice’s situation is exactly that of the player in the normal Monty Hall problem. She should switch, which means that she should pick Bob’s door. Since Bob’s door is the best one to pick, he should stay with it.
No, it’s not the same. In the original, Alice picks a door, then Monty has to open an empty door. Since it’s more likely that Alice’s pick was wrong, it’s also more likely that Monty was forced to open a particular door (he couldn’t open Alice’s door, and he couldn’t open the door with the prize behind it). This gives us additional information, and we can reason that Alice is twice as likely to win if she switches.
In the Alice and Bob one, Alice and Bob are clearly picking out of equal ignorance. Monty is still forced to open a particular door, but that action is not forced by where the prize is–it’s only forced by Alice and Bob’s picks (he has to open the remaining one). All the information we get from that is that the prize isn’t behind the third door–so either Alice or Bob picked correctly, with a 50% chance of success each.