Note: In the following comments, I assume that Monty can choose whether to offer a switch, but that if he does offer a switch he must first reveal an incorrect door other than the one that was picked.
Let’s say Monty offers the switch X fraction of the time when you initially picked right, and Y fraction of the time when you initially picked wrong.
1/3 of the time you initially picked right, meaning that X/3 of the time Monty offers you a switch that will screw you. (1-X)/3 of the time Monty just gives you the prize without a switch.
2/3 of the time you initially picked wrong, meaning that 2Y/3 of the time Monty offers a switch that will save you. 2(1-Y)/3 of the time Monty lets you lose without offering a switch.
Total odds of Monty offering a switch: X/3 + 2Y/3 = (X+2Y)/3
Fraction of switches that benefit you: (2Y/3) / (X+2Y)/3 = 2Y/(X+2Y)
This brings me to your first question:
(1) If we want 50% of the switches to benefit you, then 2Y/X+2Y = 1/2
Thus, X = 2Y. Monty should offer the switch twice as often when you picked right as when you pick wrong. E.g., If Monty always offers the switch when you picked right (X = 1), he should offer it half the time when you picked wrong.
As for what fraction of players win:
Suppose that Z is the fraction of players who take a switch when it is offered.
Thus, Z(X+2Y)/3 of the time the player is offered a switch and takes it. This results in victory for the player 2Y/(X+2Y) of those times.
(1-Z)(X+2Y)/3 of the time the player is offered a switch and declines it. This results in victory for the player X/(X+2Y) of those times
Multiplying those probabilities to get the odds of each outcome occuring, we have:
Z(2Y)/3 of the time the player switches and wins
Z(X)/3 of the time the player switches and loses.
(1-Z)X/3 of the time the player refuses a switch and wins
(1-Z)(2Y)/3 of the time the player refuses a switch and loses
(1-X)/3 of the time the player wins without being offered a switch
2(1-Y)/3 of the time the player loses without being offered a switch
So in total the player wins Z(2Y)/3 + (1-Z)X/3 + (1-X)/3 = 1/3 + (2Y-X)Z/3
Note that Monty controls X and Y, but the player controls Z. Note that the player maximizes their chances of winning by always switching (Z=1) if 2Y-X > 0, and by never switching Z=0 if 2Y-X < 0. (For 2Y-X = 0, the player has a 1/3 chance of winning no matter what they do. This is the same as our conclusion above: If X = 2Y, the switch is equally likely to hurt as it is to help, so the player can’t improve on the initial 1/3 odds.)
Now, the question is: would the player know whether 2Y-X is positive or negative? If they’ve watched a lot of games and paid close attention, then they should know whether switching usually pays off, so the answer is yes (with some caveats that I’ll return to in a second.)
Now for your second question:
(2) If the player has correctly determined whether switching is beneficial (and assuming the player is rational), then they will only switch if it helps them. If more than half of the switches that Monty offers benefit the player (2Y>X), then the rational player will always take a switch, and will win (1+2Y-X)/3 > 1/3 of the time. If more than half of the switches that Monty offers hurt the player (2Y < X), then the rational player will never take a switch, and will win 1/3 of the time. So Monty can’t make the rational player lose more than 1/3 of the time. His best strategy is to make no more than half the switches good switches (for the player).
Of course, the game is more entertaining when there’s the option to switch, so Monty will probably offer as many switches as possible without giving the player better than a 1/3 chance of winning. That means maximizing (X+2Y)/3, subject to the constraint that 2Y is less than or equal to X. Thus, Monty should set X = 1 and 2Y = X, meaning he’d offer a switch all the time when the player initially picked right, and half the time when the player initially picked wrong, as described above.
Of course, all this assumes that the player has correctly determined whether switching is beneficial, and that they are making the rational choice in light of this knowledge. In practice, some players will choose irrationally, and some simply won’t have watched enough games to know if switching helps them. So if Monty wants to have as few winners as possible, he should actually make switching pay off slightly less than half the time. That way, a few suckers will switch (which usually hurts them), and lower the odds of winning below 1/3. There’s no risk to Monty in this strategy, since even if the players do the smart thing and never switch, their odds of winning can’t go above 1/3. (In contrast, making switches pay off more than half the time means that the worst players won’t ever switch, and win 1/3 of the time, and the best players will always switch, and win more than 1/3 of the time. So Monty has no incentive to make switching pay off more than half the time – however, see my addendum at the end of this post.)
In practice, Monty needs to balance two competing factors: Making switching pay off less often lets him take advantage of a greater fraction of the switchers, but making it pay off too infrequently will reduce the number of people who switch (by allowing them to catch on to his strategy). Thus, Monty is really trying to minimize the odds of winning given by 1/3 + (2Y-X)Z/3, where Z is a decreasing function of the ratio of bad switches to good switches X/2Y.
To restate this a little clearer, let’s create a variable b, which represents the fraction of switches that are bad for the player. b = X/2Y
Thus, the chance of the player winning (I’ll call it W) is:
W = 1/3 + (2Y)(1-b)Z(b)/3 = 1/3 + (X/b)(1-b)Z(b)/3 = 1/3 + X(1/b-1)Z(b)/3
where Z(b) means “Z as a function of b”, not “Z times b”.
Note that the odds are worse than 1/3 for b > 1. Furthermore, the odds are 1/3 at b = 1 and at b = infinity (if all the switches are bad, virtually no one will accept a switch). Thus, for any given X there’s some b>1 that minimizes the odds of winning. That value depends on the behavior of the players Z(b), but after enough games Monty should be able to pretty much figure out what value of b is best. For that value of b, the odds of the player winning are worst if X is maximized (X=1). (Here I’m assuming that the players behavior depends only on what proportion of the switches are bad, not on what proportion of switches are offered. However, Monty could hope that offering fewer total switches will make the players less quick to catch on to whether switching is good or not, which would give him some incentive to reduce X. Also, some players might notice that Monty always offers switches to those who initially picked correctly (X=1). They might incorrectly conclude that when he does let them switch, it means their initial pick was probably right and they should stick with it. That’s not rational, since Monty can make switching beneficial for the player more than half the time (2Y > X), while still always offering switches to those who picked correctly (X=1). However, if Monty has chosen to make switching bad (2Y < X), he may want to sometimes give the player their correct choice right away (X<1) to make these irrational players more likely to take a switch.)
So, to answer your third question:
(3) If the players have seen a lot of games and worked out the odds, they’ll all know whether to switch or not. In that case, Monty can’t make less than 1/3 of the players win, which means he might as well offer the offer the switch as often as possible to make the game more interesting to the viewers. If, on the other hand, some of the players haven’t figured out if switching is a good idea, Monty should try to screw as many of those players as possible by always letting them switch away from their correct picks. Either way, it makes sense for Monty to always offer a switch when they picked the right door.
However, if some players are irrationally switching based on X instead of 2Y/X (as I mentioned above), then there might be some benefit for Monty to make X < 1 to trick them.
So, in summary:
**(1) If Monty offers a switch twice as often to players who chose correctly as to players who chose incorrectly, then switching is equally likely to help the player as it is to hurt them.
(2)(a) If the players are completely rational and have watched a lot of games, they’ll figure out whether switching is good or not, and always do that. In that case, Monty can’t make there odds of winning worse than 1/3 (since they can guarantee those odds by always refusing to switch).
(2)(b) However, if the players are irrational or inexperienced, Monty can make less than 1/3 of them win if he makes it slightly disadvantageous to switch. In that case, the never-switchers still win 1/3 of the time, but the ones who do switch have worse odds.
(3)(a) If the players are completely rational and have watched many games, they’ll win 1/3 of the time no matter what, so giving away the prize without offering a switch won’t benefit Monty.
(3)(b) If they players are irrational about their choice, but still rational about what factors they consider in making their decision, then they won’t care if Monty sometimes gives the prize away without a switch – they’ll only care about how often switching is beneficial. In that case, giving away the prize without offering a switch won’t benefit Monty.
(3)© If the players are irrational about what factors they consider in making their decision, then perhaps they will take a bad switch based on the specious reasoning that a former player who wasn’t offered a switch ended up winning. So maybe it would benefit Monty.**
Summary of summary: Ultimately, what Monty should do to screw over the players the most depends on how knowledgeable/rational they are.
Addendum:
In the above, I assumed that the players have a 1/3 chance of guessing the right door to begin with. But if Monty knows that players tend to pick one of the doors (say, door number 3) more than 1/3 of the time, he can make their odds worse than 1/3 by disproportionately putting the prize behind a door they don’t pick. Of course, if he makes it too disproportionate people will eventually catch on, and he’ll need to change his strategy.
If Monty makes switching bad to try to trick the players that don’t catch on, and then after a while he notices that most players are on to him, he could temporarily change strategies and make switching good. In the short run this would make him lose slightly more than 1/3 of the time (but not too bad since most people weren’t switching anyway). But eventually this could cause more players to be willing to switch, meaning they’d get screwed when Monty returns to his original strategy.
In conclusion: If you think Monty’s out to get you, use a random number generator to determine your choice in advance, reject all chances to switch, and accept the fact that 2/3 of the time you’re going to lose anyway. If Monty wants to cap your odds at 1/3, he can. (This was obvious from the start, since if Monty offers only bad switches, then your odds are capped at 1/3.) The only way to get better than 1/3 odds is if Monty is employing a less than optimal strategy, such as always offering a switch (or perhaps if he’s varying his strategy in a non-random way, such as I described in the previous paragraph, in which case you could potentially benefit if you were able to correctly predict when he changed the odds.)