If the other guy has a car behind his door then Monty has to open yours. You lose. So let’s put that possibility away, since we want to know what happens after Monty opens the door and you’re still in the running.
So - the other guy does not have a car behind his door, and Monty opens his door 'cuz he hates him. He will never open yours if that is the case 'cuz he loves you. That can happen whether you have a car behind yours or not. So the chance that you have a car behind your door is now (after Monty opened the other guy’s door) 50%.
But in both the original scenario and the “Monty loves you” scenario we’re already assuming that the other guy has been eliminated, and you’re the one left.
If Monty is “meh, whatever” about you, you know the following:
You both picked a door at the beginning. The other guy is out. His door had no prize. The car is now either behind your door or behind the third door.
If Monty loves you, you know the following:
You both picked a door at the beginning. The other guy is out. His door had no prize. The car is now either behind your door or behind the third door.
See, I just copy-pasted that, because they’re the exact same scenario.
As I hinted at earlier in thread, the probability of winning when you stick is 1/(2-A) where A is the probability that Monty reveals your door in the case that both contestants pick the wrong door.
So if Monty loves you (A=0) the probability of winning by sticking is 1/2, if Monty hates you (A=1), the probability of winning by sticking is 1 and when Monty is fair (A=1/2) the probability of winning by switching is 2/3
If Monty hates you, and you know he hates you, and Monty knows you know that he hates you, then Monty may eliminate the other guy when the car is behind the unchosen door. You will be certain you are winning, and therefore will not switch even when given the opportunity. Monty will trumph over you by having given you the chance to switch and by tricking you into thinking you had won with certainty. This is a devilish Monty indeed.
When I worked out the answer to the problem I left A as its value isn’t fixed by the problem as stated.
The point I made earlier was that when A is unknown the best strategy is still to stick. You don’t know if sticking gives you an advantage, but you do know that you won’t be disadvantaged by sticking.
This is true. So, I guess my added sentence from earlier is not needed.
Since the question asks “Should this contestant switch?” The answer will always be no.
ETA: Also, apologies for apparently skipping over your posts before.
Discussing probabilities about which you have no information doesn’t do much for me. Equally as uninteresting would be, “If the car is behind the 3rd curtain, but you didn’t know it, your best strategy would be to pick curtain 3, although you wouldn’t know that.”
Let’s look at the named-goat scenario with a spiteful but dumb Monty, who will always get rid of you if he can, OR will always get rid of your opponent if he can.
Things change in these circumstances:
In scenario 1 and 2, Monty is gonna reveal Bob’s goat. Right? You’ll get a chance to switch.
In scenario 3 and 4, Monty is gonna reveal your goat. Right? You’ll be out of the game.
In scenario 5 and 6, Monty has to choose which goat to reveal. Here is where his hatred comes in: if he hates you, he’s always going to eliminate you, and if he loves you, he never will.
Do you agree with me so far?
If he hates you, the only scenarios where you stay in the game are 1 and 2: in either scenario, you MUST stay or you lose.
If he hates you, you stay in the game in scenarios 1 and 2 and 5 and 6. If it’s 1 or 2, you should stay; if it’s 5 or 6, you should switch. This gives you a 50/50 chance of success.
I don’t really understand the point you are making. You don’t “know” that Monty is choosing randomly in the OP. Unless you add my sentence from before.
To understand why this different from the original Monty Hall problem:
Your chance of picking the right door first time is 1/3.
What Monty tells you (in this variation) by revealing your opponents door is either:
i) Nothing (when Monty loves you A=0)
ii) Your first pick is correct (when Monty hates you A=1)
iii) Your first pick was probably correct (0<A<1)
In i) as Monty tells you nothing switching doesn’t alter your chances of winning, in ii) and iii) what Monty tells you is that you should either definitely or either probably stick.
Whatever happens as long as you employ the strategy of sticking you will win the game 1/3 of the time and you can only disadvantage yourself by choosing to switch if Monty is telling you should definitely or probably stick.
In the original Monty Hall problem your chance of picking the right door first time is 1/3 and what Monty tells you by revealing one of the other doors is that what the right door is if you picked the wrong door first time (which has a probability of 2/3). I.e. by switching you are making the sensible gamble that you picked the wrong door first time.
Just to be clearer: in this variation you win 1/3 of the time by sticking including those times where Monty reveals your door (and hence you lose before being given the option to switch or stick).
I think there’s a mistake here. In 2b, you’re eliminated as well, so the only remaining option is 1. If Monty hates you, always stay.
I agree with the formula, but I think the very last statement was accidentally switched. Where A=1/2, the probability of winning by **sticking **is 2/3.
Given that, it doesn’t matter whether Monty loves you or hates you. As long as he HAS to open a door picked by a contestant that doesn’t have a prize (even if he gets to pick which one if there’s a tie), you should always stick with your first door (because at best switching is neutral, at worst it’s a guaranteed loss and if things are random it lowers your odds).
