I wasn’t tricked, but may have confused some by claiming it was the “same as the classic problem.” Although I’ve argued (correctly!) the Monty Hall problem off-and-on for decades, I guess I never memorized the Monty–>Switch shortcut and reason it out from first principles each time.
I regard the problems as the same (even though Switch transmutes to Don’t Switch) because in each case Monty exposes a non-prize from one of two specified doors, with the 3rd door retaining its original 33% chance. The unopened of the Monty-openable doors then has 67% chance.
My sister gave me this version last weekend during our family gathering. She teaches at Appalachian State and had used the original Monty Hall riddle as part of a problem solving exercise with one of her classes. However, she discovered it was all too easy for her students to google the old problem, so she changed the riddle just a bit. Now she says it is not at all uncommon for her students to turn in work that explains the original problem instead of the one she actually included as part of the assignment.
My sister also notes that on the old LMAD TV game show, there were always two contestants at the end vying for the “Big Deal”, so the revised problem has a tenuous connection to the old show format.
I think that had I hinted in the orignial post that the problem would be a variation, SDMB solvers would have not just assumed after a few words that it was the old riddle …and the poll would have skewed much more heavily towards the correct answer. I regret not posting it that way. But at our family gatherings, where most of us are quite experienced in riddles (including knowledge of the old Monty problem) we all have to be on our toes for traps.
Though come on folks, I have been at this site for fifteen years. Would I really try to post the old Monty Hall problem as something fresh? Jeesh!
I have one last riddle from this past weekend. I will try to post it some time in the next few days.
Good job to bup who first answered that refusing to trade was the right answer.
Am I correct in saying that,
In the event that neither contestant has the prize behind the door, Monty always chooses so that say Contestant 1 is the one remaining…
Because Contestant 1 is his buddy.
then it doesn’t matter whether Contestant 1 switches or not because at this point the odds are 50/50?.
Correct. Again it is enough just to focus on the information you’ve acquired from Monty’s action.
In your new scenario, all you’ve learned is that the prize isn’t behind door #2. In OP’s scenario you also learned that, if the prize is behind #3, then Monty’s mental coin flip came up Heads.
I’m not making any excuses for my obtuseness earlier, because this wasn’t a factor.
But the riddle would be slightly less ambiguous if it included a sentence that was something like. In the event that the prize is behind the unchosen door, Monty randomly selects one of the contestants to eliminate.
Say the two of us are on the show again. Monty has a secret crush on you, while he hates my guts. We each pick a door.
There are now, again, two sets of doors. The doors we’ve picked, and the door that hasn’t been picked. The chance of the prize being in our set is 2/3.
Monty hates me, so unless the prize is behind my door, I’m out. He opens my door, no prize. Now you know that the prize wasn’t behind my door. It’s behind yours, or behind the unpicked door.
I don’t see why this changes the probability of the two of us picking the set of doors with the prize to begin with.
Maybe I’m wrong. But someone will need to tell me why.
But it has to be absolute. If Monty only mostly hates your guts and 1 out of a hundred times, he would still let you play for a chance to win, then if I get to choose, there is a very slight advantage to me staying pat rather than switching.
I don’t consider it a trick question, exactly. It’s our fault, and I’ll include myself in that, for not properly analysing the question asked and determining the correct answer. Something that took me basically 3 minutes to do, but was too lazy upon the first attempt.
But I do have a problem with it for one reason. Most people who have not heard the Monty Hall problem will say to stick to the door they chose, even though that’s wrong. They’ll do so automatically and without thinking and many won’t change their minds even in the face of evidence to the contrary.
Now we have a new problem where the answer is to stand pat. Which means that people unfamiliar will still choose to stick with their door for no reason other than their gut told them so. Only this time they’ll be right.
That doesn’t seem like the hallmark of a good logic problem where the unthinking luck themselves into the correct answer.
I will repost my earlier post, modified for “Monty hates me” twist:
There are three possible initial scenarios:
You: car, Bob: nothing, third door: nothing - probability 1/3
You: nothing, Bob: nothing, third door: car - probability 1/3
You: nothing, Bob: car, third door: nothing - probability 1/3
Let’s see what the scenarios become when Monty opens the door:
You: car, Bob: nothing, third door: nothing, Monty HAS to open Bob’s door- probability still 1/3
2a. You: nothing, Bob: nothing, third door: car, Monty picks Bob’s door to open - probability is 0 since Monty hates you
2b. You: nothing, Bob: nothing, third door: car, Monty picks your door - probability is 1/3
You: nothing, Bob: car, third door: nothing, Monty HAS to open your door - probability still 1/3
From the four above, 2a and 3 are removed, since you are eliminated in those scenarios, and the original question presupposes that you stay to the next round. So what is left is 1 and 2b. The chance of 1 happening is the same as that of 2b. So it doesn’t matter if you switch or not.
No, if Monty hates me, I only get to keep playing if the car is behind my door. Otherwise, my door is always opened revealing no prize, and Bob gets to proceed to the next round.
So, if Monty hates me, I should most certainly always stick with my door.
Now, the question is what the situation is for Bob, who is Monty’s crush. If he’s the one who gets to keep playing, I contend that he still has a 2/3 chance if he stays put.
I think Terr’s post needs to be rewritten as a “Monty’s my buddy” scenario. But it should still work. I don’t have time right this second.
If Monty hates me and I know it. Then 100% time I should stay pat because he’ll only let me play if I have the car.
I still don’t get why. How does Monty’s mechanism for picking the final contestant change the initial probability of the two contestants picking the set of doors with the prize at the beginning?
If I’m just being dense, feel free to kick me. That’s what my friends and family always do.
Like I said, it doesn’t affect the *initial *probability. It just affects the probability of whether you should switch or not and makes it uneven for the two players.
Monty hates me guy. - 100% chance if you don’t switch.
Monty loves me guy. - 50% chance if you don’t switch.
This averages out to the answer if Monty neither loves or hates.
