Huh… this has morphed into Monty Hall of the Mountain King.
It seems irrelevant but it isn’t because the door you pick initially has some bearing on which door Monty will reveal (because he always reveals a loser).
I saw what you did there.
I think people stick because they assume a far simpler mechanism is at work – Monty opens a door to try to get you to switch. He wouldn’t do that if you had picked the wrong door in the first place (except for a few to throw off suspicion). The only way to assess the accuracy of this hypothesis is to watch hundreds of games and determine whether picking the right door the first time is correlated with the reveal-and-offer.
Guy A picks Door #1.
Guy B picks Door #2.
Monty throws open Door #3 and shows it’s empty.
Guy A and Guy B get asked if they want to switch.
Should they both say yes?
Okay, another way to visualize it for those having problems seeing the logic.
Standard set-up of one prize and three doors. You pick a door at random. But Monty’s lazy in this game and doesn’t feel like opening doors. He just offers you a choice - you can keep the door you have or you can switch and take both of the other doors.
Now do you stay with the one door you have or do you switch and take two doors?
Mathematically this game is identical to the standard game.
No, different situation. This scenario is only possible if one of the players picked the prize door - if both had picked losing doors then the unpicked door would be the prize door and Monty couldn’t have opened it without revealing the prize. So here the odds are equal and each player has a 50/50 chance of winning.
The important fact to note here is that Monty can’t do this if neither picked the winning door. That happens 1/3 of the time, and that accounts for the chance of a particular contestant winning being only 50/50, instead of 2/3.
Well, yeah: in the two-guy scenario, Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.
But make it a one-guy scenario by removing Guy B: Guy A picks Door #1 – and Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.
Or make it a one-guy scenario by removing Guy A: Guy B picks Door #2 – and Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.
The only time you lose by switching is if you chose the winning door to begin with.
Let’s see if I got this: Now you’re looking at two instances of the original problem being played out, and in both the prize is, say, behind door 2. It’s behind door 2 when guy A plays, he switches, and he wins. It’s also behind door 2 when guy B plays, he also switches, and he loses.
So what? Keep playing the game enough times, and switching will still have you win two out of three times. So yes, both of the guys should switch, even though one of them end up losing and one of them end up winning, because all they can do, short of using telepathy or time travel, is to use what information they have to pick the best strategy in order max out their chances.
So in those last two sentences, you’re looking at a case where the prize is behind Door #1 or Door #2, but never behind Door #3. And then Monty reveals there is no prize behind Door #3 (which you already knew). So yes, in that case Door #1 and Door #2 have equal chances of winning, and there’s no benefit to switching.
ETA: If this isn’t what you mean, you need to specify what happens if the prize is behind Door #3.
Hang on. How could anyone possibly know that there’s no prize behind door 3 before the game starts? Because if you do know, then it’s a different game, with only two doors, and not the original problem.
Well, no; I’m just saying it happened not to be behind #3 when folks picked #1 and #2. As to this:
I hadn’t gotten that far; I’m feeling this out as I go. So let me twist it further out of shape:
- Figure there are four doors: three empty, one prize.
- Guy A picks one door.
- Guy B picks another door.
- Monty opens one of the two remaining doors.
- A and B are then given the chance to switch with each other. Should they?
Look, just to make one thing clear: What you shouldn’t do is overthink the problem, or imagine that there’s some deep paradox at work or anything profound going on.
There isn’t. It’s dead simple. Your cat could understand it.
Sure, you can also make up new scenarios with various numbers of doors or contestants. Those scenarios will have different optimal strategies depending on various factors.
That doesn’t have anything to do with the original problem, which has very clearly stated rules, and a definite answer.
ETA: Sorry, didn’t notice you specified four doors. Deleted my post.
It was a bit disappointing. Jamie actually said at one point that they didn’t care about the Math or proofs or some such. (But they did give a brief explanation later.)
They didn’t quote the numbers of their simulated run, but the boards looked like Adam won far more than 2/3 of the time and Jamie far less than 1/3. Anybody know what their data was?
OTOH, they were mostly (but not completely) clear on the rules of the game. Failure to precisely state the conditions is extremely common. So people fill in with their own assumptions, and the 2/3, 1/3 result does not apply in all these cases. Cf. “Ask Marilyn”.
The idea that just because there are two options, the odds have to be 1:2 is a damning indictment of the failure to teach people basic probability. I guess that’s why lotteries, casinos, and bookies stay in business.
I assume Monty knows where the prize is, and always opens an empty door. In this case, the prize has a 25% chance to be behind each of the doors one of the Guys picked, and a 50% chance to be behind the remaining door. Switching with each other won’t help, but switching to the other remaining door will.
If you make your choice and Monte offers you to keep your door or trade for the other two doors would you trade? That is essentially what you are being offered with the non-winning goats removed.
The breakthrough for me that helped me understand the math is when I realized that by switching I am actually getting the other two doors. It’s just that I know for sure one of the two is definitely a loser (which is a given anyway).