The Monty Hall Problem is a common mathematical fallacy based on Monty Hall's game show Let's Make a Deal. It works like this:
It's counterintuitive to many people, but switching doors will double your chances of winning:
The bad news is that according to a scientific study, pigeons are better as this task than we are:
Link via The Presurfer | Photo: Library of Congress
Imagine that you’re in a game show and your host shows you three doors. Behind one of them is a shiny car and behind the others are far less lustrous goats. You pick one of the doors and get whatever lies within. After making your choice, your host opens one of the other two doors, which inevitably reveals a goat. He then asks you if you want to stick with your original pick, or swap to the other remaining door. What do you do?
It's counterintuitive to many people, but switching doors will double your chances of winning:
The problem is that most people assume that with two doors left, the odds of a car lying behind each one are 50/50. But that’s not the case – the actions of the host beforehand have shifted the odds, and engineered it so that the chosen door is half as likely to hide the car.
At the very start, the contestant has a one in three chance of picking the right door. If that’s the case, they should stick. They also have a two in three chance of picking a goat door. In these situations, the host, not wanting to reveal the car, will always pick the other goat door. The final door hides the car, so the contestant should swap. This means that there are two trials when the contestant should swap for every one trial when they should stick. The best strategy is to always swap – that way they have a two in three chance of driving off, happy and goatless.
The bad news is that according to a scientific study, pigeons are better as this task than we are:
Each pigeon was faced with three lit keys, one of which could be pecked for food. At the first peck, all three keys switched off and after a second, two came back on including the bird’s first choice. The computer, playing the part of Monty Hall, had selected one of the unpecked keys to deactivate. If the pigeon pecked the right key of the remaining two, it earned some grain. On the first day of testing, the pigeons switched on just a third of the trials. But after a month, all six birds switched almost every time, earning virtually the maximum grainy reward.
Link via The Presurfer | Photo: Library of Congress
In reality, the Monty Hall problem doesn't necessarily make the remaining unchosen door as the one with the car behind it.
As far as the Monty Hall Problem goes, the doors are randomly assigned and not changed after the goat is revealed.
The choice remaining is still 50/50, unless the result (where the car is) had been altered when the goat has been shown.
What happens if the goat is behind the door that the contestant had chosen in the beginning?
Maybe the reproduction above isn't accurate, but what it describes is not what you describe. The computer turned off the light on one of the unpecked keys, just like Monty revealed one of the unchosen doors to be worthless. I don't see any reason to believe that "the one remaining option that was remaining was made to be the correct answer, with no chance that the first choice was the right one."
The way it was explained on the linked website is that the first choice was never the correct one.
The pigeons learned to switch because they didn't want to go hungry.
If the chance of finding the prize first go was truly random, then the pigeons should have been fed 33.3 percent (1 in 3) of the time without switching and by pure chance alone.
Biasing the result by switcheroo-ing the prize just gives the result that the "researcher" was looking for in the first place.
If the goat was behind the first door you chose, then the host would pick the other door with a goat. Because there are three doors, two with goats and one with a car.
G/G/C
If you look at that you have a 66% chance of choosing a goat door. Then the host reveals a goat. It doesn't magically change the original chance of choosing a goat to 50%. SO it it's more likely that you will have chosen a goat in the beginning, then it's more likely that you have not chosen the car, meaning that switching=a higher chance of winning a car.
I think the first choice was the correct one in 1/3 of the trials in the linked web page. There is nothing to indicate otherwise. And if they were trying to reproduce the Monty Hall problem, 1/3 of the time the first choice would be correct. Why do you think the first choice is never right?
"Each pigeon was faced with three lit keys, one of which could be pecked for food. At the first peck, all three keys switched off and after a second, two came back on including the bird’s first choice. The computer, playing the part of Monty Hall, had selected one of the unpecked keys to deactivate. If the pigeon pecked the right key of the remaining two, it earned some grain. On the first day of testing, the pigeons switched on just a third of the trials. But after a month, all six birds switched almost every time, earning virtually the maximum grainy reward."
You are right in that the pigeon is switching so as to not go hungry, but we need to assume some motivation for the experiment to matter. If you assume the pigeon doesn't want food, the whole setup is meaningless.
I think the article was poorly worded. The first time I read it, I read this sentence: "If the pigeon pecked the right key of the remaining two, it earned some grain." to mean "If the pigeon pecked the key on the right hand side, it earned grain. Which isn't a correct model of the MH problem. I think it's safe to assume the researchers did correctly understand and model the MH problem, and it is the article that is ambiguous in the description, it should have stated "If the pigeon pecked the correct key of the remaining two, it earned some grain."
Of course, if the game is being manipulated in response to the player's choice, (ie, the car switches doors, the player is only given this choice if they chose correctly, etc) then questions of probability are irrelevant.
When you first choose, there is a 66% chance you picked the wrong door. This means there is a 66% chance that the correct door is one of the other two.
Then one of those two is eliminated. This does not change the fact that there is a 66% chance that one of those two is correct, except now you know one of them that isn't the correct door. So there is a 66% chance that the other door is the right one. You should switch.
I think the hang up for most people (certainly for me) is that it seems like the choice between two doors is independent of the original choice between three doors. What I just realized is that they aren't independent, because your first choice helps determine what doors will be available for the second choice.