I have read through around 10 articles and webpages on this problem, and still don’t understand. I’ve run simulations and yes, switching does get you better odds, but why?
In: Mathematics
Let’s say they don’t open any doors. They just say “Do you want to switch doors? If you choose to switch, we’ll switch to the right door! Well, unless you already got it right, of course, because then there won’t be any right door to switch you to.”
You’ll probably say “Sounds great! In that case, I’m probably going to benefit from switching.”
And they might ask “Want us to tell us which door that is? Which door is the right one, unless you already got it right?”
And you’ll probably say “I guess so, but it doesn’t really matter, does it? It’ll only tell me which door *would* be the right one if I picked the wrong one to start with. It won’t tell me what the odds are that I did pick the wrong one.”
They open one door, making sure to pick one that you didn’t choose and that is not the winner door. “Now that we’ve made it clear which door is the winner if yours isn’t, wanna switch?” they ask.
You might reply “Sure, but it doesn’t matter that you opened the door, because the odds that I picked the right one isn’t dependent on *which* other door might be the right one.”
**You gotta remember:** A lot of people will insist that if you pick randomly between the door you picked to start with and the only other remaining door, then the odds are 50%, right? And the answer is yes, but that’s not what the question is about. It’s about what the odds are if you *know* that you picked one specific door to start with, and you also *know* that the other door is guaranteed to be right if your first door was wrong.
**Shorter explanation:**
When should you switch?
When you picked the wrong door to start with.
What are the odds that you picked the wrong door to start with?
66%.
So the odds that switching works are also 66%, right?
Of course.
There are 3 doors(A/B/C), one with a car and other two with goats.
Points you should remember
1) The host knows which door has the car
2) The host will never open the door with the car
Let’s say you pick door A, at this point you have 1/3 chance of picking the car. Meaning there is a 66.66% chance you will pick a goat.
Now the host revels one door which has a goat and gives you a choice to change your initial choice. Since you have a much higher change of picking a goat in the initial choice and the host will always reveal the door with the goat, switching is your best chance of wining. (You have 2/3 chance of winning if you switch)
The key to the difference (and how it suddenly started making sense to me) is that Monty Hall, the game show host, knows where the prize is, and he’ll *always* open a door that leads to a goat. He’s not picking randomly, which is why the odds change from the first pick to the second pick.
The other replies go into the specifics of the calculation, but this explains *why* there’s a difference.
What worked for me is reducing “3 doors, only one has the prize you want*”, to the two outcomes you care about: P(car) (initially 1/3) and P(goat) (initially 2/3),
In other words, there are two sets of doors Cars={car}, Goats={goat, goat}. You are hoping to select the first set/avoid the second*. It should be clear you initially had a probability of 1/3 selecting a door in the Cars set, and 2/3 for a door in Goats set.
Now, once you make your initial selection and the host pauses to open a door to show you a goat, it’s tempting to think that the problem has become Cars={car}, Goats={goat} (so 50/50 for selecting either).
However, we know the host was never going to open a door in the Cars set (if they were selecting randomly and may have shown you a car, then the previous logic does actually apply and it is 50/50 now **). Essentially, what they have done is identified a door from the initial Goat set. They have, in fact, removed all of the doors in Goat set but one. You now have two doors—one ‘representing’ the Cars set and the other representing the Goats set. And because the host knew what they were doing, the initial probabilities apply. So, your initial guess had a 1/3 chance of being correct, and still does! This means there is still is a 2/3 chance that the car is in ‘one of the other’ doors (which in this case, is only one door), which is why you’re twice as better off switching.
I find the problem becomes a lot more obvious if you make it say, 100 doors, and the host opens all but yours and one other (so always ending w/ one door from Cars and one from Goats). What’s more likely—you got a 1/100 choice right from the start, or it was any of the other 99 doors (remember that host knew to make sure none of the 98 doors were a car).
You can even play with the conditions more (e.g. add more doors w/ or w/o a prize, change the number of doors the host opens, etc.) and work out the difference probabilities once you understand the framing.
* Assuming you want a car, not a goat, which everyone seems to ignore.
** Assuming you don’t get shown a car, in which case you can achieve 100% probability of success.
Here’s a fresh point of view that usually doesn’t get mentioned.
Let’s change the rules slightly. After the host opens one of the doors, you can either keep the one door or switch to the two other doors. So you can choose two doors instead of one. Choosing an open losing door is meaningless, so these rules are equivalent to the original ones.
Let’s make another small change; the host doesn’t open a door. Even if the host doesn’t open a door, you already know that one of the other two doors is a losing door. The host opening the losing door doesn’t change anything, so these rules are equivalent to the original ones.
The Monty Hall problem is just a confusing way to ask whether you want to open one or two doors.
Here’s an explanation that might help. A sort of simile.
You’re playing roulette at a casino, and bet money on number 4 winning out of the 37 numbers on the wheel. You turn away, because you’re too nervous to look.
“Did I win? Did number 4 win?” you ask your friend who’s there with you.
“What do you think?” your friend teases.
“Well,” you say, “the odds are 1/37 that I picked the right one, so I’m not positive. Did I win?”
“I won’t say!” your friend says. “I’ll just tell you that it was *either* number 4 that won, or number 28. Now what do you think your odds are?”
“My odds are the same,” you say. “Because I still only won if number 4 won, and the odds of that are still 1/37. That doesn’t change just because I know exactly which number *might* have won.”
The key piece of info many have a hard time with is that Monty knows which door has the prize at all times.
He will never eliminate the car. That makes it so what Monty does DOESN’T MATTER. He will never remove the car from the equation.
The bit that makes you think it’s a 50:50 is that the swapping confuses people. The entire thing is a 2 step problem. You pick -> other door is revealed to be goat. BUT you initially had 3 choices, so the outcomes branch from the initial 3. The fact that is is 2 steps cause people to see it as your door vs remaining door, thus the 50:50 until you actually map out all possibilities.
I know you pretty much got the gist from the other comments, I wanted to note why I think the 50:50 is so common.
We can’t really help you if you don’t explain what specifically you have an issue with. You probably heard all these explanations already.
Well two CHOICES, don’t automatically make both choices of equal probability. If instead we roll a die, and choice A is “roll a 1” and choice B is “roll anything but a 1”. You need to choose A or B before the die is rolled. Do you assign 50/50 chance of success to both choices?
Monty Hall is essentially saying, choose a door. Now we group the other two doors into the other choice. Do you stick to your choice or switch? You are given the option to switch your CHOICE. Because Monty Hall always opens a door (not randomly) that does not have the prize, he is essentially saying stay with your door choice or switch to choose the other two doors.
With 100 doors the odd for a single door is 0.01,
the host removes 98 wrong doors,
you have still a door with an 0.01 odd.
Only in 1 out of 100 cases your door is the right door
but in 99 cases your door was and is the wrong one.
Taking out 98 wrong doors after your choice did not change the quality of the first choice.
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