Why does the probability “lock in” and not change if you remove a variable but doubles for the other one?
I read it multiple times and it still doesnt make any logical sense to me.
In: Mathematics
imagine you have to choose a correct door out of a million doors and then after you choose, all the doors but the door you chose and only one other door remain
It’s sometimes easier with problems like this to actually run through all the possible scenarios. Luckily with 3 doors there are only 9
1) prize behind 1, you choose 1. Monty opens either 2 or 3, doesn’t matter, correct coarse of action (COA) **stay**
2) prize behind 1, you choose 2. Monty opens 3, correct COA **change**
3) prize behind 1, you choose 3. Monty opens 2, correct COA **change**
And so on for the others. Of the 9 possible scenarios you’ll find that in 3 of them the best choice is to stay, and in 6 the best choice is to change.
The reason *why* it’s not 50:50 is because Monty is acting on outside information, he changes the rules. If Monty did not know where the prize was then 1/3 of the time you would guess right at the beginning, 1/3 of the time you’d want to change to the other door, and 1/3 of the time *Monty would accidentally open the prize door after your first guess*. That doesn’t happen because he knows where the prize is and intentionally avoids it. The result is that you still guess right 1/3 of the time, but in BOTH of the times you guess wrong he tells you where the prize is.
It works because the host always reveals a door without the prize, and always a door you didn’t pick to start.
Doors that cannot be opened by the host don’t have their probabilities change.
Doors that can be opened have the same combined odds before and after the reveal, but how it’s distributed within that category does change.
All the math involved with this just makes me confused tbh, the way I like to think about is this. Let’s take away the host opening one of the doors. After you chose a door the host offers you the chance to either stick to your one door, or switch to the other two doors you didn’t pick, and if the prize is behind one of them you still win it. I hope it’s obvious that in this scenario it’s a much better idea to switch to the two doors instead of sticking to just one. And if you think about it this scenario isn’t actually any different from the scenario where the host opens one of the doors, opening that door is there simply to obscure the fact that you’re picking between opening one door and opening two.
If you stick with your first choice, the only way you can WIN is if you guessed right which is one out of three. If you switch, the only way you can LOSE is if you guessed right to begin with which is still one out of three. That makes your odds of winning two out of three if you switch.
I found the easiest way to envision how this works is to extend this to a point where it is more intuitively obvious what is going on.
Imagine that instead of 3 door there are 100. Still only one prize though.
You pick one at random.
You have a 1 in 100 chance of having picked the right door.
Monty, knowing which door is the right one reveal 98 door that he knows did not contain the prize.
That means, If you picked a dud (99 out of 100 chance) he reveals all the door except the one that does contain the prize.
If you were lucky originally and picked the correct one (1 out of 100) he will reveal 98 of the 99 remaining duds leaving one dud.
So you are left with two doors one that you originally picked and has 1 in 100 chance of containing the prize and the one Monty left that has a 99 in 100 chance of containing a prize.
You would obviously want to switch at that point.
The basic thing to remember is that however many doors there are, Monty knows which door has the prize. If you didn’t pick the right right away he will show you which one does. If you did pick the correct one right away Monty shows you a dud.
So the single door Monty leaves has the prize if you guess wrong initially.
The chance that Monty’s door is the correct one is the same as the chance that you were incorrect in your initial guess.
With three door you have a 1 in 3 chance of being right with your first guess and the door Monty leaves has 2 in 3 chance that you were wrong initially and it now contains the door.
The more doors there are the more obvious it gets.
If you pick a lottery ticket you have a very low chance of guess right, but if someone were to offer you to switch tickets with you with one that represented all the numbers you didn’t pick and assured you that either the ticket they were holding or the one you were holding was a winner, that would be good deal.
Let’s take this idea a bit further.
Let’s briefly assume that you know if your first choice was correct or incorrect. Now, if your first choice was correct you would want to stay with your first choice, right? And if your first choice was wrong you’d want to switch.
With that in mind, what is the chance that your first choice was right?
It’s easier to understand if you imagine that there’s a **billion doors**, and only one prize.
You pick a door randomly. There’s a billion doors, so you doubt you have any real chance of winning.
The host **shows you what’s behind every door except your door (all 999,999,998 or them)**, and the last door. No prizes behind any other one. The host knows where the prize is. Under this scenario, it’s pretty intuitive that the prize is almost certainly in the other door.
Because switching doors is guaranteed to switch whether or not you have the prize. If your chosen door had the prize, Monty will open one of the goat doors at random, leaving one goat door for you to switch to. If you chosen door had a goat, Monty will open the other goat door, leaving only the prize door for you to switch to.
Because you had a 1/3 chance of picking the prize door at the beginning, and switching doors is guaranteed to switch whether or not you have the prize, then switching turns the chances of winning from 1/3 to 2/3.
An intuitive way of getting it is making the number of doors way higher.
Imagine you have 10,000 doors in front of you. You choose one. Then someone opens 9,998 doors, revealing that they are empty.
Two doors now remain:
– the one you chose facing 10,000 closed doors,
– and one out of the 9,999 other doors, all of the others being empty.
Do you keep your initial door, and trust your unreal luck (1 chance out of 10,000) or do you change to the only door that is left closed?
Pretty instinctively, you’ll want to change, right?
Think of it this way :
When you choose you door, the probability that it contains the prize is 1 out of 10,000.
Correlatively the probability that the prize is in the set {all other doors} is 9,999 out of 10,000.
When the doors are opened, the set {all other doors} still exists, you just have the information that 9,998 doors in it are empty, except for one.
It is as if the remaining closed door captures the 9,999/10,000 probability by itself. It has “become” the set {all other doors}.
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