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:
1. You are a contestant on a game show, and there are three doors in front of you.
2. Behind one of the doors is a car (the prize you want), and behind the other two doors are goats.
3. You pick a door, let’s say Door #1.
4. The host, Monty Hall, who knows what’s behind each door, opens another door (let’s say Door #3) revealing a goat.
5. Now, there are two doors left: the one you initially chose (Door #1) and the other unopened door (Door #2).
Should you stick with your original choice (Door #1) or switch to the remaining unopened door (Door #2) to maximize your chances of winning the car?
The counterintuitive answer is that you should always switch. Switching gives you a 2/3 chance of winning the car, while sticking with your initial choice only gives you a 1/3 chance. It may seem odd, but the probabilities work out in favor of switching doors.
The probability doesn’t change because nothing about the situation has changed.
When you pick a door, you know you have a 1/3 chance of being right, and a 2/3 chance of being wrong. In other words, there is a 2/3 chance that the prize is *somewhere in the other two doors that you didn’t pick.*
You also know something for certain about those other two doors: at least one of them is empty. If you picked the right door to begin with, then the other two doors are both empty, but if you picked wrong, then one of the other two doors is empty and one has the prize. Either way, there’s at least one empty door.
When the host opens one of those other two doors, it will always be empty, because he knows where the prize is.
He has shown you at least one door that is empty. You already knew that door was there…you just didn’t know which one it was. The dope showed it to you.
So there are still two sets of doors: the door you picked, and the sum total of the other two doors, with the known property that at least one is empty. There was a 1/3 chance the prize was in your door, and a 2/3 chance the prize was somewhere in the other two doors. That hasn’t changed! But since you now know which door is definitely empty, you now know which door has the 2/3 chance of the prize.
The probability would only change if the doors were shuffled each round.
The classic explanation changes the scenario from 3 doors to 100. If you pick 1 door, then I open 98 wrong doors, would you stay with your initial 1/100 pick or switch to the only other remaining closed door?
If you picked wrong door, the host tells you outright which of the other 2 is the right one.
If you picked right door, he opens a random one and provides no extra information.
2/3 of times, you’ve picked wrong one to begin with, so the first scenario applies.
Some people find it easier to grasp if you imagine 100 doors instead of 3, with the host always opening all but 1 (+ the one you chose)
The key part to remember is that the person removing the options knows where the prize is and is not removing it.
Imagine there are 100 options. You pick one at random. Your chance of winning is 1 in 100.
Monty Hall has a 99 in 100 chance of having the prize. He then removes options from his pool, purposefully not removing the prize.
He still has a 99 in 100 chance of having the prize, because he knows where the prize is and specifically hasn’t removed it. But now there is only one choice left.
[removed]
You have a one in three chance to be correct with your guess.
The prize has a two in three chance of being behind one of the two doors you didn’t pick.
The host does not open one of these doors randomly, he only opens one there is no prize behind.
Therefore the door he invites you to switch to has a two in three chance of having the prize.
It may help you to look up a tree diagram of the problem. They are available online and break down the probability of the different outcomes.
You need to think about it in terms of the chance you were wrong.
When you pick one of three doors, there’s a 2/3 chance you chose a door without the prize behind it.
The game master then opens one of the doors you didn’t choose, to show you there wasn’t a prize behind it. But this doesn’t tell you anything you didn’t already know – you knew there was only one door with a prize, so at least one of the two you didn’t choose would have to also not have a prize. Your chance of having picked the wrong door is still 2/3, so you should switch, turning that into a 2/3 chance of winning.
When they reveal one of the doors and ask if you would like to switch, the probability doesn’t become 50/50 because they aren’t asking you what door the goat is behind.
Instead they are asking you if you believe you picked right in the beginning.
Because there were 3 doors, you had a 1/3 chance of getting it right (staying) and a 2/3 chance of getting it wrong (switching)
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