Game One: You are given a choice of three doors. You pick number one. The host opens one of the other two doors, having been given instructions that, if you pick the car, the host is to open one of the other doors, and if you pick a goat, the host opens the other door with a goat. Stalemate. It is a predetermined outcome.
Game Two: The prior game’s outcome stands. The new choice you have is do you keep door number one, or do you switch?
How do you have a 2/3 chance of winning if you switch?
In: 107
You only have a 1/3 chance of picking the right door the first time making it a 2/3 chance that this door is wrong and you should switch
No, it is not two separate games. Monty has information from the start which the player does not have. When you have the opportunity to switch you have the opportunity to act based on some of Monty’s information, which improves your odds of success.
Because you had a 1/3 chance of picking the car on your first pick. Those odds do not change, and you don’t gain new information when the host opens a goat door because you already knew one of the doors had a goat.
The second choice is not an independent event – the chances that you are standing in front of a goat door when the “second game” starts is twice as likely as that you start in front of a car door.
You can also just draw out all the possibilities.
Pick goat A -> Stick -> Goat A
pick goat A -> Swap -> Car
Pick goat B -> Stick -> Goat B
Pick goat B -> Swap -> Car
Pick Car -> Stick -> Car
Pick Car -> Swap -> Goat A
Two out of three times swapping will end you up with a car.
When you first pick, you only have a 1/3 chance of beeing right. So you know that 2/3 of the time, the car will be behind one of the other doors. By switching you basically get to choose both other doors at once, because the host gives you one for free by telling you it’s not the car.
You can also just simply list all possibilities:
1. You pick the door with the car, you switch and loose.
2. You pick the door with goat number 1, you switch and win.
3. You pick the door with goat number 2, you switch and win.
So in 2 out of 3 scenarios you win.
They’re not entirely separate; Game One gave you information that you can use in Game Two.
Let’s imagine a different game. Suppose there are two coins: one is a normal fair coin, the other is a two-headed coin. The host picks one of the two coins at random and then flips it. Your job is to guess whether it’s the two-headed coin.
If the flipped coin lands “tails”, then the answer is obvious: it’s not the two-headed coin. But what if it lands “heads”? In this case, you should guess that it probably *is* the two-headed coin, because the two-headed coin is twice as likely to land “heads” as the fair coin. You have a two-thirds chance of being right.
(25% of the time, the host will pick the fair coin, it’ll land “tails”, and you’ll correctly guess that it’s the fair coin. 25% of the time, the host will pick the fair coin, it’ll land “heads”, and you’ll wrongly guess that it’s the two-headed coin. 25% of the time, the host will pick the two-headed coin, it’ll land “heads”, and you’ll rightly guess that it’s the two-headed coin. And 25% of the time, the host will pick the two-headed coin, it’ll land on the other “heads”, and you’ll rightly guess that it’s the two-headed coin. So you’ll be right 50/75 of the times that you guess “it’s the two-headed coin”. 50/75 = 2/3.)
This is what’s going on in the Monty Hall problem. Suppose you picked Door #1 and the host opened Door #3. You now know that the car is not behind Door #3. It was originally equally like to be behind Door #1 or Door #2. However, if it were behind Door #1, there would have been only a 50% chance of the host opening Door #3; if it were behind Door #2, there would have been a 100% chance of the host opening Door #3. So, just as you should guess that the coin which landed “heads” is the two-headed coin, you should guess that the car is behind Door #2.
It’s only two separate games ***if*** the position of the car can change between round 1 and round 2.
The way that helped me think about this was what if there were a million doors?
You pick on, and then the host opens 999,998 more leaving you with the one you picked, or one door from 999,999. Obviously you didn’t pick the right door out of the million, but in opening those other 999,998 doors the host knows exactly where the prize is and obviously avoids that. So of course you should switch.
It all comes down to the fact that the host knows which door has the car behind it and would never accidentally open that door.
When picking the first door, you have a 1/3 chance of picking correctly. Then the host eliminates a door that he knows has a goat behind it. The odds that you initially picked correctly don’t change from 1/3 as the host reveals a door with a goat.
When given the option to switch, you still only have a 1/3 chance of being right initially, meaning that the other door has a 2/3 chance of having the car.
The important piece of information is that the host knows the correct answer. They aren’t picking randomly.
Your first choice is 1/3 to be correct, 2/3 to be wrong.
Then the host removes one of the incorrect answers.
Your first choice is still 2/3 chance to be wrong. But now there is only one other choice, so logically it must be 1/3 to be wrong, or 2/3 to be correct.
Imagine if there are 100 doors. You pick one. Monty opens them all up but one. Do you switch?
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