Suppose you're on a game show, and you're given the choice of three doors:
Behind one door is a car; behind the others, goats.
You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat.
He then says to you, "Do you want to pick door No. 2?"
Is it to your advantage to switch your choice?
Suppose there are N (let's pick 1000) doors: Behind one door is a car; behind the others N-1 (999), goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens N-2 (998) doors, excluding No. 1, which have have goats.
Now there are only two doors closed, the one you picked No. 1, and one that was part of N-1 (999) doors. He then says to you, "Do you want to pick door No. 1 or the other one?" Is it to your advantage to switch your choice?
When it clicked for me is when I thought of it as being 1/3 probability the car is behind your door, and 2/3 it isn't. That doesn't change whether they show you one of the other two doors or not, so switching to the 2/3 option is better.
Say you pick one of the two non-prize doors initially. The host now opens the other non-prize door, leaving only the prize door. If you switch you get the prize door.
So switching always changes the prize door to a non-prize door and a non-prize door to the prize door. You start on the prize door 1/3 of the time and a non-prize door 2/3, so you should always switch.
The version that helped you is the version that helps nearly everyone in my experience.
"If you stay with the door you picked initially you succeed if the initial door has a car, which has a chance of 1/3. If your strategy is to switch then you succeed if your initial pick is a goat, which has a chance of 2/3."
1) Your 1-in-a-trillion chance original choice just happened to be correct
2) The omniscient host picked the correct door
Clearly, you should switch.
Draw three rows of three rectangles each. Now draw a star (representing the prize) in row one, box one, another star in row two box two, and finally a star in row three box three. This depicts all possible "game states". Now we will play the game three times, but in each case you will always choose the first box in each row as your first chosen door. Now the host will open one of the other doors. In the first row, you have two choices to open, so pick one and draw a circle in it. This door is open revealing nothing. In the second and third rows, you must not open door with the prize, so you choose the empty box in each row.
Now go through and tally up how many times you would have won if you had not switched doors and stayed with your original guess (1/3). The tally how many times you would have won had you switched (2/3).
Only one of the three doors has the car. So you have a 1/3 chance of correctly picking it and 2/3 of the time you'll pick a wrong door.
Monty opens a door that he knows doesn't have the car. Since 2/3 of the time, you've picked a goat, knowing where the other goat is means switching will guarantee you get the car in those 2/3 of the trials in which you picked a goat initially.
So if you always stay with your initial pick, you keep a 1/3 success rate over n trials. By always switching, you get a 2/3 success rate over n trials.
But here's the intuition: because the host never opens a door with a car behind it, the fact that he didn't open a particular door suggests that the door in question is the winner. Compare that to the initial door, which was selected blindly.
You picked a door. It has either a goat or a car behind it, but the goat is more likely. In which case, the remaining doors have a goat and a car behind them. And the host has just kindly revealed which one has the goat.
Which part of three (doors) are you having snags with?
However, another possibility is that: (i) if both of the remaining doors contain goats then the host reveals one of them; but (ii) if one of the remaining doors contains the car then the host doesn't open a door at all.
Many I told were just about ready to argue their view to death and asserted the switch your choice conclusion is wrong despite them having no formal learning on the subject.
This and the birthday problem effectively illustrate how intuitively bad we are at probability and statistics in general.
The other thing I find interesting about this puzzle is wondering whether the people who find the arguments in favor of switching convincing are actually reasoning better than I am. If so, how are they doing this? What is it that makes those (allegedly superior) arguments convincing to them?
The thing that kind of convinced me that the pro-switch arguments must be correct is that people have run simulations of this problem on computers and the switching strategy turns out to win in line with the pro-switch argument probabilities. But the tricky thing about probabilities and randomness is that such successes could be "just a matter of luck", as it were. Maybe they were wrong, but just got lucky. It's improbable, but still possible.
Only mathematical proof independent of experimentation should be rock-solid correct. But mathematical proof seems to be a matter of being convincing enough to mathematicians, and no so independent after all, and that seems disappointing somehow. There's a feeling that mathematical proof should be above mere argumentative skill and the audience's susceptibility to being convinced.
Well, no, that's the point of this, it's not 50%. Before you pick there is a 1/3 chance of the car being behind any of the doors, so there is a 2/3 chance of it cumulatively being behind the other two. But there is only one car, so the set of doors you didn't pick must include at least one non-car. The host knows the location of the car so can always open a non-car door from this set. But still as a set they have a 2/3 chance of containing the car. By swapping you can effectively think of it as choosing both other doors. An easy way to get your head around this is increase the number of doors, say to 100. You pick 1 of those 100, the host opens 98 other doors, do you swap?
Notice that all of the externally visible actions are identical to the original game.
I don't get this. Why does't the player insist that they open door no 1. Or is it part of the game that the host will open a different door other than the one the player picked...
If the host knows what is behind the doors, and you are only given two chances, and the host can open a different door from what you picked, isn't your chance of winning zero?
I don't get this...
Turns out the host just ends up giving you more information that you can use to make a better decision.
>If the host knows what is behind the doors, and you are only given two chances, and the host can open a different door from what you picked, isn't your chance of winning zero?
Also the answer becomes intuitive once this scales to more than 3 doors (someone below already mentioned).