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If you see this problem as a special case of another problem the answer becomes pretty intuitive. Suppose there are N (let's pick 1000) doors: Behind one door
by casta 10y ago
If you see this problem as a special case of another problem the answer becomes pretty intuitive.
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?
- std_throwaway 10y agoSuppose there's two doors. You can't decide. The host opens one. There is no goat behind that one. Now you know for certain.
- p1esk 10y agoIt did not get any more intuitive for me.
- bmay 10y agoSay there are a trillion doors and the host closes 999,999,999,998 of them. Which scenario is more plausible: 1) Your 1-in-a-trillion chance original choice just happened to be correct 2) The omniscient host picked the correct door
- nerdponx 10y agoAgreed. 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.
- Jedd 10y agoThe most succinct description I've ever found of why you switch was so good I made a note of it, though sadly not the provenance: "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."
- justinpombrio 10y agoHow about this: 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.
- yellowstuff 10y agoLet me try. Say you pick the prize door initially. Obviously, if you switch you will now have a non-prize door. 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.
- p1esk 10y agoActually, this makes sense! Thanks.
- deong 10y agoIn teaching a bunch of discrete math classes over the years, I've very rarely found someone for whom the "imagine it's 1000 doors" version helps at all. The only way that gives any insight is if you've already internalized the correct answer. It makes the numbers more extreme; it does nothing to make them make intuitive sense. The version that helped you is the version that helps nearly everyone in my experience.
- diogofranco 10y agoI think the 1000 doors version does help, if you clearly explain the point that "now, if you switch, the only way you lose is if you had guessed the right door out of the 1000!"
- deleted 10y ago[deleted]
- casta 10y agoYou can pick one door, or all the other doors. What do you pick?
- hellogoodbyeeee 10y agoI've thought about this problem a long time, and the only thing that allowed me to wrap my head around it was to draw it out on a small piece of paper. 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).
- criddell 10y agoThe explanation that clicked for me was doing this with a deck of cards. If I put them all face down and tell you to pick the ace of spades. You choose your card randomly and I flip over all the others except one. So now the ace is either the one you picked or the one card I didn't flip over. Clearly, you should switch.
- sn9 10y agoThe thing that made it click for me is to think about having multiple trials. 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.
- kaoD 10y agoIt didn't get any more intuitive for me like that. If anything, less. 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.
- gerdesj 10y agoYou are only a small step away from "assume a spherical cow" or something involving Hilbert's Hotel. Which part of three (doors) are you having snags with?