5 ms·
Most of the subtlety of the problem lies in this small sentence in the Side Notes: "He deliberately chooses to show you goats." This is not made as explicit in
by tromp 6y ago
Most of the subtlety of the problem lies in this small sentence in the Side Notes:
"He deliberately chooses to show you goats."
This is not made as explicit in the standard formulation
"the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat."
Which could be read as "which happens to have a goat".
It's the ambiguity in Monte's door opening strategy that leads to different answers.
- lcuff 6y agoOne of the realities we face here is that different people have different brains which work differently. Thus different phraseology is going to help. Adding variant phrases such as "He will never show you a car", will probably help for some.
- tromp 6y agoIt's also possible that Monty only gives you the option to switch if you chose the car. Or only if you chose door No 1. All of which affect the answer. Therefore, it must be made fully explicit what Monty can and cannot do.
- basch 6y agoWhether or not he knew or whether it was luck, it doesnt exactly matter, although traditional stats knowledge would make you think it does. The important thing to understand is that the premise of the question says he will show you a goat. If you rerun the experiment 10,000 times, he will show you a goat 100% of the time, either through peaking, premonition, or consistent luck. The problem gets trickier because people start applying domain knowledge of stats, and treating it as a simulation with random events. The goat being chosen is not random, it is an event that occurs 100% of the time in the premise of the thought experiment. Thinking about the random chance of him choosing the car is outside the bounds of the axiom/postulate we start with. tldr: it doesnt matter how he opened a goat door, all that matters is that he did.
- tromp 6y agoYou said it well: "he will show you a goat." But the standard formulation just says Monte opened a door and it had a goat behind it. No explicit mention of intention. It matters not that he happened to do. It matters that he will.
- spullara 6y agoIt doesn't, as long as he always shows a goat.
- roywiggins 6y agoWhat if he didn't know and just happened to open the goat door? Should you still switch?
- pxx 6y ago
- joppy 6y agoWhat door Monty opens is irrelevant. He shows you a door with a goat, you’re better off switching. He shows you a door with a car, you’re still better off switching. What matters is that when you first choose a door, you have most likely chosen a goat.
- Tomminn 6y agoFirst caveat: If he shows you a door with a car, there is literally no point switching. There's a goat behind both doors you're allowed to choose from, which are the remaining two. (Below I use non-chosen to mean non-chosen by the contestants initial choice.) Second point: If we are in the subset of all possible histories where Monty picked randomly revealed a goat, then we will have 50% of histories where both non-chosen doors contain 1 goat selected by our history subset, and 100% of histories where both non-chosen doors contain 2 goats selected by our history subset. Since there are twice as many possible histories where the non-chosen doors contains 1 goat vs 2 goats, after selection, we have an equal number of histories in our sample where we have 1 goat or 2 goats behind the non-chosen doors. Or equivalently, we have a 50% chance that the non-chosen doors contain a car. Therefore it is irrelevant whether you switch. Monty needs to make an intelligent selection to change the game.
- joppy 6y agoI interpreted the original game as follows: you can freely choose any of the three doors. Monty then opens one of the doors that you didn't pick, and then you may again freely choose any of the three doors. (In the standard problem, there is no reason why you would choose Monty's door). The question is: does the strategy "choose a new door" have better odds of winning? We can work out odds for this "always switch away from original door" strategy: suppose you initially chose a door with the car (1/3 probability). Then choosing a new door makes you surely lose regardless. On the other hand, suppose you initially chose a door with a goat (2/3 probability). Then regardless of which of the two doors Monty opens, you can choose the car (if he revealed the car, choose that. If he revealed the goat, choose the other door). So our odds of winning with this strategy are still 2/3. So it's up to the interpretation of the modified game I guess.