3 ms·

First 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 ar

First 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 ago
I 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.
- deleted 6y ago
  [deleted]
- Tomminn 6y ago
  Okay, let's go with your interpretation. Monty has now randomly chosen a door, and not revealed a car. If he had revealed a car you would have changed doors to it and won. But he didn't.[0] Now you have a choice to make. His probability of not revealing a car in the case that both doors you didn't pick had goats behind them, is 100%. His probability of not revealing a car in the case that one door you didn't pick had a car behind it, is 50%. We can reason then, that of the three possible scenarios for the two non picked doors {g,c}, {c,g}, {g,g}, that it is equally likely now that since he didn't randomly reveal a car, {g,g} must be weighted twice as much as the other two. Therefore, we must be as likely to have a goat behind the remaining door as we didn't. In conclusion, Monty only changes the problem if he selects with information. [0] Notice that at this point, we have to throw away from the decision tree 1/3 of the histories, and in all of them switching was beneficial. In what remains then, switching is less beneficial than it was before this trimming.
- kgwgk 6y ago
  > So our odds of winning with this strategy are still 2/3. That’s true before he opens a door. Then either A) he shows a car and the odds of winning with this strategy are 100% or B) he shows a goat and the odds of winning with this strategy are between 1/2 and 2/3 depending on how the choice of door was made If the choice is random, he shows a car with probability 1/3. The probability of winning is 1/3 x 1 + 2/3 x 1/2 = 2/3 But that analysis is for the unconditional problem and what we’re asked is what to do in case B, after a goat has been unveiled. If he shows always a goat, the probability of winning is 0 x 1 + 1 x 2/3 = 2/3. Here there is no difference between the unconditional and the conditional problems because A never happens.

joppy 6y ago

I 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.

deleted 6y ago
[deleted]
Tomminn 6y ago
Okay, let's go with your interpretation. Monty has now randomly chosen a door, and not revealed a car. If he had revealed a car you would have changed doors to it and won. But he didn't.[0] Now you have a choice to make. His probability of not revealing a car in the case that both doors you didn't pick had goats behind them, is 100%. His probability of not revealing a car in the case that one door you didn't pick had a car behind it, is 50%. We can reason then, that of the three possible scenarios for the two non picked doors {g,c}, {c,g}, {g,g}, that it is equally likely now that since he didn't randomly reveal a car, {g,g} must be weighted twice as much as the other two. Therefore, we must be as likely to have a goat behind the remaining door as we didn't. In conclusion, Monty only changes the problem if he selects with information. [0] Notice that at this point, we have to throw away from the decision tree 1/3 of the histories, and in all of them switching was beneficial. In what remains then, switching is less beneficial than it was before this trimming.
kgwgk 6y ago
> So our odds of winning with this strategy are still 2/3. That’s true before he opens a door. Then either A) he shows a car and the odds of winning with this strategy are 100% or B) he shows a goat and the odds of winning with this strategy are between 1/2 and 2/3 depending on how the choice of door was made If the choice is random, he shows a car with probability 1/3. The probability of winning is 1/3 x 1 + 2/3 x 1/2 = 2/3 But that analysis is for the unconditional problem and what we’re asked is what to do in case B, after a goat has been unveiled. If he shows always a goat, the probability of winning is 0 x 1 + 1 x 2/3 = 2/3. Here there is no difference between the unconditional and the conditional problems because A never happens.