3 ms·

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 m

by 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.

by joppy 6y ago

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.