3 ms·

Indeed. Looking at this from an information point of view, the fact that the host can open a door tells you nothing as to whether your first choice was right.

Indeed. Looking at this from an information point of view, the fact that the host can open a door tells you nothing as to whether your first choice was right. So the chance you picked the right door is always 1/3. The information you get is which door to open if you are allowed to switch.

Jensson 5y ago
The confusing part is that him opening the door doesn't reveal any new information. In most cases knowing that an event happened changes other probabilities. Or in other words, probability of A is equal to probability of A given B in this case, where A is that your initial pick is correct and B is that he opened a door with a goat. For example, lets say that he opened two doors with goats instead, then what is the probability of A, ie your first guess being correct? It is 100% of course, since it is the only door left, him opening two doors with goats revealed new information about your door. So the argument "your first pick is always 1/3, so you should always switch" is wrong. That doesn't normally apply. It is true in this case, but not true in the general case of problems like this since probabilities change as new information is revealed.
- ColinWright 5y ago
  > The confusing part is that him opening the door doesn't reveal any new information. It does reveal new information, just not about the door you chose. > lets say that he opened two doors with goats instead, then what is the probability of A, ie your first guess being correct? But you can't guarantee that he can do that. If you do guarantee that he can do that then you are giving me more information. It's always the way, if you change the problem you change the analysis. > So the argument "your first pick is always 1/3, so you should always switch" is wrong. That doesn't normally apply. Just because it doesn't normally apply, that doesn't mean it's wrong in this case. It does apply in this case (as you say) and therefore it's right. Knowing when it's right is hard, I agree.

Jensson 5y ago

The confusing part is that him opening the door doesn't reveal any new information. In most cases knowing that an event happened changes other probabilities. Or in other words, probability of A is equal to probability of A given B in this case, where A is that your initial pick is correct and B is that he opened a door with a goat. For example, lets say that he opened two doors with goats instead, then what is the probability of A, ie your first guess being correct? It is 100% of course, since it is the only door left, him opening two doors with goats revealed new information about your door. So the argument "your first pick is always 1/3, so you should always switch" is wrong. That doesn't normally apply. It is true in this case, but not true in the general case of problems like this since probabilities change as new information is revealed.

ColinWright 5y ago
> The confusing part is that him opening the door doesn't reveal any new information. It does reveal new information, just not about the door you chose. > lets say that he opened two doors with goats instead, then what is the probability of A, ie your first guess being correct? But you can't guarantee that he can do that. If you do guarantee that he can do that then you are giving me more information. It's always the way, if you change the problem you change the analysis. > So the argument "your first pick is always 1/3, so you should always switch" is wrong. That doesn't normally apply. Just because it doesn't normally apply, that doesn't mean it's wrong in this case. It does apply in this case (as you say) and therefore it's right. Knowing when it's right is hard, I agree.