The Sniper and the Porta-Potty Problem

[Apple of Eris]

If you interpret it that way Vekkers, then you're right.

I just think the situation as described in the original question leaves enough ambiguity that the argumant could be made either way. It wasn't clear if everytime a sniper showed up a port a potty not being aimed at would open, or if this open potty would always be one not containing the officer. 

That I think is where the hang up remains, and that is all I'm arguing :)

[Xillen]

Okay, you have a set of random possibilities for the host to make. If the host picked a door completely at random, then you would have the distribution as I gave above. But, he doesn't, he puts constraints on it.

Likewise, the description of the problem makes these exact same constraints. For your purposes, the text of the problem itself acts as the 'sentient host' - the same criteria are followed.

Imagine it like this. Monty Hall decides to be a jerk, and he does not reveal a door. But instead, one of the goats - which just happens to be in a door that the contestant does not pick (note that the described situation here stands in for the conscious decision of the host), goes nutso and kicks its door open.

I still disagree.

If the contestant chooses the left door, there's a 50% chance that the host opens the right door. After all, the host won't open the left door, since the contestant is aiming at that.

Let's assume for a moment that the Contestant picks the left door. (chances for the middle and right door are equal to that).

33% chance that the car is behind the left door. 50% chance that the middle door is opened. The contestant opens the right door and loses (50% of 33% = 16%).
50% chance that the right door is opened. The contestant opens the middle door and loses (50% of 33% = 16%).

33% chance that the car is behind the middle door. 100% chance that the right door is opened. The contestant opens the middle door and wins (100% of 33% = 33%).

33% chance that the car is behind the right door. 100% chance that the middle door is opened. The contestant opens the right door and wins (100% of 33% = 33%).

So when the host makes sure not to open the door that the contestant picks, and not the door the car's behind, the contestant has a 67% chance of winning. I think we all agree by now.


Now let's assume that the door is opened completely at random. It's given in the story that a soldier opens the door, and the sniper knows that, so we'll open one of the two goat doors at random. Again, I'm going to assume that the contestant picks the left door at first, and that he switches door afterwards.

33% chance that the car is behind the left door. 50% chance that the middle door is opened. The contestant opens the right door and loses (50% of 33% = 16%).
50% chance that the right door is opened. The contestant opens the middle door and loses (50% of 33% = 16%).

33% chance that the car is behind the middle door. 50% chance that the left door is opened. The contestant is baffled and has to pick one of the other doors at random (50% of 33% = 16%). 50% chance that the right door is opened. the contestant opens the middle door and wins (50% of 33% = 16%).

33% chance that the car is behind the right door. 50% chance that the left door is opened. The contestant is baffled and has to pick one of the other doors at random (50% of 33% = 16%). 50% chance that the middle door is opened. the contestant opens the right door and wins (50% of 33% = 16%).

There is a total of 33% chance that the contestant switches door and is right. There is a total of 33% chance that the contestant switches door and is wrong. There's a total of 33% chance that the contestant HAS to switch door, leaving him of course with a 50% chance of that 33% chance is a total of 16% chance.

We already know this last 33% is not the case, since the text specifies that another door is opened, meaning only the first two 33% chances applying. They outweight each other, making the total chance of success equal to 50%.

[MagicalPen]

 :o  ???