[datruth29]

On the initial choice yes. But on the second choice, that other door is a single door that is the sum of the odds of the other 99 doors. So you're second choice would be to keep the door you initially chose (1/100) or select the other door (99/100).

Remember, the host always knows which is the correct door, and if you selected incorrectly on the initial choice they will ALWAYS select the correct door for the second choice.

[ricardobeat]

I thought it would be obvious that I’m not arguing the statistical facts, but the idea that “it is easier to think about” the 100 doors scenario. There is simply no straightforward explanation that works for laypeople.

[coderenegade]

I think the issue most lay people have is that the host opening a door changes the odds of winning, because he knows where the prize is.

I think the easiest way to demonstrate that this is true is to play the same game with two doors, except the host doesn't open the other door if it has the prize behind it. This makes it obvious that the act of opening the door changes the probability of winning, because if the host opens the other door, you now have 100% chance of winning if you don't switch. Similarly, if they don't open the other door, you have a 0% chance of winning, and should switch. It's the fact that the host knows and chooses that is important.

It's only once you get over that initial hurdle that the 100 door game becomes "obvious". You know from the two door example that the answer isn't 50/50, and so the only answer that makes sense is that the probability mass gets concentrated in the other door.

[TeaBrain]

It's probably easier for most people to not think of them as two remaining doors, but two remaining sets. Originally, with one hundred doors, if the goal object is only behind one of them, then there would be a 1/100 probability it would be behind the initially chosen door, which comprises one set, while there's a 99/100 probability that the goal object is behind one of the doors in the set of not originally chosen doors. If 98/99 of the doors in the not originally chosen doors set are excluded as having the goal object, then this does not change that there's a 99/100 probability that the goal object is behind a door in this set, it just means it wasn't one of the other doors in the set.

[iasondemiros]

It is easier to think about this in the case of a very large number of doors. It is unlikely that you could have picked the right door from, say, 1B.