Erik, I presume what you call the Monty Hall problem is in summary this:

- You’re in a game and you have to choose one of three doors behind which there may be a prize (apparently in this case a donkey, hmm, maybe not a prize then ;-)
- After you have chosen a door, the game show host opens one of the other doors behind which there is no prize.
- You get the chance to change your choice of doors.

The correct answer is that you should switch your choice without thinking, because it doubles your chance to get the prize. Is this the problem we are talking about?

Most people will intuitively feel that there is now a new situation where you have a 50% chance of getting the prize, because there are two doors left and one of them has the prize.

The reason this is not the case is in rule number 2 above. The important part is that the host chooses one of the **other** doors and never the door you have chosen. Now, this gives you two possibilities:

- You initially chose a door with the prize: (33% chance) -> This is easy for the game show host, he can chose any of the other doors. The other door has 0% chance to contain the prize. Switching will give you the wrong door.
- You initially chose a door with no prize (66% chance) -> The game show host has no choice. He has to choose the remaining empty door. Now this is interesting. The other door has 100% chance of containing the prize. Switching will give you the right door.

See, because the chance to choose a room with no prize initially is twice as high, you have a bigger chance that the game show host is forced to open the other empty room, which gives a bigger chance for the third room to contain the prize.

Does that help? Or were you talking about something else entirely? ;-)

#### Update:

Maybe this is a better summary:

If you choose right the first time, switching will never give you the prize. If you choose wrong the first time, switching will *always* give you the prize. There’s a much bigger chance you choose wrong (2 out of 3).