I may be missing something (not a mathematician, probabilist) but doesn’t the fact that the host would never reveal the car change the calculation? In the table shown in the article game 3 & 6 are nonsensical so with the remaining 4 games the odds are even between switch & stay.
The fact that the host never shows the car is critical to the solution. You choose a door. There's a 2/3rd chance that the car is behind one of the other doors. Monty is kindly going to show which of the other two doors not to choose. ... The other way to think of it is, you choose a door, and then you're asked "would you like to keep what's behind your door, or would you like to have the sum of the other two doors?"
If the host opens the door with car, and asks you if you'd prefer to switch to the unopened door, then you have a 100% chance of being wrong, as the door with the car isn't one of the choices. But, sure, my second point still stands, that it's effectively asking if you want your door or the sum of the other two doors.
But, to be clear, the question states that Monty opens the door with the goat. If you were to write a monte Carlo to test this empirically, you'd have to have your "Monty" choose the door with the goat.
You appear to have completely edited the comment I was replying to. My "not at all" does not apply to your comment as it currently reads, which is correct.
On HN it is considered polite to append corrections rather than replace content that has become part of the discussion.
No, it's correct. The easiest explanation, IMHO, is:
1. If you already selected the correct door, you should not switch, obviously. Equally obviously, the odds of being in this state are 1/3, because your initial choice was random.
2. If you did not select the right door, then the host clearly revealed the only one of the two remaining doors without a prize. In this state, you clearly should switch to the only door remaining, which contains the prize.
That is, the likelihood that switching will get you the prize is 2/3, because it corresponds to the states where you initially guessed wrong.
This is a very good explanation. I hope you won't mind if I say the exact same thing you did in just a slightly different way that might help it click for someone still not getting it.
Your initial chances of choosing correctly are 1 in 3. So the car is behind the door you chose 1/3 of the time. After the host reveals which of the 3 doors definitely does not have the car it's still the case that your initial choice will have the car 1/3 of the time [0]. That means the other door must contain the car 2/3 of the time. So you should switch.
[0] The haters are correct that the host hasn't changed the odds that your initial choice is correct. They just aren't tracking the implications of that fact.
The fact that Montey only reveals a goat is key to the probability calculation. When the player selects a door, they have a 1/3 probability that they have chosen the correct door, and Montey has a 2/3 probability of having the car.
When he reveals the goat (which he always had behind one of his doors) that 2/3 probability shifts to the remaining door. There is no chance in Montey's selections, he knows where everything is to begin with.
Thanks for the helpful examples, makes sense. I had assumed it was accurate and I just wasn’t seeing it so good to understand why (also sleeping 4 hours a night with a new puppy so that doesn’t help).
