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I agree with the parent poster that the participant's awareness of whether the host acted randomly or made an informed decision is critical for the participant to decide whether set B's 2/3 probability shifted/concentrated into the one unopened door in set B or whether set B's overall probability got reduced.

I can illustrate this with a variation to demonstrate that revealing a goat in the door is not that important compared to whether the host knowingly opened that door. For example, say the host blasted the door (and it's contents) instead of opening and revealing what's inside. Now it becomes critical to know whether the host randomly blasted it or whether it is guaranteed that he would never blast a door with car inside it. That knowledge rather than the 'reveal' of what's inside the door he selected (to open or blast) is what influences my decision to recalculate or keep the probability of set B.

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This. However, the alternative scenarios aren't the "Monty Hall Problem" - - the host will never open/blast a car door.

I usually find this problem annoying, not because it's all that difficult, in fact it's quite intuitive - when you're told the exact parameters defining the Monty Hall Problem and systematically work through them.

In my experience though, it's used more often as an exercise in diminution, a sick wet dream of probability teachers, where the learning party isn't aware of the problem, and usually either hasn't been explained, or doesn't quite grasp, the exact circumstances around whether the host's choice is random or decided.

There are lots of "it depends" moments that can be applied to incomplete descriptions of the problem, including (amazingly) whether the host offers a choice at all - this is the one that seems to trip up most people, as they might start to question the "motives" of the host (which are irrelevant in the actual statistical problem).

See my response in a sibling. The host revealing does matter, in that otherwise you don't get to act on the reveal. If the host just blasts it away as you said, then the probabilities don't change. If the host reveals, then there is a chance you don't even get to the swap before you lose. But if you do get to the swap, you have better odds of winning.

This is wrong. Whether "host reveals car" leads to win, loss, or replay, once you are faced with the choice the odds are 50/50 if the host picked randomly.

Isn't that what I said? Your overall odds of winning the game are not 50/50, but once you are at the choice, if you are choosing between two doors at the end, you are then at a 50/50 chance if you swap.

Hmm. You said, "if you do get to swap, you have a better chance of winning." Rereading, it seems you might have meant "better than 1/3, namely 1/2 - same as if you do not switch", which does seem correct. If he blasts the door, your odds of winning with any strategy (that does not involve cheating) is 1/3.

To be clear, if the host picks randomly (whatever happens to the door he picks) your odds are the same whether you switch or not.

So it does sound like we disagree some. If the host picks randomly and does not reveal, it makes no difference. If the host picks randomly, revels, and you are still in the game. It should make a difference. (After all, you now know more than you did before your first pick. Namely, that a 1/2 chance after a 1/3 exclusion did not remove the winning door.)

Will try and run a simulation tonight or this weekend.

I haven't had a chance to do the simulation yet, but the symmetry of the odds here finally dawned on me.

Specifically, if you have a 50% chance of winning on swap, you have a 50% chance of losing by not swapping. So, yeah, by the time you get to the swap, no matter what, you are at a 50% chance of winning. Swap or not.

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