[skizm]

The premises of the riddle in both cases is that the host reveals a goat. There is no randomness. The fact that there was a chance the host might show a car doesn't matter. He showed a goat. The probability of the host showing a goat in both cases is 1.

I understand the math. I know what the op was trying to say. But they worded it wrong. The 2nd riddle is the same as the first.

[dllthomas]

There are two separate things here. First, the premise of the riddle. Second, the procedure of the game show (as fictionalized within the riddle).

If "... and shows you a goat" is a part of the procedure of the game show, there is additional information imparted by the host's selection compared to the situation where "... and shows you a goat" is an artifact of the particular play through the game that is being described in the riddle and the procedure by which that situation was arrived at did not incorporate knowledge of where the car was.

[skizm]

The question is phrased ad-hoc. We know the host picked a goat in both scenarios. Decision time. Do we switch or no? It didn't ask: "On an infinite timeline where the host might not always pick the goat, should we switch and if so when?" Which I think is what op was trying to say.

[dllthomas]

It matters why the host picked the goat. (Or, more precisely, what the odds were of him doing otherwise when he picked). That we're looking at a goat doesn't tell us everything.

Again, write the simulation. Either the switch strategy will win 2/3 of the time or 1/2 of the time, depending on how you write Monty. Only one of those approaches matches the results we saw IRL - Monty never revealed the car (so far as I'm aware) - but either is consistent with many phrasings of the question, and OP correctly specified the other case above.

[skizm]

you keep telling me to write a simulation. It won't matter. the code is the same for both questions since they are asking the same thing.

[dllthomas]

That's not true, but we'll pick it up under the code jayvanguard supplied.

[Cushman]

No. Even in the case of a single trial, it matters.

If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point.

If the host always chooses a goat, and the host picks a goat, and you switch, your odds are 2/3 in favor of now having a car. You should always switch.

Intuition suggests this. Statistics proves it. If you implement this in software, you will see it happening to you in black and white, right before your eyes.

[dllthomas]

"If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point."

1/2, once you see the goat. One of 1/3, 2/3, and 1/2 beforehand, depending on what happens when Monty shows a car: whether you automatically win, automatically lose, or start over from the top, respectively.

[Cushman]

Right, to be clear, that's your total chance of walking away from the show with a car. Switching doesn't change it.

[dllthomas]

Right.

[Cushman]

> The probability of the host showing a goat in both cases is 1.

Are you familiar with the concept of prior probability? The premise of the meta-riddle is that both hosts showed a goat, but one host could have not shown a goat. So there must be a difference in the information conveyed by that goat.

Again, there isn't anything special about this problem, it's just how probabilities work. Respectfully-- there are a number of different people trying to explain this to you in different ways, and you can verify it on Wikipedia. Are you positive you do understand the math?