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Comment by jjnoakes | Hacker News Reader

The original situation covers all scenarios. Any choice of door you make and all possible doors the host might open. The odds are the same every time.

The proposed 50/50 scenario only makes sense for some possible situations - those where the host draws a door at random AND the door isn't the same one I chose AND the door opened isn't the car... once you put that situation on equal footing with the original one (all possible doors chosen and all possible doors opened) then it isn't 50/50 any more. Some of those situations are 100% wins (I chose door 3, host opens door 3, car is behind door 3) and some are 100% losses (I chose door 2, host opens door 1, car is behind door 1). Once you look at all outcomes, 50/50 doesn't apply. It becomes 33/66.

If your point is that you can cherry-pick a subset of outcomes and get 50/50 odds then I can't really argue, but I'm also afraid I don't how that's useful.

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> If your point is that you can cherry-pick a subset of outcomes and get 50/50 odds then I can't really argue,

Why do you say “cherry-pick a subset of outcomes”? It’s the subset of outcomes relevant for the question.

Consider the simple problem that I proposed: “You're given the choice between two envelopes. One contains $1 and the other $10. You open an envelope and find $1. Is it to your advantage to switch to the other envelope?”

Do you agree that the answer is “yes” or would you argue that “once you look at all outcomes” the 100% probability of getting $10 by switching doesn’t apply?

> but I'm also afraid I don't how that's useful.

It’s useful to answer the following question:

“You pick door #1. The host picks a number between 1 and 3 from a hat and opens the corresponding door #3. There is a goat behind the door. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?”

What’s your answer?

That's about as relevant as this:

"I claim the odds of flipping a fair coin 100 times in a row and getting heads every time is 50/50, but only in the situation where you happened to flip 99 heads in a row already".

Of course it is true, but again, not useful in my opinion.

What about the envelopes problem? Do you think that claiming that you should switch the envelopes “but only in the situation where you happened to” find $1 in your first pick is not relevant? The question asked is precisely what to do in that particular situation! Knowing that if you switch you get more that if you don’t seems useful…

I’m also curious about your answer to the following question, really.

What do you think that the probabilities would be if you you were to find yourself in that situation? A hint: the probability that the car is behind the open door is zero. What about the others?

Don’t you have any answer at all?

I'm not sure why you keep asking the same question. The answer to the second one is here: https://news.ycombinator.com/item?id=36752870 and the answer to the first one is basically alluded to here: https://news.ycombinator.com/item?id=36758795

In your hypothetical situation, the rules are something like:

"Pick a door, then the host will randomly select one of the other doors. If the host doesn't select your door (1/3 odds) AND the host doesn't open a door with a car behind it (2/3 odds) then you can choose to switch or keep your door."

The odds of winning _after_ the host already beats 1/3 and 2/3 odds is indeed 50/50, but the odds of playing the game _overall_ is 33/66.

The only real way I can see of making this situation analogous to the Monty Hall problem is to consider the odds of the whole game (like Monty Hall does, which is always 66% chance of winning by switching, every time, no matter the path the game takes from start to end). If you want to cherry-pick just the situation where the host ALREADY beat some odds, then you do get 50/50, but it really is a non-sequitur in this thread (or in any thread about Monty Hall).

> The odds of winning _after_ the host already beats 1/3 and 2/3 odds is indeed 50/50, but the odds of playing the game _overall_ is 33/66.

The problem is not about the game overall anymore than the question about whether you'd like to switch the envelope after you got $1 is about the game overall. The question is whether you want to switch conditional on the situation where you got $1!. That's not cherry-picking - it's considering the exact situation described in the problem.

> The only real way I can see of making this situation analogous to the Monty Hall problem is to consider the odds of the whole game

The Monty Hall problem is not about "the whole game". The Monty Hall problem is quite explicitely about what would you do in the following situation:

1) you were given the choice of three doors: one car and two goats

2) you picked a door

3) the host opened another door

4) there was a goat behind that door

After all those things have happened you're being offered to switch from your initial pick to the remaining door.

What is the probability - after all those things have happened - that the car is behind each of the doors?

If you don't agree that this describes the problem there is no point in reading further. We can just agree that we have very different understandings of what the problem is about.

-----

If you agree with the description above the answer to the question depends your assumptions about how the host's choice in step 3) was done.

There are many different assumptions that could be made.

If he was avoiding the door you picked and the door with the car - as the original formulation of the problem implies - the probababilities are 1/3 and 2/3 (for the door you picked and the remaining one respectively).

If he was completely ignoring the location of the car when he made the choice the probabilities are 1/2 and 1/2 (it doesn't matter if he was avoiding your door or not).

> The problem is not about the game overall > The Monty Hall problem is not about "the whole game".

The Monty Hall problem is about the whole game. That's the difference, and that's why the 50/50 claim doesn't make much sense here.

> The Monty Hall problem is quite explicitely about what would you do in the following situation: [...] After all those things have happened you're being offered to switch from your initial pick to the remaining door.

Steps (3) and (4) in the Monty Hall problem are a bit more nuanced - the host opens a door guaranteed to be a goat, because the host knows where the goat is.

At this point, when you choose to switch or stay with your door, you have a 1/3 chance of winning if you stay, and a 2/3 chance of winning if you switch.

But the key point that you are missing is that this is the same for every single time you play the entire game. It doesn't matter where the car is in step (1), or which door you choose in step (2), or which door the host opens is step (3) - if you always switch doors, no matter what, every single game you double your odds.

And in the 50/50 scenario posed, that just doesn't hold. It isn't analogous.

> But the key point that you are missing is that this is the same for every single time you play the entire game. It doesn't matter where the car is in step (1), or which door you choose in step (2), or which door the host opens is step (3) - if you always switch doors, no matter what, every single game you double your odds.

The key point that you are missing is that it doesn't need to be "the same for every single time you play the entire game" to have a well-defined solution for the particular realization of the game described in the problem statement.

And the solution for the problem where (1),(2),(3),(4) happen with the host picking a door at random is 50/50. That's what cman1444 said.

You're asked a question conditional on (1),(2),(3),(4) happening and the things that didn't happen are irrelevant.

Just like if you're asked if you want to switch the envelopes when you got $1 you will say "yes" and it's completely irrelevant that you could have got $10 and you would have said "no" in that case - because you didn't.

In conclusion, some people find the problem where (1),(2),(3),(4) have happened and the host was picking a door at random (and its 50/50 solution) interesting and you don't. De gustibus non est disputandum.

I addressed this above. See the coin flipping example.

> I addressed this above.

I'm not sure what is "this".

> See the coin flipping example.

"I claim the odds of flipping a fair coin 100 times in a row and getting heads every time is 50/50, but only in the situation where you happened to flip 99 heads in a row already".

If I'm told that I'm in a coin flipping contest and (1) in my first flip I got heads, (2) in my second flip I got heads, ..., (n) in my n-th flip I got heads, ..., (99) in my 99-th flip I got heads and then I'm asked what's the probability that I get to 100 heads in a row there are different answers that I could give.

I could assume that the probability of every single flip is 50/50 and then my answer will by 50%. Or I could give an answer higher than 50% if I'm able to control the outcome - or if by now I suspect that the coin has two heads. Or I could give an answer lower than 50% if I suspect the game is rigged and they are going to make me lose now.

Whetever my assumptions about the upcoming flip, my answer will be conditional on (1),(2),...,(99) having happened already. There is no point in thinking about the "whole game" at this point in time. I already got heads 99 times - just like in the Monty Hall problem I already picked a door and he already picked another door with a goat.

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