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Probability Problem

Thermal said:
It's a relativistic game. The odds are fiddy fiddy for any observer who did not know which door you picked, whether they were there at the beginning or walked in when there were only two left. But you have the additional information of the process of elimination which changes the odds from your perspective.
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I'm not sure that is correct?
For instance, if you write down all the permutations for each door you start with, objectively in 99 of those permutations you end up on a goat and only in 1 permutation you are on a car, so swapping is objectively the correct thing to do.
Any observer who walks in at the end and just sees your door and one other, if they think it's 50/50 they're wrong, and it can be demonstrated they're wrong by again just writing down all the permutations, including the observer picking your door 50% of the time and picking the other door 50% of the time.

If their odds were indeed 50/50 then they would end on the car half the time, but they obviously won't for the same reason the player doesn't end on the car half the time, but on a goat 99 times out of the 100.
 
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p0lka said:
I'm not sure that is correct?
For instance, if you write down all the permutations for each door you start with, objectively in 99 of those permutations you end up on a goat and only in 1 permutation you are on a car, so swapping is objectively the correct thing to do.
Any observer who walks in at the end and just sees your door and one other, if they think it's 50/50 they're wrong, and it can be demonstrated they're wrong by again just writing down all the permutations, including the observer picking your door 50% of the time and picking the other door 50% of the time.

If their odds were indeed 50/50 then they would end on the car half the time, but they obviously won't for the same reason the player doesn't end on the car half the time, but on a goat 99 times out of the 100.
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The independent observer would choose the car half the time, assuming he knew nothing of the process of elimination. He has a one or the other choice, I'm pretty sure.
 
Thermal said:
The independent observer would choose the car half the time, assuming he knew nothing of the process of elimination. He has a one or the other choice, I'm pretty sure.
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We already know that out of the 100 permutations, the player is on a goat 99 times and on the car 1 time.
The observer picks the players door in 50 of those permutations and the other door in the other 50 permutations, it's not possible for the observer to be on the car half the time, when we also know that the observer has picked the players door half the time.
 
p0lka said:
We already know that out of the 100 permutations, the player is on a goat 99 times and on the car 1 time.
The observer picks the players door in 50 of those permutations and the other door in the other 50 permutations, it's not possible for the observer to be on the car half the time, when we also know that the observer has picked the players door half the time.
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I think you're missing that the newcomer does not know how the other 98 doors were opened. IOW, it could be that the player got 99 picks at doors, and the one he's on now is his last.

The odds only change if you have the knowledge of how the other doors were opened.

ETA: Or more accurately, the newcomer can only evaluate the odds on the basis of information he has; if he doesn;t know how the process worked that got them down to 2 doors, he can't evaluate the odds accurately.
 
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SumDooder said:
So I've been going back and forth with a friend of a friend online regarding the Monte Hall Problem. It has devolved into disagreements about probabilities and I'm having a hard time. He has gone into 'actual' and 'perceived' odds. My latest example is 'When you dive by my house, I'm either sitting on the porch or I'm not. What is the probability I am sitting on my porch when I drive by.' He says 'each time there will be a 1 in 2 chance'. Which I kind of get, given there are two possible outcomes (sitting on my porch or not) to one event (him driving by my house). He calls those 'actual' odds while 'perceived' odds can't be calculated until my parch sitting habits are examined. I assume there is some sort of fallacy involved saying out of two options they both have an equal chance of happening, but google is failing me. Does anyone have any good examples that demonstrate this fallacy?
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It is not a fallacy. If you have no information about the outcome of a trial whatsoever except that it may or may not happen then it is valid to assign a probability of 0.5 to that outcome.

The thing is that we actually have a lot of information about people sitting on porches and others have mentioned some of the factors that may influence the probability of seeing somebody sitting on a porch. We may not be able to give a reliable estimate of the probability but from what we can deduce, we can reliably say the the probability would be less than 0.5.

"Perceived" and "actual" odds is not a thing. What we have is "conditional" probability. That means the probability of an outcome depends (usually) on the conditions of the trial. In fact, there is no such thing as "unconditional" probability. A common condition is that there is a number of outcomes of equal probability and that the outcome is random. In a coin toss for example, we say that heads or tails is equally likely and don't consider things like the coin landing on its edge or snatched mid-air or being tossed by a cheat etc.
 
Thermal said:
The independent observer would choose the car half the time, assuming he knew nothing of the process of elimination. He has a one or the other choice, I'm pretty sure.
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Ah, I think you are correct, I just had another think about it, doh, sorry.





Hellbound said:
I think you're missing that the newcomer does not know how the other 98 doors were opened. IOW, it could be that the player got 99 picks at doors, and the one he's on now is his last.

The odds only change if you have the knowledge of how the other doors were opened.

ETA: Or more accurately, the newcomer can only evaluate the odds on the basis of information he has; if he doesn;t know how the process worked that got them down to 2 doors, he can't evaluate the odds accurately.
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Yeah, I confused myself, sorry. :o
 
SumDooder said:
So I've been going back and forth with a friend of a friend online regarding the Monte Hall Problem. It has devolved into disagreements about probabilities and I'm having a hard time. He has gone into 'actual' and 'perceived' odds. My latest example is 'When you dive by my house, I'm either sitting on the porch or I'm not. What is the probability I am sitting on my porch when I drive by.'
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There is no chance that you are sitting on your porch if you are driving by, because you are obviously in your car.

As for your problem with the Monte Hall puzzle, the simplest way to prove it's better to switch is to sit down with your friend and play it out a few times.
 
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SumDooder said:
So I've been going back and forth with a friend of a friend online regarding the Monte Hall Problem. It has devolved into disagreements about probabilities and I'm having a hard time. He has gone into 'actual' and 'perceived' odds. My latest example is 'When you dive by my house, I'm either sitting on the porch or I'm not. What is the probability I am sitting on my porch when I drive by.' He says 'each time there will be a 1 in 2 chance'. Which I kind of get, given there are two possible outcomes (sitting on my porch or not) to one event (him driving by my house). He calls those 'actual' odds while 'perceived' odds can't be calculated until my parch sitting habits are examined. I assume there is some sort of fallacy involved saying out of two options they both have an equal chance of happening, but google is failing me. Does anyone have any good examples that demonstrate this fallacy?
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There are 2 ways of understanding probabilities: Frequentist and Bayesian.

The frequentist view is that a probability is information about how often something happens; or the frequency at which something happens. If you told a frequentist that you sit on the porch 1 hour each day, then he would say that the probability of finding you on the porch at any time during the day is 1/24.
Without any such information, the frequentist would deny that the question makes sense: You either sit on the porch or you don't. The probability is either 0 or 1 and we simply don't know either way.

The Bayesian view is more abstract. Probability is a statement about your knowledge or lack thereof.
Under this view you could just pull a number out of your ass. You could say: At this time of day, almost no one sits on the porch so the probability is 5%.
Or you could say:He knows I'm coming so he's probably waiting on the porch. The probability is 80%.
If you really don't know either way you might as well chose 50/50.
Neither number is right or wrong in any mathematical sense. Bayesian statistics provides mathematical methods to refine the initial guess (called prior probability) with empirical data. Every time you drive past the porch and see someone sitting there or not, your guess is nudged closer to the true value, regardless of how right or wrong your initial guess was.
How to best come up with a prior is subject to contentious debate. https://en.wikipedia.org/wiki/Prior_probability#Uninformative_priors
 
theprestige said:
What I want to know is, is it even possible to have a "raw" likelihood, without any modifying information?
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Just use the maximum entropy distribution where the given information is the empty set (ie "without any modifying information" to put it in your terms) - it works just as well when the information set is empty as when it isn't.
 
So your friend is either on the porch or not, so it's 1 of 2.

So your friend is either on the porch, or on the sofa or on neither, so it's 1 of 3.

So your friend is either on the porch, or on the sofa or in the kitchen cooking dinner or none of those, so 1 in 4.
 
I'm surprised no one has brought up the classic example:

When visiting Egypt, poster shemp is either dry-humping the Sphinx, or he is not. 50/50 odds?
 
bobdroege7 said:
So your friend is either on the porch or not, so it's 1 of 2.

So your friend is either on the porch, or on the sofa or on neither, so it's 1 of 3.

So your friend is either on the porch, or on the sofa or in the kitchen cooking dinner or none of those, so 1 in 4.
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That is similar to the fallacy used to prove that it is almost certain that there is life on planet X.

We have no knowledge of whether there are horses on planet X or not so the probability is 50%
We have no knowledge of whether there are cows on planet X or not so the probability is 50%
So the probability that there are neither horses nor cows on planet X is 50% x 50% = 25%.

Add all the other species of life on Earth in the mix and we can conclude that the probability that none of them exists on planet X -> 0% so it is almost certain that some form of life on planet X exists.

The flaw is that these are all conditional probabilities but we are not expressly told what conditions have been assumed.

The starting point is that the probability that life can be supported on planet X is 50%. The probability that life exists on planet X given that life can be supported on planet X is 50% etc etc.
 
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