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Probability Problem

SumDooder

Student
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Jun 28, 2018
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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?
 
The easiest way to illustrate Monty Hall is to say there are a hundred doors, not three.
Monty opens all but two.
The chance has clearly switched from 1:100 to 1:2.
 
There are several threads dedicated to this problem. I suggest searching them. Rehashing the same arguments can be tiresome.
 
There being two possible outcomes doesn't necessarily mean those outcomes are equally probable. Roll two dice and you either will or will not roll 6-6. The chances of rolling 6-6 are 1 in 36.

If you, however, spend a random 50% of your time on the porch then he'd be right in calling the chances 1 in 2.
 
lionking said:
There are several threads dedicated to this problem. I suggest searching them. Rehashing the same arguments can be tiresome.
Click to expand...

I don't think he's talking about Monty Hall specifically.
 
GlennB said:
There being two possible outcomes doesn't necessarily mean those outcomes are equally probable. Roll two dice and you either will or will not roll 6-6. The chances of rolling 6-6 are 1 in 36.

If you, however, spend a random 50% of your time on the porch then he'd be right in calling the chances 1 in 2.
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This.

Stating that there are two possibilities simply provide no useful information on the odds. There also is or is not an elephant in your front yard when he drives by.

Hans
 
There is either a flying saucer crew from the Omicron Persei system doing an interplanetary survey in my back yard right now, or there isn't. I haven't checked.

Fifty-fifty?
 
Myriad said:
There is either a flying saucer crew from the Omicron Persei system doing an interplanetary survey in my back yard right now, or there isn't. I haven't checked.

Fifty-fifty?
Click to expand...

Well, in fact, there's information there that can be used to form at least a vague estimate of the probabilities involved. Since none of us has ever seen a f.s.c.f.t.O.P.s, and history doesn't record one ever being seen, we would typically form an initial estimate that the odds were vanishingly small. On the other hand, since people are known to sit on porches, an initial estimate might be that the odds against are not too great - unless, of course, SumDooder's house doesn't actually have a porch, in which case we go back to vanishingly small. The point is that probability is about dealing with the level of uncertainty in the truth value of a statement, and it's very difficult to make a statement that conveys no information whatsoever as to its probability.

Dave
 
Again this is why I think the Three Prisoner Problem demonstrates it better then the Monty Hall Problem.

In the Monty Hall Problem people tend to get hung up on two factors; whether or not Monty is being honest and the fact that the individual can change their choice midway into the problem.

The Three Prisoner Problem removes those variables and still retains the same counter-intuitive nature.
 
It's just Bayes, isn't it?

You modify your estimate based on other assumptions and new information.

Either I'm a dog or I'm not. If you don't know a lot about dogs, you might provisionally put the probability of me being a dog much higher than someone who does know a lot about dogs.

And that's okay.
 
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?
Click to expand...
Tossing a bent coin. It will either land heads or tails, but the probability is not 0.5 heads and 0.5 tails.

In fact most texts will introduce coin examples with "assume a fair coin". My old uni statistics text calls statistics a "semi-empirical" branch of mathematics for this reason.
 
p0lka said:
The chance you are on the car is still 1/100, the chance the other door left is the car is 99/100, so you should swap. Where did the 1/2 come from?
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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.
 
JoeMorgue said:
Again this is why I think the Three Prisoner Problem demonstrates it better then the Monty Hall Problem.

In the Monty Hall Problem people tend to get hung up on two factors; whether or not Monty is being honest and the fact that the individual can change their choice midway into the problem.

The Three Prisoner Problem removes those variables and still retains the same counter-intuitive nature.
Click to expand...

With the Monty Hall problem you can stipulate that the rules state that Monty has to offer the chance to change and this makes it irrelevant as to whether or not Monty is honest.
 
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?
Click to expand...

If your house is underwater, I propose that the odds are slim you will be sitting on the porch.

If you mean your neighborhood is a dive, the odds are slim your buddy wants to go there anyway.

QED
 
Here's a money-making opportunity: When you cut a deck of cards, the result will either be the six of spades or not. So, according to the friend, it's 50-50. See if you can get him to bet.
 
Trebuchet said:
Here's a money-making opportunity: When you cut a deck of cards, the result will either be the six of spades or not. So, according to the friend, it's 50-50. See if you can get him to bet.
Click to expand...

Pretty sure that you just proved immortality.
 
Trebuchet said:
Here's a money-making opportunity: When you cut a deck of cards, the result will either be the six of spades or not. So, according to the friend, it's 50-50. See if you can get him to bet.
Click to expand...

His friend isn't stupid, so he probably won't agree to that bet. From the OP:

He calls those 'actual' odds while 'perceived' odds can't be calculated until my parch sitting habits are examined.​

The friend clearly understands the need to account for additional information when assessing probability. His argument is that when given an binary choice with no additional information, or with information that does not further clarify the probability, the starting estimate is going to be 50/50. Then that estimate will change as more information is added. This seems perfectly cromulent to me.

In reality, of course, more information is almost always present - or assumed to be present - at the start anyway.

I already know a lot about decks of cards, such as tarot cards, or the kind of playing cards that are customarily used for this kind of activity. So I would not take the bet. Not because the friend's reasoning is flawed, but because the friend's reasoning provides for adjusting the estimated probability based on additional information.

Trebuchet's money-making "opportunity" rests on taking only the first half of the friend's argument, and ignoring the second.

This isn't really a question about probability. It's a question of how we assign likelihood as a cognitive heuristic. What I want to know is, is it even possible to have a "raw" likelihood, without any modifying information? The friend's argument seems correct but largely academic. In the real world, phenomena carry too much baggage to admit a baggage-free starting probability.
 
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