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re: Let's discuss the Monty Hall problem (probabilities, odds)
Posted on 8/12/17 at 1:37 pm to TigrrrDad
Posted on 8/12/17 at 1:37 pm to TigrrrDad
quote:
Exactly. The odds change because you get to reselect (i.e., start over) - not because you switch.
No that is incorrect and this is because you know that the car stays in the same place. Each door maintains a 1/3 chance of having a car. By switching you get to choose both doors 2 and 3 as being in your favor. By staying you only maintain 1 of 3 doors.
If it is easier to visualize you can say that by switching, you are increasing your odds to 50% in your selection (not technically correct but maybe easier to visualize); however, by staying you are still technically playing against two doors and thus still at 33% - then only difference from the beginning is that one of the answers has been revealed.
0
Posted on 8/12/17 at 1:37 pm to SlapahoeTribe
Most of the idiots around here have proven that they can't grasp something as simple as the order of operations. I don't need to read this thread to know that it's ten pages of stupid frickers who cannot grasp the subtle probability question in play, but aren't going to let that stop them from arguing as loudly as they can about it.
How close am I?
How close am I?
This post was edited on 8/12/17 at 4:11 pm
Posted on 8/12/17 at 1:39 pm to TigrrrDad
quote:
Exactly. The odds change because you get to reselect (i.e., start over) - not because you switch.
Then why in real life simulations aren't the results 50/50 instead of 2/3?
Posted on 8/12/17 at 1:39 pm to TigrrrDad
quote:
quote:
In the case you're talking about, your odds have improved from 1/n to 1/2. It doesn't matter if you switch; it doesn't help.
Exactly. The odds change because you get to reselect (i.e., start over) - not because you switch.
No a start over would be if the car was moved, or was able to be moved, to a different door.
Posted on 8/12/17 at 1:39 pm to Spock's Eyebrow
quote:OK, let me rephrase.
Exactly quite. Revealing all goats by chance until one door is left is the same as picking that one door without revealing any goats and offering the chance to switch. If there are 10 doors, then the player had a 1/10 chance at the start of the game, and the other door had a 1/10 chance, just like all the doors had. 1/10 = 1/10, so there's no benefit in switching.
Your odds of winning the game based on a decision you made can never change from 1/3, assuming Monty picks at random.
Monty will knock you out 1/3 of the time, you initial pick will be right 1/3 of the time, and your switch pick will be right 1/3 of the time.
The only way for your odds to be 1/2 is if Monty opens a goat door before you choose. But, in this case, if you consider the game to have "started" before he opens, then you only have 2/3 odds to even get to play.
Posted on 8/12/17 at 1:40 pm to KosmoCramer
quote:
Then why in real life simulations aren't the results 50/50 instead of 2/3?
Information bias.
This post was edited on 8/12/17 at 1:42 pm
Posted on 8/12/17 at 1:41 pm to TigrrrDad
quote:
Information bias.
Results bias?
Posted on 8/12/17 at 1:41 pm to TigrrrDad
quote:There are 4 possible outcomes, but those outcomes are not weighted equally.
Precisely. And in that ONE scenario, there are four possible outcomes. 2 win and 2 lose. 50/50
When you chose door #1, there was only a 33% chance that you picked the right door from the start. There is a 67% chance that one of the other two doors holds the car.
So if you played 100 times and stayed with door 1 every time, you'd win the car 33 times.
If you played 100 times and switched every time, you'd win the car 67 times.
If there were 1,000 doors and you chose a door, you'd have a 0.1% chance of choosing the right door. If the host eliminated 998 other doors and you were only left with your door and the one other door, there is still a 0.1% chance you picked the right door, and a 99.9% chance the car is in the one remaining door.
The odds would be heavily in favor of switching.
This post was edited on 8/12/17 at 1:43 pm
Posted on 8/12/17 at 1:43 pm to PrimeTime Money
quote:You should probably stop arguing with Dad, he isn't talking about the Monty Hall problem.
PrimeTime Money
He is discussing a totally unrelated problem that only has 2 doors.
Posted on 8/12/17 at 1:44 pm to ShortyRob
It isn't helping that some of you who know the right answer are explaining it wrong.
It's really a very simple problem if you don't pretend to be an ignoramus and I'd like the problem started the after you pick so we'll ignore those me
For everyone else it's simple. When you pick there's a one-in-three chance you are correct. So this means you will be wrong twice.
In the circumstance where you are correct switching will lose
In the circumstance where you were wrong the host will reveal your other Miss.
That means every time you pick wrong switching will win because the host is never going to reveal the car
They call it the Monty Hall problem because it's basically asking how do you play a 3-door game in the answer is you pick and switch
It's really a very simple problem if you don't pretend to be an ignoramus and I'd like the problem started the after you pick so we'll ignore those me
For everyone else it's simple. When you pick there's a one-in-three chance you are correct. So this means you will be wrong twice.
In the circumstance where you are correct switching will lose
In the circumstance where you were wrong the host will reveal your other Miss.
That means every time you pick wrong switching will win because the host is never going to reveal the car
They call it the Monty Hall problem because it's basically asking how do you play a 3-door game in the answer is you pick and switch
Posted on 8/12/17 at 1:44 pm to Korkstand
quote:
You should probably stop arguing with Dad, he isn't talking about the Monty Hall problem.
He is discussing a totally unrelated problem that only has 2 doors.
...otherwise known as "the OP."
If the OP (and Marilyn Vos Savant) had simply stated this as "you select a door, and the host opens one of the other doors to reveal a goat," this whole debate would be a non-issue.
This post was edited on 8/12/17 at 1:47 pm
Posted on 8/12/17 at 1:47 pm to Korkstand
quote:The math still works out the way he's describing it.
You should probably stop arguing with Dad, he isn't talking about the Monty Hall problem.
He is discussing a totally unrelated problem that only has 2 doors.
Posted on 8/12/17 at 1:47 pm to Korkstand
quote:
OK, let me rephrase.
Your odds of winning the game based on a decision you made can never change from 1/3, assuming Monty picks at random.
Monty will knock you out 1/3 of the time, you initial pick will be right 1/3 of the time, and your switch pick will be right 1/3 of the time.
The only way for your odds to be 1/2 is if Monty opens a goat door before you choose. But, in this case, if you consider the game to have "started" before he opens, then you only have 2/3 odds to even get to play.
No, listen, what you're talking about is like the end game for "Deal or No Deal", where the prizes are $1 million, $1, and everything in between is goats.
Posted on 8/12/17 at 1:47 pm to TigrrrDad
I think you need to work on reading comprehension and basic math - perhaps your kids can let you watch as they do homework...
Posted on 8/12/17 at 1:47 pm to TigrrrDad
quote:
...otherwise known as "the OP."
Posted on 8/12/17 at 1:48 pm to PrimeTime Money
quote:
quote:
You should probably stop arguing with Dad, he isn't talking about the Monty Hall problem.
He is discussing a totally unrelated problem that only has 2 doors.
The math still works out the way he's describing it.
Well he is describing it in a way that isn't in line with the Monty Hall Problem or the OP...
Your odds cannot change unless the problem restarts.
The only way for it to be 50 / 50 is for host to say "here are three doors, one has a car behind it"
Then he opens a door
Then you get to choose a door.
This post was edited on 8/12/17 at 1:51 pm
Posted on 8/12/17 at 1:48 pm to TigrrrDad
quote:Jesus, give it a rest.
...otherwise known as "the OP."
If you think the OP is phrased poorly, then say it is worded in a way that takes door #3 out of play, so it is only a 2 door problem.
Don't show your arse by arguing a very well-known problem from a flawed perspective and think you're smarter than everyone else.
Posted on 8/12/17 at 1:50 pm to TigrrrDad
quote:
If the OP (and Marilyn Vos Savant) had simply stated this as "you select a door, and the host opens one of the other doors to reveal a goat," this whole debate would be a non-issue.
Even in your scenario where the host doesn't know the answer, you still only win 1/3 of the time, unless you assume he opens a goat door every time (which he wouldn't if he didn't know).
This post was edited on 8/12/17 at 1:55 pm
Posted on 8/12/17 at 1:52 pm to Sneaky__Sally
quote:
Your odds cannot change unless the problem restarts.
The only way for it to be 50 / 50 is for host to say "here are three doors, one has a car behind it"
Then he opens a door
Then you get to choose a door.
The odds of winning that game aren't 50/50.
Posted on 8/12/17 at 1:53 pm to KosmoCramer
yes they are. You don't choose until there are only two doors, one with a car and one without a car. - sorry i guess i should have said that the first door he opens has a goat.
But i think we should probably focus on the current problem first.
But i think we should probably focus on the current problem first.
This post was edited on 8/12/17 at 1:54 pm
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