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Women, math, and the Monty Hall problem
- Thread starter IndigoChild5559
- Start date
Not entirely sure how helpful this is, in the context of a rather more trivial problem. But the answer remains the same (even if not in this case, in which you've previewed the outcome): To switch your choice will be correct 2 times out of 3. That means that in 100 such situations, you will have killed about 33 people, and saved about 330.
It was just a crossover between the two hypotheticals, since we had a recent thread about the trolley problem. I wasn't really making any point, I just thought it might be interesting in the context of this topic.
Heyo
Veteran Member
It seems to be a very severe case of the Dunning-Kruger effect, which itself is a special case of a delusion, which is one sign of schizophrenia.
(If you disagree, please look up the definitions before posting.)
Disclaimer: I'm not a psychologist, and even if I were, it would be unethical to diagnose without a personal meeting and further tests. Also, for a pathological case, suffering, of either the patient or people in their vicinity, must be established. I'm just pointing out how I think that this case matches the definitions.
Dunning-Kruger: Clearly overestimating own capabilities in Bayesian probability.
Delusion: Not changing the position in light of contrary evidence, disregarding or disparaging evidence.
I'm not posting this to mock @PureX. He is just an example of a trend I see.
We have always had trolls and unreasonable people, especially on the net. But we see them more and more IRL, and they gain popularity. We should be wary about people who declare, "I probably know more about [subject X] than about anybody else". They most likely suffer from the Dunning-Kruger effect. People who reject evidence as "fake news" or "scientism" are likely to be delusional about the subject.
We should not let this go easily. We should not let delusions be acceptable in society.
As with religious beliefs, I am not one to convince people of their errors or bad choices - but I would have expected better of some over this problem - given the evidence.
PureX
Veteran Member
There are no chances involved in selecting 1 of 3 doors because it will not be opened whichever one you select. So the selection is simply irrelevant. Why don’t you address this?
You expect evidence to make a difference?As with religious beliefs, I am not one to convince people of their errors or bad choices - but I would have expected better of some over this problem - given the evidence.
The most unexpected error yet.
Koldo
Outstanding Member
What's actually relevant is that if you switch (rather than stick with your original pick) you increase your odds. That's what actually matters.
But since I'll never be on that show
the odds of picking the car are zero
whether I switch or not.
PureX
Veteran Member
No, it isn't. You had no chance of selecting any outcome. As it was 100% certain that Monty would ignore whatever you said and open whatever door he chose.
Yet you still refuse to see it.
It's high time to stop digging that hole deeper,
& to carefully consider the proofs shown you.
No longer going to respond -- you've sunk into madness. You don't need a teacher, but a therapist, and that isn't me.
No. Information has been added, which changes the situation.
The odds of the car being behind any given door don't change. If you pick door #1, you have a 1 in 3 chance you picked right and a 2 in 3 chance you picked wrong.
This is still the case when Monty opens a door you didn't pick. You still have a 2 in 3 chance that your initial guess of door #1 was wrong. Effectively, Monty showing the goat means that you now have a way to choose "not door #1" as a single option... and the odds of "not door #1" being correct are 2 in 3.
I assume that you are referring to after the reveal compared to before. If so, I would word that differently.
The odds that you have chosen the right door don't change after the reveal, but the odds for the two doors not chosen DO change after the reveal.
Chances of any given door concealing the car before the reveal:
[1] 1/3
[2] 1/3
[3] 1/3
Let's say you chose door [1] and Monte reveals door [3] and the goat it concealed.
Chances of any given door concealing the car after the reveal:
[1] 1/3
[2] 2/3
[3] 0
*******
For completeness's sake, let's change it to door [3] was chosen by somebody who did NOT know where the care was, and revealed a goat
Chances of any given door concealing the car before the reveal:
[1] 1/3
[2] 1/3
[3] 1/3
Let's say you still chose door [1] and that new somebody revealed door [3] and the goat it concealed.
Chances of any given door concealing the car after the reveal:
[1] 1/2
[2] 1/2
[3] 0
Unlike the first example, where the reveal changes the odds only for doors [2] and [3], in this case, the odds change for all three doors after the reveal. And the difference resides entirely in the fact that Monte knew where the car was before his reveal and had no chance of revealing the car, but the new somebody didn't know what he or she was revealing when choosing door [3].
I discovered that looking for examples of the Big Deal from the 70's and 80's yesterday for this thread and realized then that the Monte Hall problem as stated doesn't seem to have much to do with Monte Hall or Let's Make A Deal, which doesn't diminish the instructive value of the problem, but suggests that perhaps it should be renamed.
Here are the episodes I reviewed if anybody would like to see them. These are both full episodes, so skip to near the end to see the Big Deal, which as Stevicus noted, doesn't resemble the scenario in the problem as stated:
Whateverist
Active Member
Perhaps it would help to spell out a few facts that are implied more than stated in this problem;
1 - the prize is behind only one of the doors and isn’t moved during the show;
2 - Monty knows all along which door has the prize;
3 - using that knowledge Monte will always open a door to show you a goat after you have chosen a door..
But imagine the same game was played with ten doors but still only one good prize. After you have chosen Monte shows you eight doors with goats and asks will you keep you initial choice or switch to one remaining unopened door?
Is it still 50/50 between your initial pick out ten doors and the one other unopened door?
Was it more likely that the prize would be behind the door you picked or behind one of nine you didn’t pick?
Should you stay , switch or doesn’t it matter?
1 - the prize is behind only one of the doors and isn’t moved during the show;
2 - Monty knows all along which door has the prize;
3 - using that knowledge Monte will always open a door to show you a goat after you have chosen a door..
But imagine the same game was played with ten doors but still only one good prize. After you have chosen Monte shows you eight doors with goats and asks will you keep you initial choice or switch to one remaining unopened door?
Is it still 50/50 between your initial pick out ten doors and the one other unopened door?
Was it more likely that the prize would be behind the door you picked or behind one of nine you didn’t pick?
Should you stay , switch or doesn’t it matter?
The odds don't change in the sense that the original selected door still has a 1 in 3 chance of being correct and a 2 in 3 chance of not being correct.
Showing the goat doesn't increase the probability that your initial pick of door #1 was correct.
PureX
Veteran Member
The whole point here is to assess the odds of WINNING THE CAR.
The odds of WINNING THE CAR when facing the three closed doors and being asked to choose a door are ZERO because to win the car, Monty has to open the door you chose and find a car lurking behind it. And that is NOT GOING TO HAPPEN. And it does not happen. Proving that the odds were, in fact, ZERO.
Why are so many of you insisting that the odds are initially 1 in 3 of winning the car when they very clearly were ZERO? And they were shown to be zero by Monty refusing to open whichever door you pick? It couldn't possibly be more obvious that at this stage in the "game" you're chances of winning that car were zero.
Then Monty removes one of the doors from the game by opening it and revealing that there is no car behind it, leaving the remaining 2 doors to contain either the car, or the goat. And then he asks you to choose one of these 2 doors, and this time he will open the door you choose and he will give you whatever is behind it. So NOW, the actual odds of you winning the car are 50/50.
The mistake many of you are making is that you are so obsessed with picking the right door by using the numbers that you are ignoring the reality that the numbers were supposed to be helping you assess. The reality of winning a car on a game show.
The odds of WINNING THE CAR when facing the three closed doors and being asked to choose a door are ZERO because to win the car, Monty has to open the door you chose and find a car lurking behind it. And that is NOT GOING TO HAPPEN. And it does not happen. Proving that the odds were, in fact, ZERO.
Why are so many of you insisting that the odds are initially 1 in 3 of winning the car when they very clearly were ZERO? And they were shown to be zero by Monty refusing to open whichever door you pick? It couldn't possibly be more obvious that at this stage in the "game" you're chances of winning that car were zero.
Then Monty removes one of the doors from the game by opening it and revealing that there is no car behind it, leaving the remaining 2 doors to contain either the car, or the goat. And then he asks you to choose one of these 2 doors, and this time he will open the door you choose and he will give you whatever is behind it. So NOW, the actual odds of you winning the car are 50/50.
The mistake many of you are making is that you are so obsessed with picking the right door by using the numbers that you are ignoring the reality that the numbers were supposed to be helping you assess. The reality of winning a car on a game show.
PureX
Veteran Member
But those are not the odds that matter. Because they will not get us the car.
But you are not going to win the car. And those are the odds that we are supposedly determining. Those odds are ZERO at this point. Not 1 in 3.
No, you still have a ZERO odds chance of winning the car at this point.
You are trying to calculate the odds of "guessing right". But "guessing right" is not the same as winning the car. The whole point here is not to "guess right", it's to win the car. If guessing right were the object of the game, your odds theory would be valid. But it's not. The object of the game is to win the car.
Those odds are delusional, because you never had a 1 in 3 chance of winning the car in the first place. You had NO CHANCE of winning it. You're focused on the numbers and ignoring the reality that the numbers are supposed to be representing.
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Heyo
Veteran Member
Without picking a door in phase one, Monty won't open a door and start phase 2.
Assume the following game: you don't get to make a first choice. There are three doors, Monty opens one of the doors and reveals a goat. Then you are asked to make a choice. In this scenario, the chances are 50/50.
What you don't realize is that that is not the game, that is played.
By making a first choice out of three, and Monty then opening a door, the first choice and the second one are related, not independent. You have to apply Bayes Theorem to calculate the chances.
The mistake you make is that you assume the second choice to be independent, it isn't.