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Thread: Do you know the correct answer?

  1. #631
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    Default Re: Do you know the correct answer?

    .
    The fact that Monty has a 66.66% chance that one of his two doors has a car DOES NOT MEAN that each door in Monty's group of two has 33.33% of containing the car.

    If neither door contains the car (Monty has only goats) then each door has a probability of 0% of having the car (the car is behind the contestant's door).

    If Monty's group of two has a a car and a goat, then one of those doors has a 0% probability of the car and the other has a 66.66% probability of the car (the contestant does not have the car). Revealing the door with the goat tells us which of the doors had the 0% probability of car and which door has the 66.66% probability of car.

    The contestant should accept the offer to switch doors.
    .
    Last edited by Tom Montgomery; 06-14-2022 at 04:09 PM.
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  2. #632
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa
    That is three unsubstantiated statements. I say they are wrong, can you prove them right? Why not move that 1/3 to the contestant's door, that would be equally mathematically valid. Contestant's first pick, 2/3. Other door 1/3. Why not that?
    Because the probabilities cannot be changed. They can only be exchanged.

    Monty does not have a door with a 33.33% probability of car. Only the contestant does. Monty has a door with a 0% probability of car and a door with a 66.66% probability of car adding up to a total probability of 66.66% that Monty has one of two doors with a car.
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  3. #633
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Tom Montgomery View Post
    .
    The fact of the matter is that each door in Monty's group of two do not have a 33.33% of containing the car.

    If neither door contains the car (only goats) then each door has a probability of 0% of having the car (the car is behind the contestant's door). If Monty's group of two has a a car and a goat, then one of those doors has a 0% probability of the car and the other has a 66.66% probability of the car (the contestant does not have the car).

    The contestant should accept the offer to switch doors.
    How can 3 x 1/3, become 1/3 and 2/3 when the unknowns are reduced to two from three? What is the mathematical proof in probability theory that make this so?

    Every keeps on restating the unsubstantiated statement from the whiteboard "statistician" where did he show that this is true?
    He ignored that the unknown doors had reduced to two by the reveal. He should have recognized that the probabilities for two UNKNOWNS have to be equal. In effect, he fitted his explanation to what he wanted to prove, rather than following the math.
    It really is quite difficult to build an ugly wooden boat.

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  4. #634
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    Default Re: Do you know the correct answer?

    This thread is just circling the plug hole.
    No one has come up with anything from the math of probability that justifies adding the two probabilities together. Nor has anybody addressed my facts or logic to show where I am in error.

    Tom, you may as well close the thread.
    It really is quite difficult to build an ugly wooden boat.

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  5. #635
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    Not really, and it is an unworkable solution. If the host at random reveals the prize, that is game over. The contestant is sent home, the director has a swearing fit, and they have to start over with a new contestant.
    The game works with the host knowing where the prize is.
    The difference between the two analyses are:


    • The 1/3, 1/3, 1/3 analysis assumes that you can add the probability from the door revealed to have zero probability to the "3rd" door. This is a fallacy, as it implies that opening the door affects the location of the price, which is impossible.
    • No, you have that backward. The location of the prize drives the decision of which door to open. It is NOT opened randomly.
    • My analysis, that is based on sharing the "revealed" 1/3 probability, actually a zero probability, equally between the other two closed doors.
    • No, the doors are not equal. The contestant's door represents a random pick among three doors. The offered door represents a pre-sorted "best chance of two" doors. Again, the sorting was NOT done randomly.
    • This is valid because we will not know which door has the prize until the end of the game.

    In effect, asking "Do you want to switch" is repeating the "Chose a door" request using different words.
    The contestant is then picking from two unknowns.
    Yes, two unknowns, but not two equals. The odds are better with the pre-sorted door.

    If picking from three gives him 1/3, 1/3, 1/3 probabilities, picking from 2 must give him/her 1/2, 1/2 probabilities. That is called "consistency"
    At this point, I think it might be better called "stubbornness". "Best of two" has better odds than "One of three."

    p.s. I'd really like to see a comment from you or CWSmith about Peb's observation first made in post 123 and repeated or copied a couple of times since. It's pretty hard to reconcile his observation with your viewpoint.


  6. #636
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    This thread is just circling the plug hole.
    No one has come up with anything from the math of probability that justifies adding the two probabilities together. Nor has anybody addressed my facts or logic to show where I am in error.

    Tom, you may as well close the thread.
    lmfao

    '
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  7. #637
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by peb View Post
    If you switch, you only lose if you had chosen correctly. You had a 33% chance of that. It's is as simple as that. Cannot be explained better.
    Quote Originally Posted by Keith Wilson View Post
    To repeat #619 in slightly different words:

    If you don't switch, your odds are one in three.

    If you switch, you are effectively getting what's behind two doors out of three. At least one of those two was certain to have a goat, and revealing that makes no difference.
    No true
    After the reveal, we have two unknown doors and one with zero probability. No other data. Nothing to bias the result. So the odds that the two closed doors are winners are now evens.
    It really is quite difficult to build an ugly wooden boat.

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  8. #638
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa
    No one has come up with anything from the math of probability that justifies adding the two probabilities together. Nor has anybody addressed my facts or logic to show where I am in error.

    Tom, you may as well close the thread.
    Nick, you simply misunderstand the statistics. I will close this thread if you first play the game for yourself 50 times and report the results to us. I trust you will not cheat.
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  9. #639
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    This thread is just circling the plug hole.
    No one has come up with anything from the math of probability that justifies adding the two probabilities together. Nor has anybody addressed my facts or logic to show where I am in error.

    Tom, you may as well close the thread.
    From the wikipedia link that I posted back somewhere in this thread.

    Initially, the car is equally likely to be behind any of the three doors: the odds on door 1, door 2, and door 3 are 1 : 1 : 1. This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given that the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is, by definition, the probability of the new piece of information (host opens door 3) under each of the hypotheses considered (location of the car). Now, since the player initially chose door 1, the chance that the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios 1/2 : 1 : 0 or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. Thus, the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given that the host opened door 3, the probability that the car is behind door 3 is zero, and it is twice as likely to be behind door 2 than door 1.

  10. #640
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    No true
    After the reveal, we have two unknown doors and one with zero probability. No other data. Nothing to bias the result. So the odds that the two closed doors are winners are now evens.
    No. Before opening anything, Monty's two doors have a 2/3 chance of containing a car, and are certain to contain at least one goat. Monty knows where his goat is from the beginning, and opening that door does nothing to change the 2/3 chance.
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  11. #641
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    Default Re: Do you know the correct answer?

    .
    Initially, the car is equally likely to be behind any of the three doors: the odds on door 1, door 2, and door 3 are 1 : 1 : 1. This remains the case after the player has chosen door 1, by independence. According to Bayes' rule, the posterior odds on the location of the car, given that the host opens door 3, are equal to the prior odds multiplied by the Bayes factor or likelihood, which is, by definition, the probability of the new piece of information (host opens door 3) under each of the hypotheses considered (location of the car). Now, since the player initially chose door 1, the chance that the host opens door 3 is 50% if the car is behind door 1, 100% if the car is behind door 2, 0% if the car is behind door 3. Thus the Bayes factor consists of the ratios 1/2 : 1 : 0 or equivalently 1 : 2 : 0, while the prior odds were 1 : 1 : 1. Thus, the posterior odds become equal to the Bayes factor 1 : 2 : 0. Given that the host opened door 3, the probability that the car is behind door 3 is zero, and it is twice as likely to be behind door 2 than door 1.
    As I said, the contestant holds door 1 (say) with a probability of 33.33% while Monty holds two doors - door 2 (say) with with a probability of 66.66% (twice that of door 1) and door 3 (say) with a probability of 0%. Those probabilities will never change. Monty will always offer to switch his door with a 66.66% chance of a car with the contestant's door with a 33.33% chance of a car.
    .
    Last edited by Tom Montgomery; 06-14-2022 at 04:40 PM.
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  12. #642
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Gordon Bartlett View Post
    At this point, I think it might be better called "stubbornness". "Best of two" has better odds than "One of three."
    Just so. One of two is even odds, best of three is a lot less than evens. You got that bit right.
    p.s. I'd really like to see a comment from you or CWSmith about Peb's observation first made in post 123 and repeated or copied a couple of times since. It's pretty hard to reconcile his observation with your viewpoint.
    We did, several times.

    Quote Originally Posted by peb View Post
    If you play my the rule to always switch
    You only lose if you picked the right door. That's a 33% chance of losing. It is very, very simple.
    33% goes straight out of the window with the reveal. It is then a pick from one of two unknowns. Show me how that can be anything other than evens if both are unknown? When the game reaches that stage, it is equivalent to flipping a coin.
    It really is quite difficult to build an ugly wooden boat.

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  13. #643
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    Default Re: Do you know the correct answer?

    This is why we are doomed as a species.

  14. #644
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Tom Montgomery View Post
    .
    As I said, the contestant holds door 1 with a probability of 33.33% while Monty holds two doors - door 2 (say) with with a probability of 66.66% (twice that of door 1) and door 3 with a probability of 0%.
    And when he opens the door, the game becomes pick one of two unknown doors. Why is that so difficult to grasp?
    Can you provide a link that shows adding the two probabilities is a valid action in these circumstances, rather than "sharing" the probability between the two, or adding it to the contestant's door?
    Don't make me post the Mrs Doyle clip again.
    It really is quite difficult to build an ugly wooden boat.

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  15. #645
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    33% goes straight out of the window with the reveal. It is then a pick from one of two unknowns.
    But those unknows were not created at random; see 663 and 664. One of the unknows, the door you picked first, has a 1/3 chance of hiding the car. The other has a 2/3 chance.
    Last edited by Keith Wilson; 06-14-2022 at 05:02 PM.
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  16. #646
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    Default Re: Do you know the correct answer?

    Nick - try coming up with a scenario where opting for the switch doesn't reverse the original choice.

    There'll always be a goat behind one of the two remaining doors, which Monty will reveal to you. If you chose a car, the other door also has a goat, so if you switch doors, you'll swap a car for a goat. If you chose a goat (doesn't matter which one) and Monty reveals the other, then switching doors will swap you goat for the car.

    Therefore if the initial pick has a 1:3 probability of being the car, then switching is the reverse of that, 2:3. It's not adding the probabilities, is the probability of picking a goat and then switching (which will always result in a car).
    'When I leave I don't know what I'm hoping to find. When I leave I don't know what I'm leaving behind...'

  17. #647
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Decourcy View Post
    This is why we are doomed as a species.
    Nah, some are trained to think clearly, as their career depended on it.
    It really is quite difficult to build an ugly wooden boat.

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  18. #648
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa
    33% goes straight out of the window with the reveal.
    It does not. This belief is erroneous. Try playing the game.
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Decourcy View Post
    This is why we are doomed as a species.
    Nah, some are trained to think clearly, as their career depends on it.
    It really is quite difficult to build an ugly wooden boat.

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  20. #650
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa
    Nah, some are trained to think clearly, as their career depends on it.
    Everyone is capable of error no matter how well trained to think clearly.

    I am asking you to actually test your hypothesis by playing the game as I did. I played 100 times. It goes quickly. And I'm sure it will take fewer plays for you to decide whether or not your hypothesis is correct.
    .
    Last edited by Tom Montgomery; 06-14-2022 at 04:54 PM.
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  21. #651
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    And when he opens the door, the game becomes pick one of two unknown doors. Why is that so difficult to grasp?
    Can you provide a link that shows adding the two probabilities is a valid action in these circumstances, rather than "sharing" the probability between the two, or adding it to the contestant's door?
    Don't make me post the Mrs Doyle clip again.
    I have provided a link a couple of times. https://en.wikipedia.org/wiki/Monty_Hall_problem

    From that link:

    When first presented with the Monty Hall problem, an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter.[9] Out of 228 subjects in one study, only 13% chose to switch.[21] In his book The Power of Logical Thinking,[22] cognitive psychologist Massimo Piattelli Palmarini [it] writes: "No other statistical puzzle comes so close to fooling all the people all the time [and] even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans.[23]

    Most statements of the problem, notably the one in Parade, do not match the rules of the actual game show [10] and do not fully specify the host's behavior or that the car's location is randomly selected.[21][4][24] Krauss and Wang conjecture that people make the standard assumptions even if they are not explicitly stated.[25]

    Although these issues are mathematically significant, even when controlling for these factors, nearly all people still think each of the two unopened doors has an equal probability and conclude that switching does not matter.[9] This "equal probability" assumption is a deeply rooted intuition.[26] People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not.[27]

    The problem continues to attract the attention of cognitive psychologists. The typical behavior of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as:

    The endowment effect,[28] in which people tend to overvalue the winning probability of the already chosen – already "owned" – door.
    The status quo bias,[29] in which people prefer to stick with the choice of door they have already made.
    The errors of omission vs. errors of commission effect,[30] in which, all other things being equal, people prefer to make errors through inaction (Stay) as opposed to action (Switch).

    Experimental evidence confirms that these are plausible explanations that do not depend on probability intuition.[31][32] Another possibility is that people's intuition simply does not deal with the textbook version of the problem, but with a real game show setting.[33] There, the possibility exists that the show master plays deceitfully by opening other doors only if a door with the car was initially chosen. A show master playing deceitfully half of the times modifies the winning chances in case one is offered to switch to "equal probability".

  22. #652
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by CWSmith View Post
    Okay, Nick, I'll see if this old dog can learn a new trick:

    Contestant Picks this Door Prize is Behind This Door Monty Shows This Door Stays Switches
    1 1 2 Win Lose
    1 1 3 Win Lose
    1 2 3 Lose Win
    1 3 2 Lose Win
    2 1 3 Lose Win
    2 2 1 Win Lose
    2 2 3 Win Lose
    2 3 1 Lose Win
    3 1 2 Lose Win
    3 2 1 Lose Win
    3 3 1 Win Lose
    3 3 2 Win Lose
    Win % 0.5 0.5

    Okay, Tom, I told you what I thought was wrong with your chart. Now I'd like you to tell me what is wrong with mine.

    Edit - Sorry. I had a typo that I corrected.
    I think I can explain the problem with this. After the contestant picks a door, there is a 33% chance of the car being behind it. That means there is a 33% chance that there will be some result, either win or lose, of a door with a car. This table shows there is a 50% chance of a result (again any result, win or lose) if the car is behind the door chosen. The chance of there being an output is greater than the chance of an input. This is impossible. The table must be wrong.
    Last edited by rgthom; 06-14-2022 at 05:07 PM.

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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    Nah, some are trained to think clearly, as their career depends on it.
    You sound exactly like Aquinian

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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    Nah, some are trained to think clearly, as their career depended on it.
    Oh really? Lot's of us have had technical careers and have seen awful mistakes come out of unquestioned technical competence.

    Maybe this is what it eventually boils down to:


    The doctor says: "I'm going to say 4 words, repeat them after me: elephant, umbrella, orange, bicycle. Now tell me, how are you feeling today Mr. Jones?

    50 minutes later the doctor asks: "Can you tell me those four words we discussed at the start of your appointment, Mr.Jones?"

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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Decourcy View Post
    You sound exactly like Aquinian
    You really have a nice line in insults.
    Whatever did I do to you?
    It really is quite difficult to build an ugly wooden boat.

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    Default Re: Do you know the correct answer?

    I must be on ignore.

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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    OK, explain the mechanism that makes 1/3 become 2/3, based on the rules of probability. Where is it written down?
    Monty has two doors(2/3)(one with a goat,for certain).
    You have one door(1/3)
    If you switch, you get his door (with no goat)(2/3)
    End of story.
    R
    Sleep with one eye open.

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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Ron Williamson View Post
    Monty has two doors(2/3)(one with a goat,for certain).
    You have one door(1/3)
    If you switch, you get his door (with no goat)(2/3)
    End of story.
    R
    If you think that is a proof, think again.
    It really is quite difficult to build an ugly wooden boat.

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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Ron Williamson View Post
    You have one door(1/3)

    End of story.
    R
    No no.

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    Default Re: Do you know the correct answer?

    So with two doors, he doesn't have an advantage over you with one door?
    When you switch you don't get his advantage?
    R
    Sleep with one eye open.

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    Default Re: Do you know the correct answer?

    Looks like there is never going to be an end to this story. At this point it seems like you all have to ask yourselves who would you accept as an authority on this. I still think it is Schrodinger’s goat. The goat is both behind the door you pick and not behind the door you pick until you open the door.

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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Peerie Maa View Post
    You really have a nice line in insults.
    Whatever did I do to you?
    Didn’t mean it as an insult. Just that you are defending the indefensible. Which I think is a good lesson for everyone. You are obviously educated and intelligent, but you’ve got yourself caught in a loop. A good lesson to keep in mind when someone that is defending the indefensible elsewhere leads to frustration.

  33. #663
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    Default Re: Do you know the correct answer?

    Quote Originally Posted by Ron Williamson View Post
    So with two doors, he doesn't have an advantage over you with one door?
    When you switch you don't get his advantage?
    R
    After he reveals the door with the goat, you have to pick between two unknowns. What are the odd on betting on a choice of two, flipping a coin for example?
    It really is quite difficult to build an ugly wooden boat.

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    Default Re: Do you know the correct answer?

    Waste of band width. Nick is clearly correct.

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    Default Re: Do you know the correct answer?

    Your post 469 shows that win/lose is always reversed when you switch.

    That's all there is to it.

    The reveal doesn't do anything to the odds, because the odds only matter in the initial pick, one of three doors.

    Form there, you can keep your 1:3 choice, or swap it for the opposite.
    'When I leave I don't know what I'm hoping to find. When I leave I don't know what I'm leaving behind...'

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