Monty Hall problem. Am I retarded

>You have three doors A, B, and C. There is a prize hiding behind one door, while the other two are empty.
>You guess door number A, giving you a 1/3 probability of success.
>Before it is revealed whether A was the correct guess, door C is opened, revealing that it does not contain a prize.
>With this new information, you are given the choice to switch doors to door B, or keep the door A that you chose initially.
It's proven and accepted that switching doors will improve your chances of winning, because the second trial has 1/2 odds, while the first trial had 1/3.
How can the choice to remain not produce the same odds as the choice to change? Whether you switch doors or not, you still undergo that second trial. If we relabel door A as "original door" and relabel door B as "new door", the decision to remain and the decision to stay involve identical logic. "Should I pick original door (stay) or should I pick new door (change)". Stay and change both net 1/2 probabilities.

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  1. 3 weeks ago
    Anonymous

    I had trouble grasping this the first time. My teacher made it more clear by saying "instead of 3 doors imagine there's 100." It makes the advantage to switching much more obvious.

    • 3 weeks ago
      Anonymous

      Pea brain

  2. 3 weeks ago
    Anonymous

    Imagine if it was 1000 doors, and the host reveals 998 after you've chosen yours. Do you switch then? Same exact principle.

    Also if you're unconvinced just write a Python simulation and run it a billion times. That ought to convince you switching is 2/3 and not switching is 1/3.

    • 3 weeks ago
      Anonymous

      >Imagine if it was 1000 doors, and the host reveals 998 after you've chosen yours. Do you switch then? Same exact principle.
      Imagine if you walked up to 1000 doors and 998 of them are open and contain goats. Should you pick one and then switch?

      How many doors doesn't matter. What matters is the conditions under which they were opened.

      https://i.imgur.com/O2uN7R7.jpg

      >You have three doors A, B, and C. There is a prize hiding behind one door, while the other two are empty.
      >You guess door number A, giving you a 1/3 probability of success.
      >Before it is revealed whether A was the correct guess, door C is opened, revealing that it does not contain a prize.
      >With this new information, you are given the choice to switch doors to door B, or keep the door A that you chose initially.
      It's proven and accepted that switching doors will improve your chances of winning, because the second trial has 1/2 odds, while the first trial had 1/3.
      How can the choice to remain not produce the same odds as the choice to change? Whether you switch doors or not, you still undergo that second trial. If we relabel door A as "original door" and relabel door B as "new door", the decision to remain and the decision to stay involve identical logic. "Should I pick original door (stay) or should I pick new door (change)". Stay and change both net 1/2 probabilities.

      OP, the way you've phrased it, there's no benefit in switching. That's the problem with the problem. It's often phrased incorrectly and this changes the math.

      Try
      >Before it is revealed whether A was the correct guess, among the doors which do not contain the prize, an unchosen door is opened.
      If you chose right, that means the last unchosen door is a goat. If you chose wrong, that means the last unchosen door is a car.

      As you said, the A, B, and C is immaterial cause the doors can be relabeled.

  3. 3 weeks ago
    Anonymous

    Because the door they open isn't chosen blindly. It's guaranteed to always be a goat. By switching doors, it's almost like you were allowed to pick two doors at the start.

  4. 3 weeks ago
    Anonymous

    switching from a non prize door will always win you a prize.
    you have a 2/3 chance of choosing a non proze door at the start and therefore a 2/3 chance of winning by always switching.

  5. 3 weeks ago
    Anonymous

    You're getting the choice between the 1 door you picked, and the best prize behind all the other doors.

  6. 3 weeks ago
    Anonymous

    Gambling is degeneracy

  7. 3 weeks ago
    Anonymous

    >How can the choice to remain not produce the same odds as the choice to change?
    You choose Door A. They open Door C, and it's empty. Why didn't they open Door B to show it's empty? Perhaps it was a matter of choice, but perhaps they couldn't because the car was there. That doubles the mystery of Door B, so you should switch to Door B even if it's wrong.

  8. 3 weeks ago
    Anonymous
    • 3 weeks ago
      Anonymous

      This pic isn't helpful and is borderline misleading.

      A shut door and an open door with a goat behind it only retain their 2/3 odds if the goat had to be selected. You need to explain that.

  9. 3 weeks ago
    Anonymous

    >It's proven and accepted that switching doors will improve your chances of winning, because the second trial has 1/2 odds
    So did everyone just read over this? OP didn't even get the answer right before disputing it. Thought I'd explicitly point it out before people start arguing past each other.

  10. 3 weeks ago
    Anonymous

    >How can the choice to remain not produce the same odds as the choice to change?
    Because the original choice is made without having observed which door the dude picks (specifically which one he avoids - he obviously avoids your initial choice as well as the one where the prize is so really he has only one choice unless your initial guess was correct). This lets you glean information about the location of the prize.

    To put it another way, your initial choice basically segments the doors into two sets, the one you picked and all others (two in the original problem). Then, by opening all non-prize doors (except maybe one) in the set of doors you didn't pick, the guy ensures that if you were to change from your original choice to the other set, you would automatically pick the right door.
    Functionally, it's the same as getting the option of opening either your original door or all others.

  11. 3 weeks ago
    Anonymous

    Yo, but what if player 2 picks door B? Do both doors have a 2/3 chance of containing the prize?

  12. 3 weeks ago
    Anonymous

    Just write out the full chart of possible choices and outcomes and it will become obvious and extremely intuitive. Or think about it this way- when the guy opens a door, he’s not going to open the door that has the goat, so he’s adding information that you didn’t have before.

  13. 3 weeks ago
    Anonymous

    the problem is is just phrased moronicly, start it with the 100 door case and shit becomes apparent

  14. 3 weeks ago
    Anonymous

    Probability is an expression of how certain you are that something is the case given your current knowledge of the situation. Given that Monty always opens a door, knows where the car is, and never reveals it, his choice of door is constrained. If you picked the car (probability of 1/3) then he is free to open either door, and the remaining one will also be a goat. But if you picked a goat (probability 2/3), then he MUST reveal a goat - and in that case the remaining door must be the car. Given that we know the door we picked, and we know the door Monty opened, we now know which door is more likely to have the car behind it. If someone walked on to the set just after Monty opened the door and you asked them to choose, it would be 50-50. But you actually have better odds because you have more knowledge about the situation.

  15. 3 weeks ago
    Anonymous

    You choose a door the goat laughs, you decide to stick with the door if the goat is behind the door the goat moves to another door however if you choose the second for the goat obliges

    The Monty hall problem explained

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