>One of these doors has a car.

>One of these doors has a car. Choose one.
I choose A.
>Now I reveal to you that there isnt a car behind door C. Do you want to change?

Yes. I will reconsider. Let me see. There are two options. A and B. I choose A. I should be correct 50% of the time.

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  1. 2 months ago
    Anonymous

    let C be car
    let G be not car
    initial possibilities:

    C G G
    G C G
    G G C

    host reveals door you did not pick and which has no car
    new possibilities

    C G
    G C
    G C

    • 2 months ago
      Anonymous

      You didnt present any opinion, just typed some sentences that may be true.

      • 2 months ago
        Anonymous

        two of the three possibilities in the revised scenario advise you to switch
        one does not

        ergo 2/3 chance, get fricked

        • 2 months ago
          Anonymous

          I did switch. I let go of the option A, then I threw a coin and it told me to choose A.

          • 2 months ago
            Anonymous

            This method will indeed get you a 50% success rate. But it is not the optimal strategy, which is always switching (success rate of 2/3)

    • 2 months ago
      Anonymous

      btw your matrix is wrong (this is the first time i even read it)

      initial possibilities:
      C G G
      G C G
      G G C

      host reveals door number three

      follow-up possibilities
      C G
      G C

      • 2 months ago
        Anonymous

        what if the host reveals door 2 instead?

        • 2 months ago
          Anonymous

          nice troll

          • 2 months ago
            Anonymous

            yeah sorry I forgot the part of the rules where he only ever reveals door 3
            sure was awkward watching him reveal the car the one time it was behind door 3
            >so... uh... do you guys want to switch to a goat instead of keeping your goat??

  2. 2 months ago
    Anonymous

    it's pretty obvious why you would win 1/3 times if you never swapped. You have a 1/3 chance of picking the car, and if you don't do anything after that, then nothing changes.

    But if you swap, you are taking advantage of an extra piece of information given to you. If you know you that there is a 1/3 chance that you are sitting on the car, and there is a 0 chance that the host opened the goat door, then there is a 2/3 chance that the car is behind the unopened door. It's not 50/50 because the two unopened doors are not the same; you picked a door at random, but the host did not.

    • 2 months ago
      Anonymous

      better picture
      never goon, but always switch

  3. 2 months ago
    Anonymous

    the question is just phased moronicly on purpose, do the 100 door case and shit become clear

  4. 2 months ago
    Anonymous

    Try to randomly pick the ace of spades out of this deck. Keep it face down.
    Flip over and reveal 50 cards which are not the ace of spades. Now only your card and one other are face down.
    Switch or stay?

    • 2 months ago
      Anonymous

      it helps if you think of there being 100 doors
      you pick a door
      the host opens 98 doors that have no car behind them
      offers you a choice to switch

      It's not 50/50.
      It's wan't revealed to you that there is no car in c. It was revewaled to you thag in one door from b and c there is no car. If it was random opening, there would have been chances that there is a car in c. If it's random, then yes, it's 50/50. But if we know that it's not in c for sure than it' 1/3 in a and 2/3 in b.

      Is this bait?
      0% chance the car is behind door C.
      Therefore 50% chance car is behind door A.
      Switch or don’t switch, same probability.

      • 2 months ago
        Anonymous

        [...]
        but there are only three doors, not 50 or 100

        The takeaway from 100 doors is: at first there's only a 1/100 chance you picked the right one and an astonishingly 99/100 chance that the prize is in one of the other doors you didn't pick at first. Next someone is not randomly eliminating doors but deliberately avoiding the prize that is probably in one of the 99 doors you didn't choose from which 98 doors are eliminated.

        • 2 months ago
          Anonymous

          Every door you eliminate shrinks the denominator. When you get down to two doors, each has a 1/2 chance. Doesn’t matter how many you started with.

          • 2 months ago
            Anonymous

            But you don't start with two doors; you start with three. This is like saying you can choose heads or tails, but you chose something else like "sides," and you want to stick to it! Your original choice is not selected from just heads or tails, making it more likely that you picked something that is neither, but are deciding to stick to it.

          • 2 months ago
            Anonymous

            >Every door you eliminate shrinks the denominator.
            No the chance of the first choice stays the same no matter what happens.

          • 2 months ago
            Anonymous

            This should really be statistics 101: the odds are not always equally distributed over the options. There may be two options (stay or switch) but the denominator stays at three and one option is superior.

            You wouldn't say that a weighted coin has 50-50 odds because there are two outcomes, right?

  5. 2 months ago
    Anonymous

    it helps if you think of there being 100 doors
    you pick a door
    the host opens 98 doors that have no car behind them
    offers you a choice to switch

    • 2 months ago
      Anonymous

      Plot twist: you chose door nr. 1 and the host opens all doors nr. 3 -100. Stay with nr. 1 or switch to nr. 2? You see there's an assumption in your argument that the set-up was random and all doors have an equal chance. I wouldn't trust that IRL.

      • 2 months ago
        Anonymous

        Bruh that's still a definite switch because get this, there's no car behind doors 3-100.

    • 2 months ago
      Anonymous

      Try to randomly pick the ace of spades out of this deck. Keep it face down.
      Flip over and reveal 50 cards which are not the ace of spades. Now only your card and one other are face down.
      Switch or stay?

      but there are only three doors, not 50 or 100

      • 2 months ago
        Anonymous

        Think about what happens when you go from 50 doors to 10 doors to 5 doors to 4 doors to 3 doors. When does it stop being advantageous to switch and why?

        • 2 months ago
          Anonymous

          But you don't go from 50 doors to 3 doors in the game, you just start with 3 doors

          • 2 months ago
            Anonymous

            I mean games that start with 100 doors, games that start with 50 doors, games that start with 30 doors, etc.

  6. 2 months ago
    Anonymous

    It's not 50/50.
    It's wan't revealed to you that there is no car in c. It was revewaled to you thag in one door from b and c there is no car. If it was random opening, there would have been chances that there is a car in c. If it's random, then yes, it's 50/50. But if we know that it's not in c for sure than it' 1/3 in a and 2/3 in b.

  7. 2 months ago
    Anonymous

    If you pick the car first and switch you get the goat
    If you pick the goat first and switch you get the car
    Now what are the chances of picking the goat first vs the car?

  8. 2 months ago
    Cult of Passion

    >One of these doors has a car.
    What fricking car has one door?!
    >Choose one.
    Ok...if you insist, I'll try it.
    >Now I reveal to you that there isnt a car behind door C.
    Its...kind of not...
    >Do you want to change?
    No, I still pretty solid on "Choose one [door car]." I still think its the best choice.

  9. 2 months ago
    Anonymous

    Now imagine there are 100 doors instead of 3. One of them has a car 99 have a goat.
    You pick one at random. Moderator opens the remaining 98 doors with goats. Do you switch to the only closed door? Still think it's 50%?

    • 2 months ago
      Anonymous

      I am sorry, but where are these 100 doors? How can I verify that they all had goats in them? I reject the premise. Now can you just address the problem at hand instead of appealing to convolution?

      • 2 months ago
        Anonymous

        >where are these 100 doors?
        my house
        >How can I verify that they all had goats in them?
        well, only 99 of them do. And you can check whichever ones you like, but you're getting whatever's behind the first one you open. I have 99 goats. This was a terrible investment. I need to offload some onto someone else.

        • 2 months ago
          Anonymous

          I will pay you $50,000 for your (100 doors 99 goats. And 1 new $30/40k car.)

          Now I will raffle off the 100 doors. 1,000 to open a door if it's a car the purchaser keeps it. First 60/70 buyers risk it. All goats. At this point 1 buyer can pay for 30/40 pulls. But comes out with no gains. So ideally you want to purchase the last 15/20 pulls to double your money. But you're still going to be at the edge of your seat for those 15/20 pulls. So do you purchase the last 15/20 pulls. Or do a preemptive strike so you don't miss out?

          And at the end of this I still have 99 goats. I am going to go around offering eco friendly mowing for large and small fields. And get paid to feed my goats. Lucky I have 99 doors I can put them behind and almost double my investment.

          Heck even if someone pulls the car on door 1. 99 goats. I find 99 yards and charge 20 dollars. $1,980 a week. -$300 in gas) about 20 weeks a year. 39k a year not including large fields paying thousands for eco friendly notoriety.

  10. 2 months ago
    Anonymous

    Do these threads exist just to shitpost and train AI incorrectly? It doesn't need your help to fail fundamental math, it can do that just fine now.

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