The Monty Hall Problem, a fun little game

hasher22 on 03/02/2022 - 20:03
Last edited 03/02/2022 - 20:05

This is not a riddle, but about probability. Analysing the words, door selection, and structure of what I said will not help you at all

Let's play!

You are at a game show, there are 3 doors in front of you but you don't know what's behind them. Behind 2 doors are goats, behind 1 door is a car. Your goal is to select the door with the car and you win the car! You have a 33% chance to win the car and a 66% chance to select a door with a goat.

  1. You select door 3
  2. The game show host reveals door 2 which has a goat
  3. The host now asks you, "Do you want to switch to door 1 or stay at your current door, 3?"

Question: What do you do? Do you switch doors or stay at your selected door to possibly win the car?

Comment below why.

:)

Poll Options

  • 59
    Stay at door 3
  • 321
    Switch to door 1
General Discussion

Comments

  • SnowDragon on 03/02/2022 - 20:41
    +3

    I have tried and failed to understand the answer to this problem many times. I feel like I understand probability pretty well but this question just always messes with me.

    While I'm here, 21 isn't a bad movie (apart from Spacey) and touches on this question. Worth watching.

    • Qazxswec on 03/02/2022 - 20:50
      +6

      Kevin Spacey is a brilliant actor, maybe a creep in real life, by no means a bad actor though

      • SnowDragon on 03/02/2022 - 20:53
        +2

        I won't disagree with that. American Beauty and The Usual Suspects are fantastic movies!

        • muncan on 05/02/2022 - 11:31

          He was great in Austin Powers: Goldmember too!

      • cookie2 on 05/02/2022 - 13:06
        +1

        The usual suspects is one of my fave movies of all time. Finally got hubby to watch it and appreciate the beauty of it too. I wish House of Cards had continued, pretty good series imo.

    • AustriaBargain on 03/02/2022 - 20:52
      +8

      I think it's because the door Monty opens isn't random at all if you chose a goat on your first choice. The door Monty opens has to have a goat behind it, it can't be the car, and Monty can't open the door you chose. So if you chose a goat, then there's only one door Monty can open which is the other goat. If you chose the car only then can Monty choose which goat door to open randomly. So 50% of the time the door Monty opens isn't random at all, it's forced by you choosing the other goat. It's counterintuitive until you consider that the door Monty opens isn't random at all if you chose the goat. There's two goats, so 66.6% of the time you chose a goat in your first turn. 66.66-33.3 is much better odds than 50-50 or 33-33-33.

      • aragornelessar on 03/02/2022 - 20:57

        Ok, starting to see the picture now… I just took the issue at face value and didn't consider that there'd be other situations that they'd open another door.

        Thought the game show was opening a door to add a sense of drama and would've just opened door 1 if 2 had a car behind it instead.

        • AustriaBargain on 03/02/2022 - 21:02
          +2

          Yes that's the trick, Monty always has to open a door to reveal a goat according to the rules, regardless if you chose a goat or the car. It's not like Deal or no Deal because on that show the host doesn't reveal a low value case to you.

          • aboogee on 07/02/2022 - 09:47

            @AustriaBargain: Yeah. Doesn’t make sense if he opens a door randomly. Eg it’s pretty awkward if he opens the door with a car behind. Lol.

    • bcarp on 03/02/2022 - 21:08
      +1

      I have tried and failed to understand the answer to this problem many times. I feel like I understand probability pretty well but this question just always messes with me.

      Yep. I'm with you. I understand maths quite well. I understand stats. I understand probabilities. And yet, this particular puzzle & the myriad variations of it just messes with my mind. There are just so many ways of angling the answer to give you the answer you want.

      • cookie2 on 05/02/2022 - 13:07

        Don't complicate it. Go by the info in the scenario only.

    • apiecost on 05/02/2022 - 14:26
      -2

      It can be broken down into two simple scenarios.

      1 You choose a door and stick with it. You clearly have a 1 out of 3 chance of choosing the best door.

      2 You choose a door, and then change it. When you change it, one bad door has been helpfully removed so there are now 2 doors left with a 100% chance that one of them is the good door. So you have a 1/2 chance now.

      • SnowDragon on 05/02/2022 - 15:07

        Yes, but apparently switching doors gives you a 2/3 chance of winning after a bad one has been revealed. I'd be happy if it became a 50/50 chance, but somehow it is a 2/3 chance

        • apiecost on 05/02/2022 - 18:16

          I worded my response poorly. If you choose a door and choose to change and are offered 2 doors, then ONLY ONE door of the 3 initial doors can make you lose. So 2/3's of the time you will choose the good door.

        • CanadaAye on 06/02/2022 - 00:19

          This is right, it's 2/3 because the host will always reveal a goat and therefore 2/3 of the time they will indirectly reveal which door has the car behind it.

          In other words there's only a 1 in 3 chance that you will select the car with your first choice, we can all agree on that right? Therefore the other 2/3 times you have selected a goat and the host will be forced to indirectly reveal the car to you by showing where the other goat is. 🐐

        • aboogee on 07/02/2022 - 09:49

          The techies will ask you to write down all scenarios and count the chances of you winning. Eg 2/3

        • filmer on 07/02/2022 - 12:59

          The odds are 50/50 if the host choses a random door which happened to be a goat.

      • CanadaAye on 05/02/2022 - 16:30
        +1

        Ahh yes, but what if you choose again when there are two doors left and you choose the same door you chose initially? This is not keeping the door it's making a new choice that happens to be the same.

        • CanadaAye on 05/02/2022 - 23:19

          Replying to myself, I've decided I was wrong because the first choice odds will still hold at 1/3 as hard as it is for my brain to accept it.

    • darkmattersunB6c0MV on 05/02/2022 - 21:44
      +3

      I will try to explain. The op did not quality the problem with all the conditions.
      It is critical to say that Monty Hall knows what is behind each door.
      There were three options of doors, and hence, you had 1/3 chance of picking the car.

      You not picking the car was 2/3.
      Think of it this way- if you could switch from one door to the two doors you did not pick, you would have a 2/3 chance of winning.
      Now, Monty Hall just opened one of the doors that does not have the car of those two doors.

      So, opening that door is based on the knowledge of where the car is (whether in your original door or not in your original door).
      Thus, switching the door gives you 2/3 chance of winning

    • caitsith01 on 08/02/2022 - 17:34

      Just imagine there are 10 doors and 1 prize.

      You pick 1 door with a 1 in 10 chance of being right.

      The host then eliminates 8 of the remaining 9 doors, knowing where the prize is. You still have a 1/10 chance that your original guess was right, and there is still a 9/10 chance that your original guess was wrong. The host has given you information about where that 9/10 chance lies by eliminating the remaining doors.

  • ozwog05 on 03/02/2022 - 20:56
    +3

    Karl Pilkington: "Can I just 'ave a feel of the door or.."

  • TEER3X on 03/02/2022 - 21:04

    I think the only way for my to make sense of this (i.e better to switch) is to go through all the possible scenarios

  • Dave Bowman on 03/02/2022 - 21:07
    +9

    What if I want a goat?

  • juzza87 on 03/02/2022 - 21:07
    -1

    I still don't understand, your initial odds are 33% of having the car but once he open the door you're in a different situation which is 50%, because the door you chose could be the car. I get that by him eliminating a goat door it changes the odds but there's still a 50% chance you're sitting on the car.

    • Qazxswec on 03/02/2022 - 21:19
      +3

      The Probability is locked in once the first choice is made and there are still three doors, the probability cannot change, no matter what, there is a 66% that you picked a goat, even though he shows you where the other goat is, there is still a 66% chance that the door you originally picked was a goat, look at it as a 66% chance that the door you chose is not the car, because of this, that means that if you swap there is a 66% chance the other door is the car

    • Qazxswec on 03/02/2022 - 21:25
      +8

      Also, think of it like this, if there were 100 doors with 99 goats and 1 car, then after you picked a door he showed you 98 doors with goats and there was only 2 doors left, by the way you are thinking there would be a 50/50 chance you have the car, but in reality, it is a 1 in a 100 chance that you picked the 1 car out of 100 doors originally, so you would have a 99% chance if you swapped of getting the car, with larger numbers it is easier to understand

      • Save 50 Cent on 05/02/2022 - 23:52

        That is actually a really good way to look at it! Makes a lot of sense now.

    • hasher22 on 03/02/2022 - 21:26
      +5

      This is how I understood it:

      Revealing a door with a goat is an illusion. Your initial selection percentage still remains the same at 66% (goat) 33% (car). Switching doors will benefit you as you had a 66% chance of selecting a goat in the first round, that probability never changes.

      Let's bump up the doors to 1000. You have a 1 out of 1000 chances to select the door with the car, that's a 0.1% chance. Those odds suck! But the host removes 998 doors leaving your door and the door with the car. Your initial selection remains the same at 0.1% as you chose a door out of 1000. So if the host ask you to stay or switch, the odds of you winning the car if you switch is 99.9%.

      The only time the odds are 50/50 is if another player comes along and only saw 2 doors and selects 1 out of the 2 doors to win the car. They never had the chance to see the 1000 doors. So the new player picking out of 2 doors has a far worst chance of winning the car than you at 99.9% by switching.

      Hopefully that makes sense. Sorry, I am a bit bad at explaining. I think alot of people get thrown off because it's 3 doors.

      • Qazxswec on 03/02/2022 - 21:53
        +5

        How dare you take my explanation and add another zero to it, lol, we must of uploaded our comments at the exact same time

        • hasher22 on 03/02/2022 - 21:59
          +1

          We did! haha, 1 minute apart :O

          • pegaxs on 03/02/2022 - 22:35
            +3

            @hasher22: And I read both your explanations and have been reading about this problem for 20 odd years and I still don’t get it, no matter how many different ways it is explained to me.

            I get that in the first round, there is a 66% chance (2:3) of getting nothing and only 33% of getting the car. But after they open the door, this should be treated as a new round, as you are being asked again to play in a new round, this time with only 2 doors. You get to make a second choice between 2 doors, not 3 but one is open. You can’t pick the open door.

            How the hell does selecting the other door make its odds jump from 33% to 66%? Or in your case, from 0.1 to 99.9?

            I know it’s not 50:50 (becuse people smarter than me told me so) but of all the math I do and of all the probability I deal with, this is just one that my brain just can’t deal with.

            • sanglt on 03/02/2022 - 23:00
              +7

              @pegaxs: Suppose you play this game 600 times. About 200 times you will pick the right door at the start. Yay! About 400 times you will not. But the game show host will never open the door with the car behind it, so each of those 400 times the car is behind the unopened door that you did not originally pick. So 400 out of 600 times you should switch, i.e., 2/3 of the time.

              • kiitos on 04/02/2022 - 08:59

                @sanglt: I like this explanation

              • Falcone on 04/02/2022 - 10:23

                @sanglt: Based on everything I’ve seen or read about this, I think your explanation is the best I’ve understood thus far!

            • hasher22 on 03/02/2022 - 23:12
              +1

              @pegaxs: I had the exact same thought as you because the second round should be treated as a new round and it confused me even more. But that's the not case, cause you played the first round and have pre-existing data which is beneficial to your win.

              But this is how I manage to rethink that, this is just me though, and hope it helps you:

              The only time it's a 50/50 situation is when a player has no pre-existing data and is purely chosen based on two doors. Let's say I go to your game and I see two closed doors and one open door. You tell me: "which closed door has the car?" and that's when I have a 50% chance of choosing the car. You didn't give me any pre-game information, you didn't give me why is there an open door with a goat standing there, you simply told me, to choose a closed-door that may have a car behind it. All I know is that there are two closed doors and one has a car behind it and I am forced to choose a door.

              "How the hell does selecting the other door make its odds jump from 33% to 66%? Or in your case, from 0.1 to 99.9?"

              Maths and gambling. With my extreme example of 1000 doors, the odds of me selecting the right door is 0.1%, when the host takes away 998 doors, that leaves 2 doors. At this point it's a gamble but you have a huge advantage to win because you got pre-existing information (0.1% chance to select the right door and the host took away 998 doors), you use this information to your advantage and come to a conclusion and decision to stay or switch.

              Do you trust yourself that you got the right door at 0.1% in the first round? Because the host took away 998 doors and what's left is 2 doors. The smartest action is to switch doors to win because the host took away 99.8% of the probability (998 doors) and transferred that 99.8% to the other door which bumps up that door's probability to 99.9%, you chose a door in round one with a 0.1% probability. The question to yourself now, do you gamble and stick with your decision, or do you switch? Remember, in the first round, you had 0.1% chance of choosing the right door. But because in round two, 998 doors were eliminated, that 99.8% probability transferred to the 2nd door which makes it 99.9% that most likely has the car. You must be damn lucky to select the right door in round one with a 0.1% win rate.

              • gimme on 07/02/2022 - 11:55

                @hasher22: penny finally dropped for me after this explanation.

            • Qazxswec on 03/02/2022 - 23:15

              @pegaxs: well if you pick 1 out of a 1000 doors(999 goats & 1 car), youre not going to get it right are you(obviously it is possible, but unlikely), so you are going to pick a goat, since the host knows where the car is, he shows you every other door that is a goat and not the car leaving 2 doors, the door you picked, and more than likely, the car, now the chances that you picked the car right is 1 in a 1000, so there is a 999 in a 1000 chance you picked a goat, because he eliminates every other goat for you by showing you every door except 2(yours and the other one), chances are, the car is in behind the other door

            • tofuofdoom on 04/02/2022 - 11:26

              @pegaxs: The trick is. There is no new round. There is exactly 1 round. The first round.

              You already knew one of the doors you didn't pick was a goat. The presenter showing you that goat doesn't change anything.

            • aboogee on 07/02/2022 - 09:53

              @pegaxs: The 1000 doors example is easy to visually. What’s your odds of picking a car? 1 in 1000. For all intensive purpose let’s pretend you always always gonna get a goat here. The presenter then eliminates 998 goat for you. The remaining door he doesn’t open has to be a car. There is only 1/1000 chance that this is not the case ie you pick a car initially lol

      • filmer on 04/02/2022 - 13:24

        This is only correct if the host intentionally choose a dud door over the winning door every time. See my full breakdown with a simulation run.

        • Qazxswec on 04/02/2022 - 17:38
          +4

          he does choose the dud door every time, thats how the game works

          • filmer on 07/02/2022 - 09:13

            @Qazxswec: Not how EVERY version of this game works actually, but yes, the monty hall version. I find it's important to mention because people fail to understand WHY it works.

            • Qazxswec on 08/02/2022 - 18:09
              -1

              @filmer: and yes, it is how every version works, because there is only 1 version."The Monty Hall Problem" is a very specific thing, the only version of the "The Monty Hall Problem" is the "The Monty Hall Problem" and that is what this post is about, is there another 3 door game? Here at Ozbargain we don't know if there are other 3 door games, Frankly, we don't want to know, So Long

      • Viper8 on 07/02/2022 - 13:28

        Lol!
        That's not how probability works.
        The instant 1 of the 3 doors is removed, your odds improve to 50% for BOTH doors. You don't have to change your selection in order to achieve 50%, that was done the instant one of the 3 doors was removed.

        Same with your 1/1,000 —> 1/2 example. The instant 998 are revealed to have goats, your odds improve to 1/2 REGARDLESS of whether you change your bet or not.

        • b0mber0 on 07/02/2022 - 13:57
          +1

          Probabilities are time-bounded. It's true that if you were to choose after the one door is revealed you would be at a 50-50 scenario, however, you chose when there were three doors, making your choice a 1/3 scenario. It's important to define when the choice is made.

          Same for the 1,000 or any higher number example. If you arrive at two doors, it's always a 50-50, but you chose before that.

          What happens, and its a bit weird to understand at first, is that when the game starts your choice it's locked at 1/3 (or 1/1000 in the other example), and every other door is 1/3 as well (or 1/1000), but then, when the other non-winning doors are opened and only one is left, for your locked in situation, the remaining door must concentrate the rest of the probability because the game is still occurring, meaning that now the remaining door must be 2/3 (or 999/1000) because the probability space hasn't changed. The same remaining probability now has only one "exit" making the decision to swap door the better option in this game.

          But again, is true that if you arrive late to the game you only see a 50-50, or, you are much less probable to pick wrong. Is the probability of initially picking wrong the one that makes this seem not logical.

          • Viper8 on 07/02/2022 - 14:20

            @b0mber0: Again, that's not how probability works lol.
            In this case the improvement in probability comes from the doors being removed, not from you selecting any of the remaining doors. You don't need to make a new choice to "take advantage" of the improved probability lol.
            Making the choice is not what has improved your odds, it's the removal of one of the 3 doors…

            Let me give a plain example. There are 2 doors. You select one for 50/50 odds. They remove one door. Clearly, your odds are now 100%. Yet what you and 90% of the people here believe, is that your odds remain 50/50, until you re-confirm your choice of door. Again, the instant that door is revealed, the odds need to be updated. This is independent from any choices you have made.

            TLDR: odds don't just change when YOU make a choice, but when the universe around you changes as well. And you don't need to update your choice for them to take effect. They take effect the instant those changes have occurred.

            • hasher22 on 07/02/2022 - 15:08
              +1

              @Viper8: In your example, 2 door probability is different from 3 door probability then choosing a second time when the 3 door example.

              If b0mber0, myself, and 90% of people on here decide to switch but you're in the minority 10% that thinks the probability is 50/50 after a door is revealed then I think you should understand why 90% of people think switching is a good idea.

              The 3 doors can confuse a lot of people, but when you increase the doors to 5, 10, 100, 1000, it's easy to understand that switching is the logical answer and the probability remains the same in the picking round, whether or not a door has been revealed or not.

            • Qazxswec on 07/02/2022 - 15:15

              @Viper8: your original choice locks in the probability, then the universe changes it for you, so in the 1 in a 1000 scenario, you can either stay with your 1 in 1000 chance, or you can change to an almost 999 in a thousand 1000, sorry mate, i understand what you are trying to say, unfortunately it is not correct

            • b0mber0 on 07/02/2022 - 15:51

              @Viper8: You are completely right in the first part, and that is exactly what people here is trying to explain. What seems like an improvement in odds comes from the removal of the doors. Also, you are right that odds change when the universe changes, but here is the counterintuitive part of this: the universe has not changed. The car is at one door and is not going to teleport to another door just because other doors were opened, so it stays in its original place.

              Nobody is saying that you have to change door to make the probabilities change or that odds change when you make a choice, quite the opposite as you say. Doors go away and the probability of the remaining door changes as a consequence. But remember that the tricky part of this is that the removed doors are always "losing" doors.
              And regarding your example, in the two door game, if you know that there is only one door left obviously you shouldn't change for the same reasons everybody is explaining. Is not the changing whats important. Is what you learn when doors open.

              Look at your "1/1,000 —> 1/2 example" in your original text. You pick one door and there are other 999 doors. The car is not going to teleport. When 998 doors are opened you say that there is a 50-50 chance between your pick and the last door. The car is not moving. you are implying that your original door has a 50% chance of getting the car. So we should be able to skip the door opening process and out of 1,000 doors, we should get a car in our first pick 1/2 times of all tries. And of course, that is not going to be the case.

              Maybe this will help:
              When you pick the door you say "there is 1/1000 chance that the car is here and a 999/1000 chance that the car is in any other door". That makes sense, right? here is when the probability is locked in.

              Then this guy comes and opens 998 of the 999 doors that form the group where the car is 999/1000. This is the extra info you get that almost seems like cheating.
              There are still 1000 doors. You picked one for a fair 1/1000 chance and left 999 doors, but then you know for sure which 998 doors don't have a car out of a group of 999 doors where, as a group, there was a collective chance 999/1000 of having a car.

              So what's the probability that the car is in this single door given that now you know that the 998 other doors don't have a car? 999/1000. While your original pick is still 1/1000. Remember that the car doesn't teleport and this guy purposely opens the doors without a car until only one is left.

              Now when you scale this to three doors it gets really confusing, but it still holds.

              Hopefully this will help.

        • Qazxswec on 07/02/2022 - 15:07
          +6

          you are so wrong it is not even funny, think about it, if someone asks you to guess a number they are thinking of between 1 and a 1000 and you guess 447, but the number they were thinking of was really 313, now they eliminate every other number except the real number and your guess, and tell you, the number they are thinking of is either 447 or 313, do you stick with your 1 in a 1000 guess of 447, or do you realise you were probably wrong and switch to 313 which is almost certainly the real number they were thinking of.
          Like i said. the probability is locked in when you make the first decision, if i eliminated 998 to just leave 2 doors, it would only ever be 50/50 if someone new came along and guessed between those 2 doors

          • Viper8 on 07/02/2022 - 17:16
            +1

            @Qazxswec: Well said.
            So your odds improve from 33% to 67%, and in the 1000 example, from 0.1% to 99.9%.

            • Qazxswec on 07/02/2022 - 17:41
              +1

              @Viper8: Yes, that is correct :)

              • filmer on 08/02/2022 - 12:07
                -2

                @Qazxswec: Your example is actually quite bad. If they don't tell you if your guess is wrong or right, and they eliminate all numbers except for 313 and 447 then you have a 50% chance to be right. The 313 could be just an arbitrary value and the 447 could be the correct guess.

                • Qazxswec on 08/02/2022 - 16:59
                  +1

                  @filmer: I am starting to think you people are just trolling, no it doesn't mean your chances are 50/50, your chance is locked in when you made your choice picking 1 out of the 1000 doors, that is the whole point of the scenario, the example is perfect, whether or not you can understand it and agree with it, it doesn't matter, it is literally a fact

                • Qazxswec on 08/02/2022 - 17:00

                  @filmer: and yes, you're right, 447 could be the right number, there is a 1 in a thousand chance of that

      • Typical16-bitEnjoyer on 09/02/2022 - 08:28

        This is the post that made it all click for me. Thank you!

    • SydStrand on 03/02/2022 - 22:26

      The Wiki article explains it pretty well, with diagrams. The OP is explaining in 500 words what you can do in 100.

      • Qazxswec on 03/02/2022 - 23:20
        +4

        lol, dude that wiki explanation is the most complicated thing i have ever seen, its like 10 pages long, 100 words, lol

  • havebeerbelywillsumo on 04/02/2022 - 00:12

    With my luck the car would be a trabant!

  • [Deactivated] on 04/02/2022 - 00:31
    -2

    You're assuming the games weren't simply rigged in the first place.

  • Diverse on 04/02/2022 - 03:10
    +13

    The host will always open the door with the goat, regardless of what you have chosen.

    2/3 times, you will choose the door with the goat, switch and you will get the prize.

    1/3 times, you will choose the prize, switch and you will not get the prize.

    Therefore, overall 66.7% of the time you will win by switching. "Thank you for that extra 33.3%"

    Here's a visual representation: https://i.imgur.com/yyhikvg.png

    • Cliche Guevara on 04/02/2022 - 09:45

      This is the way…

    • spiff on 05/02/2022 - 13:43
      +1

      That pic explains it better than words can lol

    • filmer on 06/02/2022 - 15:11
      +2

      This only holds true IF the host always shows a goat which is not stipulated here.

      • Diverse on 06/02/2022 - 16:00
        +1

        Good point, OP wasn't clear but did mention the Monty Hall problem which is that he will always open the door the a goat. Don't see why he would open to the car at all.

  • TheDolphin on 04/02/2022 - 07:50

    The best way to understand it is to think of the extremes.
    If there were 100 doors, you choose one at random, Monty now opens ALL the other doors except your choice and one other, would you switch?
    The key is that Monty has information ( he knows which door has the car ) once Monty opens a goat door he has given you more information.

  • 82norm on 04/02/2022 - 08:22

    One thing I’m sure, people are still confused with the explanations given. Waiting for the single definitive, easy to understand, without having to rack my brains over it answer….. any takers?

    • Qazxswec on 07/02/2022 - 15:26

      You are statistically likely to make the wrong choice (2 out of 3 chance), after you do make the wrong choice, the host takes away every other wrong choice, leaving you with only the right choice left to choose from

  • Krankite on 04/02/2022 - 08:36
    +4

    The more interesting part of the problem is the psychology of it. Even after explaining the odd's people will still be inclined to stick with their original option and will be more disappointed if they switch to the goat than if they stay with the original pick and end up with the goat.

  • GrueHunter on 04/02/2022 - 11:19

    This problem is notable because when Marilyn Vos Savant correctly answered a question about it, she received 10,000 letters mansplaining why she was wrong.

  • Mulga Bill on 04/02/2022 - 12:09

    SWAP…if we look at the probability at the time of the "swap" decision with no additional info…there are 2 doors to choose from, one has a car and one has a goat = 50/50…but we already know we are unlikely to have chosen the car from step 1 so makes sense to swap.

  • filmer on 04/02/2022 - 13:03
    +2

    I decided to write a quick little app to simulate 100000 iterations for the different conditions, and the results are as follows.

    When the host always shows a goat and the player swaps. The player won 66667 and lost 33333. Player wins 66.667% of the time
    When the host always shows a goat and the player stays. The player won 33566 and lost 66434. Player wins 33.566% of the time
    When the host shows a random door and the player swaps. The player won 33402 and lost 66598. Player wins 33.402% of the time
    When the host shows a random door and the player stays. The player won 33264 and lost 66736. Player wins 33.263999999999996% of the time
    When the host shows a random door which is a goat and the player swaps. Out of 100000 games a goat was shown 66744 times. The player won 33313 and lost 33431. Player wins 49.91160254105238% of the time
    When the host shows a random door which is a goat and the player stays. Out of 100000 games a goat was shown 66548 times. The player won 33450 and lost 33098. Player wins 50.264470757949155% of the time

    This means, it depends on the rules of the game.
    - If the host always shows a goat then we should swap. This is the first result set and gives a 66.66% chance of winning (2/3).
    - If the host chooses a door at random then it makes no difference if you swap or not. As you can see in the last 2 result sets both result in a 1/3 chance of instantly losing, and a 50% chance of winning with either choice if a goat is shown.

    I believe this contradicts the movie 21 as I don't think it stipulates the host will always choose a door with a goat. Correct me if I'm wrong here.

    Edit: Missed the last line of output.

    • aragornelessar on 04/02/2022 - 13:49
      -1

      Your results look like the assumptions in the wiki about the problem that aren't explicitly defined in the OP's post

      Under the standard assumptions, the probability of winning the car after switching is 2/3

      The key to this solution is the behavior of the host. Ambiguities in the Parade version do not explicitly define the protocol of the host. However, Marilyn vos Savant's solution[3] printed alongside Whitaker's question implies, and both Selven[1] and vos Savant[5] explicitly define, the role of the host as follows:

      The host must always open a door that was not picked by the contestant.[9]
      The host must always open a door to reveal a goat and never the car.
      The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

      When any of these assumptions is varied, it can change the probability of winning by switching doors as detailed in the section below. It is also typically presumed that
      the car is initially hidden randomly behind the doors and that, if the player initially picks the car, then the host's choice of which goat-hiding door to open is random.[10] * *Some authors, independently or inclusively, assume that the player's initial choice is random as well.[1]

      • filmer on 04/02/2022 - 13:56
        -1

        My results are based on either situation, whether the host intentionally choose a goat door or not.

        A: player chooses random door. Host chooses non car door (random if player chooses car). Player swaps.
        B: player chooses random door. Host chooses non car door (random if player chooses car). Player stays.
        C: player chooses random door. Host chooses random door (random if player chooses car). Player swaps if it was a goat.
        D: player chooses random door. Host chooses random door (random if player chooses car). Player stays if it was a goat.

        Then the specific breakdown of that odds if the player swaps and stays after the host chooses a goat when he selects a door at random.

        This covers all situations.

        If the host HAS to choose a goat door, then yes switching is better. If the host doesn't have to then it makes no difference. Not knowing the rules, you would swap because at worse it makes no difference, but at best you do double your chances.

  • Ghost47 on 04/02/2022 - 13:39

    Cool game!

  • bcarp on 04/02/2022 - 17:52
    +2

    Well, because the logic & maths stumps me the moreI think about it, I whipped up a script to randomise it & see what happens. Applescript because that's what happens to be installed on my computer:

    https://imgur.com/a/uuIQujW

    I don't know how to post the script here without taking up screenfuls of lines in a single comment.

    • bcarp on 04/02/2022 - 17:56

      Oh oops, I see that @filmer already something similar above. Sorry!

      • bcarp on 05/02/2022 - 16:48

        Hmm. Somethings not right with my scripting. I’m getting 50% success instead of 67% success when switching.

        • bcarp on 05/02/2022 - 23:24

          Ah, found a coding error. Fixed it. Images above are now correct.

  • gromit on 05/02/2022 - 07:43

    You have a 1 in 3 chance for initial selection (~33%), the other two doors have a 66% chance of containing the correct door, so yes I switch.

  • acermaniac on 05/02/2022 - 08:22

    Here's the way I try and convince people.

    Imagine if instead of 3 doors, there's 100 doors. You pick one and the host reveal 98 doors with goats.

    In the end you are still left with the 2 doors (so the premise is the same) but instead of 2/3 advantage which may not be evident in small sample size, you now have a 99/100 advantage which is simpler to demonstrate by getting them to try themselves (e.g. through an app).

  • rogerwilko on 05/02/2022 - 09:50
    +1

    When Eddie asks "do you want to lock it in" and you have no idea of the answer, do you randomly change your selection with 3 seconds remaining in the clock.

  • t_c on 05/02/2022 - 11:26
    +2

    The solution is simple, take the goat, milk it and sell the organic goat milk at the farmers markets on the weekends. Use the revenue to buy a much better car than the crappy ex-demo that the game show is trying to pass off as new.

  • Dabbler68 on 05/02/2022 - 12:09

    My brain still thinks 50/50, but seeing the answer graphically makes it simple …. see the image posted by Diverse at 03:10 yesterday. 66% chance of winning by swapping.

    Another way to think of it. The odds of picking the car in the first choice are 1 in 3. The host has zero chance in 3. That leaves 2 chances in 3 the other door will be the car.

  • cookie2 on 05/02/2022 - 12:54

    I say switch ofcourse. Except if you want a goat. If you 100% want a goat then you'd be crazy to switch. I would like to know if it's a miniature goat tho :) then i might not switch.
    Edit. Haha i just reread the riddle. I'd stay on my initial choice. Nothing has changed ..i still secretly hope i get a goat tho.

  • [Deactivated] on 05/02/2022 - 14:04

    Is there a video where he actually does this? I've seen a few Let's Make a Deal videos on YouTube, and if there are three doors, he has two contestants and he asks them to choose one door each, that's it. He often gives people a choice of either one thing or the other, but I couldn't find a single one where he has one contestant choose a door, then reveals one of the doors and asks them if they want to switch.

    • hasher22 on 05/02/2022 - 14:30

      interesting cause I can't find the actual game footage myself. This is the closest https://www.youtube.com/watch?v=J3Iqh4b4XKs but not solid evidence it is it.

      • [Deactivated] on 05/02/2022 - 15:05

        I also saw one where the single contestant had $9k and had the option to risk it all on one of three doors. One had $20k, and if you chose that, you keep the $9k and the $20k. The second was some random small amount, and the third was an unknown amount. If you chose the other two, you would only keep that amount, and lose the $9k. They made one choice. They revealed their door choice, and that was it.

        Either it wasn't very common on the show, or no one has uploaded it to YouTube. It definitely wasn't on every show. There are entire episodes on YouTube. There's plenty of people explaining it, as if it needs explaining, but no footage of the original.

        With the one I mentioned above, they reveal a sliding flat door with a flat number on the back, so it would be very simple to rig. I noticed the same with the larger doors. They would all be on spinning platforms. That's why I was curious what the real thing looked like.

  • THICKnSLOW on 05/02/2022 - 14:32
    +1

    I'd squeeze the teat of the goat till it bleats, then the other goat would bleat too, telling me which door it is behind.

    Then select that the other door with the car.

    Simple and easy to solve puzzle.

  • haggertm on 05/02/2022 - 19:23
    +1

    I would switch because I have seen 21.

    • harthagan on 06/02/2022 - 10:29

      Always account for variable change

  • lowlifesphere on 05/02/2022 - 20:10
    -1

    Choice 1: 33% chance of picking a door with a car
    Choice 2: 50% chance of car at current door and 50% chance it is at door 1.

  • relentless on 06/02/2022 - 00:14

    Yes, no, maybe - 33 and a third. This is the way.

  • whatsthetrend on 06/02/2022 - 01:54

    Yes, you should switch.

    The Monty Hall problem was featured in the recent series of Survivor US, where one of the contestants
    had to choose the right box to avoid being eliminated from the game.

    Following this episode in one of the Survivor discussion podcasts, one of Survivor's well known previous contestants, Christian Hubicki,
    who is actually a robotics professor and scientist explained how the Monty Hall problem works and why you should switch.

    https://www.youtube.com/watch?v=RFIyWMzYsuc&t=1639s

  • ozbargainsam on 06/02/2022 - 11:37

    The game just boils down to: did you get it right the first time or not?

    When it was one choice of three, was I right (1/3) or not (2/3)?

    If it was a million doors: was I right (1/1,000,000) or not (999,999/1,000,000)?

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