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

  • Duckie2hh on 03/02/2022 - 20:10
    +35

    Everyone knows its better to switch.

    • md333 on 04/02/2022 - 14:36
      +1

      Tried it on a bunch of people a few years ago, and it seemed like people with a PhD all insisted there was no reason to change, no matter how hard you tried to explain to them that they were wrong.

      • T1OOO on 05/02/2022 - 13:47
        +2

        So then do it with 10 doors. Pretty good to be picking the right one. Do it 10 times and ask them how they know where it is every time. Doing it with 1 out of 1m illustrates the issue easily.

        • Scrooge McDuck on 05/02/2022 - 16:27
          +2

          Before the reveal there is a 1/3 chance of winning. Whether we chose the winner or not there is always at least one remaining unchosen loser. So the revealed loser provides us no information about our first choice and thus it still has a 1/3 chance to win. However we now know that the one remaining unchosen option is within a group of only two possible winners. The reveal has provided us this information about the remaining unchosen option but not about the first choice since that option could not possibly be revealed. Since the total probability of either remaining choice winning is 1 (certainty) and we know that our first choice is 1/3, the probability of the unchosen remaining option is the remaining 2/3. Therefore the optimal strategy is to switch choices which yields a doubled return.

          • Scrooge McDuck on 05/02/2022 - 17:00

            @Scrooge McDuck: Also 2/3 expressed as a percentage is not 66 %, it is 67 % using a precision of 2 significant figures.

            • Scrooge McDuck on 05/02/2022 - 17:07
              +1

              @Scrooge McDuck: Another thing, the poll isn't valid. The previous responses are visible which biases future responses. And providing the name of the problem makes it trivial to read the solution from Google.

          • Donaldhump on 05/02/2022 - 18:11

            @Scrooge McDuck: Ignore

          • T1OOO on 07/02/2022 - 00:36
            -2

            @Scrooge McDuck: My ways simpler to explain that’s the whole issue with this problem.

      • Kangal on 05/02/2022 - 14:08
        +4

        The crux of the issue is, both Door 1 and Door 2 have a 1/3 chance of having the prize. And when Door 3 is opened to show it has nothing, well, then what happens to the 1/3 chance that Door 3 used to have, since it is now updated to be a 0/3 probability.

        A) Does it transfer equally to the other two doors?
        B) Or does it transfer fully to a single door?
        C) Or is it something in between?

        This is the crux. And depending on the circumstance of the problem it could be any of the three options. In many ecological statistics, it's actually three, since the environment is a dynamic playing field. In some computational problems, it is the first option. In the TV-Game (Monty Hall) it is the second option.

        • caitsith01 on 08/02/2022 - 12:37

          But that's just adding variables that don't exist in the problem. In the correctly expressed Monty Hall problem it is always B.

          It is extremely misleading to think of the probability as 'transferring'.

          Before opening a door, the chances of it being the door you pick are 1/3 and the chances of it being any other door are 2/3.

          After opening one of the other doors the chances of it being the door you pick are still 1/3 and the chances of it being the one remaining door is still 2/3.

          • Kangal on 08/02/2022 - 13:42

            @caitsith01: Not at all.
            It is very human and intuitive to think of things happening in real-time/dynamic, so that chances do change with events.

            Now, I'm not adding more variables. I'm explaining the crux of the issue. What happens in the Classic Monty Hall problem is that the probability of the prize is entirely transferred to the third door. So you double your chances if you switch, but that still doesn't guarantee success 100%.

            And to go one step further, let me say this.
            Imagine you're playing the same Monty Hall Game. But you have two contestants. And they choose one door each. And another door is opened to reveal it is empty. So now, should both contestants Stay or Switch?

            According to your idea, each has a 2/3 chance if they switch, which is logically broken. The answer won't change if they both Stay (one wins, another loses) or if they both Switch, so there is always a winner and "The House" loses. But if one of them Stays and the other Switches, they're both betting on the same Door. And now they either both win, or they both lose. And in that circumstance, the house can win. So what is really happening here is that the opened door (1/3) chance is transferred to another door, but we don't know which one.

            If you find this game interesting, there's a somewhat similar one called Split or Steal. And similarly there's a fairly easy solution to that game as well: https://youtu.be/S0qjK3TWZE8

            • caitsith01 on 08/02/2022 - 17:20

              @Kangal:

              According to your idea, each has a 2/3 chance if they switch, which is logically broken.

              With respect, this shows a fundamental misunderstanding of why the answer to the original problem is that you should always switch.

              And yes, both contestants should still switch, because it remains the case for both of them that there is a 2/3 chance that their original guess is wrong and a 1/3 chance that it is right, coupled with the additional information now given to them that the other door is empty. The fact that there are two of them adds nothing to the scenario.

              Of course, the major problem with your example with two contestants is that there is a 2/3rds chance that the host will be UNABLE to show you an empty door, because it is quite possible that the two contestants have selected the only two empty doors possible and the remaining door has the prize.

              Again, no chance is "transferred" anywhere. There is one prize, and it remains in the same place. Each contestant still has a 1/3 vs 2/3rds win/lose chance, and then the host revealing the empty door allows them both to switch that to a 2/3rds vs 1/3rd chance (except that, in many cases, this scenario fundamentally can't happen because the host can't show them an empty door).

              I think the problem with your reasoning is that you are treating the two doors NOT selected by the contestant as each having a 1/3rd chance of having the prize, when the better way to look at it is that those two doors together have a 2/3rds chance of having the prize. That remains the case when you reveal that one of them definitely doesn't have it.

              If you find this game interesting, there's a somewhat similar one called Split or Steal. And similarly there's a fairly easy solution to that game as well: https://youtu.be/S0qjK3TWZE8

              Respectfully, I think you should work on understanding the (empirically provable) Monty Hall problem before recommending 'easy solutions' to things.

              • Kangal on 08/02/2022 - 20:50

                @caitsith01: You're right about the 2/3 chance of winning when you have two contestants in the same game.

                I do understand the Monty Hall problem. The link to the other game wasn't to show comparisons. It was there to show that simple games like this is a common place on TV, and aren't subject to just the old days. And usually there are solutions to these games.
                …in fact, Deal or No Deal, well that is the Monty Hall game. Except there's more briefcases. There are intermediate stops. And offers made by the dealer. So its a fundamentally different game but the root is the same.

                No, I understand you, but you didn't get my point.
                The Classic Monty Hall:
                Door A has 1/3, Door B has 1/3, and Door C has 1/3 chance.
                You pick Door A. You have 1/3 chance you picked right.
                There is 2/3 chance the prize is in the other doors.
                If Door C, which has a 1/3 chance of having the prize, is opened and revealed to only have a goat. Well, we know Door C actually does not have a 1/3 chance of having the prize. It has 0/3 chance. And both Door A and Door B, both had 1/3 chance.
                …this is from the perspective of the player!
                From Monty's perspective this is NOT true, because he knows the location of all the prizes. In the game he is forced to open one of the false doors. From his perspective there's a 100% chance of choosing the correct door. But from the player's perspective if he switches, it does not guarantee that he wins. The odds from Door C, actually gets transferred to Door B. So we have Door A with 1/3 and Door B with 2/3.

                The reason why 100% of the probability (1/3) of Door C is transferred to Door B, is because Monty Hall actually KNOWS where they are.

                If the game was different, and Monty Hall, did NOT know what was where, there is a good probability that the opened door (eg Door C) may have a car. So in that scenario, both Door A and Door B no longer have a 1/3 chance each, but have 0/3 chance each. You can say their probability "transferred" to Door C which cannot be chosen. But that would make for a very awkward game. And in the other two scenarios where Monty doesn't know, and a door is opened, which has the goat, that was pure chance. So where does the 1/3 chance split to? It splits evenly to the two doors.

                • Kangal on 08/02/2022 - 21:20

                  @Kangal: Okay, let me try to explain it better.

                  Classic Let's make a Deal (Monty Hall knows everything)
                  "Car is in Door C"

                  If you never switch:
                  You pick Door A, Monty shows Door B, you stick to A and Lose.
                  You pick Door B, Monty shows Door A, you stick to B and Lose.
                  You pick Door C, Monty shows Door A/B, you stick C and Win.
                  …total chance is 1/3

                  If you always switch:
                  You pick Door A, Monty shows Door B, you switch to C and Win.
                  You pick Door B, Monty shows Door A, you switch to C and Win.
                  You pick Door C, Monty shows Door A/B, you switch to B/A (leftover) and Lose.
                  …total chance is 2/3

                  Modified "Let's make a Deal" (Monty doesn't know the Car, your door isn't opened)
                  "Car is in Door C"

                  If you never switch:
                  You pick Door A, Door B is shown, you stick to A and Lose.
                  You pick Door A, Door C is shown, prize is lost.

                  You pick Door B, Door A is shown, you stick to B and Lose.
                  You pick Door B, Door C is shown, prize is lost.

                  You pick Door C, Door A is shown, you stick to C and Win.
                  You pick Door C, Door B is shown, you stick to C and Win.
                  …total chance is 2/6… ie 1/3

                  If you always switch:
                  You pick Door A, Door B is shown, you switch to C and Win.
                  You pick Door A, Door C is shown, prize is lost.

                  You pick Door B, Door A is shown, you switch to C and Win.
                  You pick Door B, Door C is shown, prize is lost.

                  You pick Door C, Door A is shown, you switch to B and Lose.
                  You pick Door C, Door B is shown, you switch to A and Lose.
                  ….total chance is 2/6… ie 1/3

                  Total Random "Let's make a Deal" (No-one knows, your door might be opened)
                  "Car is in Door C"

                  If you never switch:
                  You pick Door A, Door A is shown, game is over and Lost.
                  You pick Door A, Door B is shown, you stick to A and Lose.
                  You pick Door A, Door C is shown, prize is lost.

                  You pick Door B, Door A is shown, you stick to B and Lose.
                  You pick Door B, Door B is shown, game is over and Lost.
                  You pick Door B, Door C is shown, prize is lost.

                  You pick Door C, Door A is shown, you stick to C and Win.
                  You pick Door C, Door B is shown, you stick to C and Win.
                  You pick Door C, Door C is shown, prize is Won.
                  ….total chance is 3/9… ie 1/3

                  If you always switch:
                  You pick Door A, Door A is shown, game is over and Lost.
                  You pick Door A, Door B is shown, you switch to C and Win.
                  You pick Door A, Door C is shown, prize is lost.

                  You pick Door B, Door A is shown, you switch to C and Win.
                  You pick Door B, Door B is shown, game is over and Lost.
                  You pick Door B, Door C is shown, prize is lost.

                  You pick Door C, Door A is shown, you switch to B and Lose.
                  You pick Door C, Door B is shown, you switch to A and Lose.
                  You pick Door C, Door C is shown, prize is Won.
                  ….total chance is 3/9… ie 1/3

                  Does that make more sense?
                  Where the probability transfers to depends on the condition of the game. If the host knows, then you effectively double your chances by switching. If he doesn't know, there is no difference between switching or sticking. The opened door splits its probability equally. If it is completely random, again there is no difference. And it is in our human nature or intuition to understand this change of events/probabilities, but in the case of Monty Hall as the host, he acts as a Control. Does that make more sense?

    • anonymous01 on 06/02/2022 - 01:23
      +4

      I disagree. Yes I've read all the arguments about why it's better to switch, but switching your opinion or decision does not switch reality. Once the host opens the door 2 and shows there is a goat, your chances between door 1 and door 3 change from 33% each to 50%-50%. The chances between door 2 and door 3 don't remain at 33% percent anymore, so the probability does not change for switching to a new door.
      That is why the greatest mathematicians fail at this. I suck at maths though, so I don't know. Happy to argue with anyone who has a PHD in maths.
      P.S. I am really dumb at maths, I really mean it.

      • Devils Advocate on 06/02/2022 - 11:42
        +15

        You need to factor in that the host was always going to select an incorrect door after your guess, then give you another chance. That means that the only way you could lose the game is if you had the door right the first time (which is 1/3).

        If you had the door wrong the first time (2/3), switching will always give you the correct door.

        • Wiired on 06/02/2022 - 19:01
          +2

          If it was actually presented that way you would be correct. But it was being presented as door selection and structure won't help and the odds of winning were 33% (which we can assume was a typo) was told to us up front.

          Due to the lack of information and the assumption that they aren't trying to give us cars with better odds in our favor (they would just use two doors for that) it's likely a second chance could be deception. Without information again (other than your chances of winning is 33%) the best social game would be to stay on your door unless you had a read and gathered more info.

          Really the odds shouldn't have been provided as it sort of implies they don't change, as well as other details should have been made clearer.

        • anonymous01 on 06/02/2022 - 20:35
          -1

          I still don't understand. Why is the maths not taking into the account that probability changes after the host opens the incorrect door with the goat. The whole equation changes, and so do the odds.

          • MrFrugalSpend on 08/02/2022 - 19:17

            @anonymous01: Because probability like you speak of is based on random events. You'd only have 50/50 chance where the result is random - like flipping a coin. Whereas in this case, the probability is NOT random, the host is not flipping a coin to select a door, the host is using their insider knowledge to deliberately select a door that has a goat behind it instead of the car.

            There is a 66.7% chance your initial choice was wrong. The host then interferes with the random odds using insider knowledge to eliminate a door and give you another choice.

            66.7% of the time your initial choice was wrong - which means 66.7% of the time the host could only open one remaining door, which also by default means the remaining door always has the prize behind it which means the remaining door has a 66.7% chance of containing the prize!!

            Amazing!

        • sir-screwball on 07/02/2022 - 13:17
          +2

          After pondering this problem since it aired on Brooklyn 99 you have just made yourself the only person who could get the concept through my thick skull.

          Thanks.

        • anonymous01 on 07/02/2022 - 19:25
          -2

          Can someone who is really good at maths, please argue with me. I am willing to lose if I can learn something new.
          Repeat of my question: Why is the maths not taking into the account that odds change to 50% each on two unopened doors, once the host opens the one incorrect door with a goat?

          I guess not everyone knows about the Schrodinger's cat phenomenon.

          Edit: I am just connecting the dots.

          • Bren20 on 07/02/2022 - 20:29

            @anonymous01: Honest question, do you agree with my explanation below? I'm just trying to work out exactly where your understanding falls over

            The reason why it doesn't reset to a 50/50 is because it is conditional probability. I'm good at maths and that's the best answer I can give.

            • anonymous01 on 07/02/2022 - 20:36

              @Bren20: Yes I do understand your maths logic. But why is it that once the host opens the door, the probability doesn't reset? What if we were to restart the whole game with only 2 doors? What if even before the host asked the question, what if the participant was told beforehand that door 2 has a goat. Here is door 1 and door 3, one contains a car, one contains a goat, pick one. Wouldn't the odds be 50% each?

              • Bren20 on 07/02/2022 - 20:54

                @anonymous01: I think the key is that Monty doesn't choose a random door, he takes away a goat. Therefore we can't use straight probability to understand the problem, as the odds are subject to the condition that Monty will always open a goat once one door is already chosen. The simple rules of probability no longer apply. It's been 10 years since I've properly understood conditional probability so I can't explain the hard maths any further.

                If Monty flipped a coin to choose a door, then the odds would remain 50/50, but Monty could get the car and then your odds are 0

                • anonymous01 on 07/02/2022 - 21:32

                  @Bren20: That does make a little sense I guess. So it depends on whether Monty picks the door before or after the question is being asked. If he picks the door before the question, then the odds are definitely 50/50, but after the question, he's eliminated a small chance of you being wrong. So it also depends on what Monty knows. He must know beforehand which door does not have the car, and his knowledge of that helps you a little. Which bring me back to "Schrodinger's Cat". Nothing can be known for sure until you open the door and examine for yourself.

                  • Bren20 on 08/02/2022 - 07:53
                    +1

                    @anonymous01: I don't think schroedingers cat is helpful to your understanding here. Schroedingers cat is about quantum mechanics where matter literally exists in two states until it is observed. Behind the door you have a tangible object, a goat or a car. Schroedingers cat is not about uncertainty or not having enough information on hand, the cat is literally dead and alive, and not just because we don't know which one it is.

                    he picks the door before the question, then the odds are definitely 50/50

                    It doesn't matter so much when he picks, just that he can't know where the car is. As soon as he has that information, all random probability goes out the window

                    • anonymous01 on 09/02/2022 - 11:52
                      +1

                      @Bren20: Thank you for being so patient with me lol.. I kind of get it from math's perspective now.

          • gimme on 08/02/2022 - 09:27
            +3

            @anonymous01: I used the same logic UNTIL I applied it to a slightly different scenario. 100 doors. You pick one and the host then opens 98. By your/my logic it's a 50/50 now but it isn't as your chance of having selected the car initially was 1/100 (i.e highly unlikely) and sticking with that same door when it's down to 2 doors would have been silly.

            • Wiired on 08/02/2022 - 16:26

              @gimme: Again, it wasn't presented as the actual game show which has other rules, none of that was explained. So in the OPs version staying on door 1 is a very valid option.

              • gimme on 08/02/2022 - 16:37

                @Wiired: never mind

                • Wiired on 08/02/2022 - 17:54

                  @gimme: It gets worse, the movie "21" screwed it up and the Lets Make a Deal game show pilot had two contestants picking doors. According to Wikipedia even the guy who called it the Monty Hall problem (Steve Selvin) didn't make it clear and was comparing it to the game show. If there weren't explicit rules, it leaves it open to interpretation.

      • diss on 07/02/2022 - 10:28
        +1

        Yeah it’s hard to get, but it is a fact rather than a debate because you can prove it’s true in practice. I wrote a script a while back to run the test thousands of times and it showed that you do indeed have a better chance if you switch.

        • Bren20 on 07/02/2022 - 12:26
          +12

          There are only three cases here, i'm not sure you need a script. There are three doors, behind which are Goat 1, Goat 2, and Car

          Case 1 - you chose Goat 1. Monty opens door to Goat 2. If you switch, you win a car
          Case 2 - you chose Goat 2. Monty opens door to Goat 1. If you switch, you win a car
          Case 3 - you chose Car. Monty opens door to Goat 1 or Goat 2. If you switch, you win a goat.

          2 out of 3 times you win a car if you switch.

          • TEER3X on 07/02/2022 - 19:44

            @Bren20: should be stickied.

          • uzz30 on 08/02/2022 - 09:50

            @Bren20: This. Answered.

          • d-hunterz on 08/02/2022 - 12:00

            @Bren20: You Cheeky!! Your response should be-

            Case 1 - you chose Goat 1. Monty opens door to Goat 2. If you switch, you win a car
            Case 2 - you chose Goat 2. Monty opens door to Goat 1. If you switch, you win a car
            Case 3 - you chose Car. Monty opens door to Goat 1. If you switch, you win a goat.
            Case 4 - you chose Car. Monty opens door to Goat 2. If you switch, you win a goat.

            2 out of 4 times you win a car if you switch. After one door is revealed probability changes to 50%. Whether it is the car or the goat, it pure luck.

            • caitsith01 on 08/02/2022 - 17:27
              +2

              @d-hunterz: This is misleading, because you are adding a bonus "you chose car" to make it look like 50/50 at the end. That's because you are ignoring the odds of each scenario happening in reality.

              Case 1 - 1 in 3 chance of happening
              Case 2 - 1 in 3 chance of happening
              Case 3 OR Case 4 - 1 in 3 chance of happening

              So in reality it's still 2 out of 3 times you win a car if you switch, and the probability DOES NOT change when Monty opens the door to show you either goat 1 or 2.

              The funniest thing about this argument is that you can run a simulation of this scenario on a computer to unequivocally demonstrate that switching is better, yet people still argue.

            • Bren20 on 09/02/2022 - 08:28

              @d-hunterz: I wanted to keep it simple. As caitsith01 states, I've rolled case 3 and 4 together as they're the same probability combined as case 1 and case 2

              Case 3 is defined by you choosing the car, it is not Monty's response.

      • abb on 07/02/2022 - 14:52
        +4

        your chances between door 1 and door 3 change from 33% each to 50%-50%.

        No, your initial door is still a 33% chance of winning. Draw out a decision tree, or run the experiment 30 times or something. It's very counter-intuitive but it is true.

        One key unspoken factor is "the host will always reveal a losing door, regardless of your choice".

        Another way to frame it is:

        You stand in a corridor with one million doors. You choose a door. Monty then opens 999,998 doors - leaving closed only the door of your initial choice and one other door. Monty always opens 999,998 losing doors and never opens the door you choose.

        Now, did you choose the right door first go (1 in a million), or is there a very high chance that it's the other door?

        • anonymous01 on 07/02/2022 - 19:36

          That actually makes much more sense, Thanks!. If that actually happens, I can see how this works, but I still think the chances shift 50% on each door until you open them. Just like the "Schrodinger's cat", you can't know for sure until you see it for yourself.

          • abb on 08/02/2022 - 11:52
            +1

            @anonymous01: Just because there are two possibilities, that does not make them both 50%.

            You could:
            A. get hit by lightning tomorrow
            B. not get hit by lightning tomorrow

            Is there therefore a coin-flip chance you will be hit by lightning each day?

          • caitsith01 on 08/02/2022 - 17:29

            @anonymous01: Maybe consider that it's never a straight "50/50" because you have additional information: Monty knows which door you chose, and he knows which door has the car, so when you see the empty door he opens that gives you additional information about the remaining door. So it is not a 'flip of a coin' situation because the coin has effectively already been flipped and then you have been given a clue about the answer.

      • caitsith01 on 08/02/2022 - 12:38

        I love how you say switching your opinion does not change reality and then immediately assert that opening a door changes the existing probability that you picked the correct door first time.

        Anyway… disagree all you like, you can empirically test this problem to demonstrate that it is always better to switch.

      • cosmos on 10/02/2022 - 13:34

        I understand what you are saying. I agree with you. I too think the new probability is 50:50, as there are now only two options to consider and one car.

  • sanglt on 03/02/2022 - 20:13
    +3

    The answer is YES, you should switch, because the probability that you will find the car by doing so is 2/3. This is because the probability that you picked the correct door in the first place does not change; it is still 1/3, regardless of the game show host’s actions.

    • Bren20 on 05/02/2022 - 12:12
      +1

      The answer is YES,

      Not necessarily, the riddle as presented here is incomplete, and in my opinion you can't be equivocal without more info.

      It depends what the rules of the game are.

      • How does the host decide which door to open?
      • Does the host have to open a door?
      • Does the host have to open a goat door?
      • Does the host know where the car is?
      • Is the host trying to sabotage you?

      Assuming the host is unaware of where the car is, and he has to open a door, then you should change doors

      Edit - the thread below says that the host knows where the car is

      • greatlamp on 06/02/2022 - 07:45
        +3

        The riddle is a bit confusing, it's based on a real game show.

        The host would always reveal a door with a goat behind it, and offer to switch.

        The reason everyone is so confident you should switch is because it's a well known case study in statistics now. It wasn't so obvious at the time, otherwise the game wouldn't have been designed that way.

      • caitsith01 on 08/02/2022 - 12:39

        The host has no strategy, he/she mechanically opens the door with a goat no matter what.

    • Wiede on 05/02/2022 - 16:12

      Understand this part, but instead of wording it as "staying" picking the same door you were on is also 2/3 right? so 2/3 door 1 and 2/3 door 3?

  • Qazxswec on 03/02/2022 - 20:14
    +19

    The OP just watched 21 on Netflix, Nine-Nine!

    • hasher22 on 03/02/2022 - 20:18

      What's 21? lol, sounds interesting. I should watch it. I went down the YouTube rabbit hole last night haha

      • dust on 03/02/2022 - 20:19
        +1

        A movie where this riddle features.

      • Qazxswec on 03/02/2022 - 20:20
        +1

        its a movie about counting cards, but on netflix, the little intro part they play to get you to watch it is a part of the movie where they dissect the monty hall problem

    • plmko on 03/02/2022 - 21:10

      I knew a family with 8 girls and 1 boy (being the youngest).

      The thing is maths will completely fail if you believe the world to be all else equal because instead it's rigged from the start.

      • StalkingIbis on 05/02/2022 - 11:20

        im one of 7 boys, we knew a family wih 7 girls….arranged marriages couldve been really convenient for our parents

    • Lastchancetosee on 04/02/2022 - 08:43
      +3

      Statistically, each time it is a 50/50 chance.

      • Niko123456 on 04/02/2022 - 08:54
        -7

        There are four combinations that can result with two kids:

        BB
        BG
        GB
        GG

        If our first kid is a boy, we can have:
        BB
        BG
        GB

        You have one set that has an additional boy, but two sets that have an additional girl.

        At least, this is my dodgy interpretation of conditional probability.

        • zrmx on 04/02/2022 - 10:35
          +6

          You can't have GB, because it is a given that your first one isn't a G, so GB is not an option.

        • gimli on 04/02/2022 - 12:46

          Lol. Love probability jokes

        • moar bargains on 04/02/2022 - 14:40
          +2

          my dodgy interpretation of conditional probability.

          Confirmed.

      • Ghost47 on 04/02/2022 - 13:35
        -1

        Depends on sperm quality. If a guy has feminine sperm the chances are higher of having a girl. If his sperm is manly the chances are higher of having a boy.

        • terrys on 04/02/2022 - 19:30

          One of the benefits of nano- technology is that we may soon see adverts for exercise equipment for your sperm - the cohort who bought from Queensland through the advertisements in the Weekly World News or Pix/People would flock to them like a sovereign citizen to a bottle of bleach.

    • cookie2 on 05/02/2022 - 13:03

      Why is it statistically higher to then have a girl? Chance would still be 50/50 (or whatever the stat is for male vs female births)

      • Niko123456 on 06/02/2022 - 09:57
        -1

        Read the explanation for the Monty Hall problem. The available outcomes for 2 children are set in stone.

        I love that people downvoted me but that's probability baby.

    • darkmattersunB6c0MV on 05/02/2022 - 15:33
      +1

      Proof?
      The two are independent events.

    • node modules on 07/02/2022 - 13:28

      That's not how probability works.

  • aragornelessar on 03/02/2022 - 20:32
    +2

    Isn't it a 50-50 chance now?

    Two out of three doors had a goat originally. Now one of the goats has been revealed. Behind the remaining two doors is one goat and one car. So 50-50 of getting the car no matter which door you pick?

    • AustriaBargain on 03/02/2022 - 20:45
      +1

      It is 50-50, but the door you chose is 33-33-33. Monty can't open your door even if you chose a goat, so the odds that Monty opened the other goat goes to 66-33, which is better than 50-50 for you to win a car if you always switch. Or something like that, it's been a while since I read the Wikipedia article.

      • aragornelessar on 03/02/2022 - 20:53
        -5

        Who's Monty? The jackass giving me a headache? lol

        The host now asks you, "Do you want to switch to door 1 or stay at your current door, 3?"

        You're basically picking one out of two doors because the third one's been eliminated right? And you know the goat was behind the eliminated door (door 2, whilst you chose door 3). I'd take my luck and flip a coin but would be interested in why otherwise!

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

          Mr Monty Hall is the host, this was a real US TV show in the 20th century iirc. It doesn't matter which goat you choose but if you chose a goat then you should switch doors because the other door WILL have a car. There is 66.6% chance you chose a goat, 33.3% chance you chose a car. So 66.6% of the time you should switch doors.

          • darkmattersunB6c0MV on 05/02/2022 - 15:42
            +2

            @AustriaBargain: Sorry, there is no "because the other door WILL have a car".
            You may still have the car in the original selection, but there is a 1/3 chance of that.

            The question is also framed incorrectly by the op.
            In the Monty Hall problem, you should switch, as there is a 2/3 chance of winning.
            Logic is pretty simple. The tv host always knows what is behind each door, and always opens the door with the goat.
            When the original door is selected, there is a 1/3 chance of the car.

            There is a 2/3 chance that the car is behind the other two doors.
            Note that the host knows the location of the car, and will never open the door with the car, so if they open the door without the car, then the door remaining has 2/3 chance.
            Always switch.
            Although you will still lose if the original door has the car, that will be 1/3 of the cases.

            • AustriaBargain on 05/02/2022 - 15:52
              +2

              @darkmattersunB6c0MV: If you chose a goat, which will happen 66.6% of the time, then after a door is opened the other door WILL always have the car. 33.3% of the time it WILL not have the car. You double your odds of getting the car by switching.

            • SummerZero on 06/02/2022 - 00:23
              +1

              @darkmattersunB6c0MV: the bit about the host knowing where the car is and always choosing one of the goat doors first is why it makes sense to switch every time. This bit of information is often left out when the problem is discussed on the internet.

    • tofuofdoom on 04/02/2022 - 11:23
      +2

      There were two things that helped me understand the problem.

      The first thing is, the presenter isn’t giving you any more information than you already know.

      What you know is, there is one car behind one of these three doors. You know that there are goats behind each of the other two.

      The presenter knows which doors have what. They will always pick a door with a goat behind it.

      Regardless of whether you picked a goat or a car, the presenter will simply show a goat.

      So we can safely ignore what the presenter does, as irrelevant information.

      Now the question is simply. Do you think the car is behind your door (1/3), or behind one of the other 2 doors (2/3). You already know one of the other two doors is a goat, so that information isn’t helpful, and therefore you always take the 2/3 chance over the 1/3

      If it still feels a little strange, what if we increase the number of doors.

      There are 100 doors. 1 door has a car, 99 have goats.

      If you pick 1 door at random, you have a 1% chance of hitting the car, meaning 99% of the time, the car is behind one of the other doors.

      If the presenter tells you that at least 98 of the remaining doors have goats, then does that change your decision? Of course not, you already knew that.

      If the presenter instead shows you that 98 of the remaining doors have goats, does that change your decision? No, it’s exactly the same.

      Therefore, you should always choose to swap when offered, because it’s 99% vs 1%

      • dracothraxus on 05/02/2022 - 07:59
        +4

        "The presenter knows which doors have what. They will always pick a door with a goat behind it."

        Good stuff - too often this is left out of the description and peoples' confusion / outrage is actually justified.

        It is not "and then the host picks a door which happens to have a goat behind it".

      • vi9ilante on 05/02/2022 - 10:41

        Respectfully - you have the right answer but incorrectly explained the fundamentals of this problem. The numbers contradict the qualitative statements:

        1) You can't ignore what the host does, it is relevant information.
        2) And so knowing one of the other two doors is a goat IS helpful and that's the basis of the change.

        Again for 100 doors, you did NOT already know anything - the host did and showed you.

        So thank the host and remember to send him a Christmas card every year.

        • tofuofdoom on 05/02/2022 - 15:22
          +1

          I guess I should be more specific.

          There are 3 essential actions the presenter does

          1. Explain the rules of the game.

          2. Reveal that one of the doors has a goat.

          3. Offer the swap.

          What I'm saying is, the second step, where the presenter reveals that one of the doors is a goat is redundant information. You already knew that at least one of the two doors you didn't pick has a goat, because there are 2 goat doors, and you can only pick 1 door.

          When you break it down, the game simplifies down to: Do you think the car is behind door 1, or is it behind one of the other 2 doors.

          The presenter has provided exactly 0 new information, by revealing a goat is behind an unpicked door.

          • CanadaAye on 06/02/2022 - 00:13
            +2

            @tofuofdoom: I know what you're saying but it is still new information because prior to the revelation you did not know what was behind the door. Until the point it is revealed it had a 1/3 chance of having the car.

            Your initial selection still has a 1/3 chance of having the car (i.e. 2/3 chance of being a goat). Therefore when presented with the next choice, the odds on it being the car will be better than your initial odds. If you choose to stay with the initial choice you will still have a 1/3 chance of getting the car, but switching to the other option increases your odds to 2/3, because the goat has been revealed and you can only select the other option. Again, this will be right 66% of the time versus 33% for your initial choice.

            It's key to remember that the host can only reveal a goat and therefore 2/3 of the time they are effectively showing you where the car is.

            • filmer on 07/02/2022 - 12:57
              +1

              @CanadaAye: Yeah, I'm not sure how they can say that the fact the host always shows a goat is not relevant information as this is the whole reason you end up wanting to swap.

              If instead the host shows a random door, and in this case happens to be goat whether you swap or not you have exactly the same odds of winning. As proof I simulated this (in a comment here). This fact changes everything.

              • abb on 07/02/2022 - 14:57

                @filmer: It's irrelevant in the sense that you already know "one of the two doors I did not choose contains a goat" to be true.

                The host could, instead of opening a door and offering to swap doors, equivalently offer "you can open either your door, or both the other two doors and keep the prize if it is behind either one".

                That's what tofu is saying.

                • filmer on 08/02/2022 - 11:55

                  @abb: That is not true at all. If the host is showing a random door, which happens to be a goat then you have a 50% chance of winning whether you swap or not.

                  Edit: It's not true in the sense that the fact that the host knows and shows the goat is relevant information.

                  • abb on 08/02/2022 - 12:10

                    @filmer:

                    If the host is showing a random door

                    Ahh, this is a different assumption to the usual (that the host always opens a non-winning non-chosen door).

                    But I don't think it changes the outcome.

                    P(winning first choice) = 33%
                    therefore P(didn't win on first choice) = 1 - 33% = 66%

                    P(any winning move given that host randomly reveals prize) = 0
                    so we can ignore that branch of possibilities, stay or swap are both losing moves

                    P(win by staying given host randomly reveals goat) = 33%
                    nothing has changed about your door - there is always at least one other losing door regardless of your choice

                    P(win by moving given host randomly reveals goat) = P(didn't win on first choice) = 66%
                    you're now effectively choosing the best of the 2 doors you didn't originally choose

                    you have a 50% chance of winning whether you swap or not

                    I don't think this is the case but I'm open to being convinced by a statistical simulation, worked example, or otherwise.

                    If the host opens a losing door before you make your first choice, that's a 50:50

                    It may help you to consider a related puzzle:

                    You stand in a corridor with one million doors. You choose a door. Monty then opens 999,998 doors - leaving closed only the door of your initial choice and one other door. Monty always opens 999,998 losing doors and never opens the door you choose.

                    Now, did you choose the right door first go (1 in a million), or is there a very high chance that it's the other door?

                    • filmer on 08/02/2022 - 12:22

                      @abb: I have the output in this thread of the simulation of both host always choosing goat, and random.

                      Given there is a 2/3 chance of the car being in one of the non player doors, and a 1/2 chance of the host choosing it if the car is there it means there is a 1/3 chance of instantly losing if the host choosing a random door (2/3 * 1/2)

                      If the door chosen is a goat you have the following scenarios based on player choosing door A.

                      Player A, Car B.
                      - Host shows door C, you swap and win.
                      - Host shows door C, you stay and lose.

                      Player A, Car C.
                      - Host shows door B, you swap and win.
                      - Host shows door B, you stay and lose.

                      Player A, Car A.
                      - Host shows door B, you swap and lose.
                      - Host shows door B, you stay and win.
                      - Host shows door C, you swap and lose.
                      - Host shows door C, you stay and win.

                      As you can see, if it gets to this stage half of the outcomes the player swaps they lose, and half they win.

                      The fact the host choose a goat on purpose is a very important fact. The same with your 1,000,000 door example. This only works if the host always opens the wrong door.

                      • abb on 08/02/2022 - 13:02

                        @filmer: Cool, I'll have to look up your sim. What you say sounds plausible but the devil is in the details :)

      • t_c on 05/02/2022 - 11:21

        OP doesn't say presenter knows whats behind the doors, so your assumtion is flawed, maybe the producer knows, to keep the game show fair the presenter may be in the dark as is the contestant.

        • tofuofdoom on 05/02/2022 - 15:17

          I mean sure, but OP also didn't tell us what a car is, or what a goat is. The idea that the presenter knows where the car is, is essential to the show. Otherwise there is a 1/3 chance that the presenter shows us the car behind a door we didn't pick, and the game just… ends there.

      • legendary-noob on 06/02/2022 - 23:48

        The presenters action does provide you more info. Put simply the odds was against you at the start, 2/3 chance of getting a goat. He just eliminated one goat after the reveal. By allowing you to switch simply allows you to turn your odds to picking what would have been 1/3 chance of getting a car.

        • filmer on 07/02/2022 - 12:57

          Yes, but this action wouldn't help you unless it was true that he would always show a goat.

      • caitsith01 on 08/02/2022 - 17:32

        The first thing is, the presenter isn’t giving you any more information than you already know.

        Of course they are.

        Initially, you know there's a 2/3rds chance that one of the two doors you didn't pick has the prize.

        By opening one of them, the presenter is clearly giving you information that the prize is not behind that door and so the 2/3rds chance remains, but applies only to the remaining door. That is clearly more information than you had when you started. Indeed, the only reason the problem works the way it does is that the presenter gives you information you didn't have at the start.

    • darkmattersunB6c0MV on 05/02/2022 - 21:41

      It would be 50-50 if there were two options.
      There were three options.
      Hence, you had 1/3 chance of picking the car.

      You not picking the car was 2/3.
      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).
      This switching the door gives you 2/3 chance of winning.

    • trongy on 06/02/2022 - 10:49
      -1

      Just because there are only two choices doesn't mean they are of equal probability.

      Would you expect the chance of heads and tails to be even when flipping a coin where the heads side is made of lead and the tails side is made of aluminium?

      The rules of this game mean you can't assume the choices are of equal probability.

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