Will These Brain Teasers Stump You?
- Thread starter Rhykker
- Start date
No no, there is a single correct answer to these problems. You can get to that answer a number of different ways, but the switching of point of view only comes into play due to an incorrect initial point of view based on not properly understanding how to apply the probabilistic arguments.DonTsetsi said:
Indeed, this is all straight up probabilities, and it not quite matching human intuition. The Monty Hall problem in particular is well known as it goes against your basic instincts which tell you that switching makes no difference.Rhykker said:
There is indeed exactly one correct answer, no room for argument if you understand probability and statistics
I don't know, it still looks like the gambler fallacy. All 3 cases have the same idea:Alfador_VII said:
You take a certain probability. Then you introduce new variables while sticking with the old probability. It's like saying you have a higher chance of getting Heads after Tails.
The important difference there is for a coin, the previous tosses have no effect on later ones. Here in these problems (at least 2 and 3) you're getting more information about a situation partway through, so the probabilities have to change.DonTsetsi said:
A extremely simple example of this is if there is a white ball and a black ball, and you select one at random, you have a 50% of it being white. If the other ball is revealed and is black, then clearly it's 100% Clearly things are way more complex here, but the same principle applies, the problem changes, or you get more information.
The first Box puzzle is different though, we just tend to analyse it wrongly. I'm not totally sure how that works, but I think I know where they're coming from, after reading it a few times.
You are incorrect. In the Three Prisoners problem and the Monty Hall problem it is not your point of view that changes, it is your question. When Alfred says "If I'm to be pardoned..." what he is really asking is "If Alfred is to be pardoned...." If Charlie went to the warden and asked: "If Bart is to be pardoned, tell me I will be executed. If I am to be pardoned, tell me Bart is to be executed. And if Alfred is to be pardoned, then flip a coin to decide whether to tell me that either Bart or I am being executed." it would be functionally the same question Alfred asked, and the results would be the same.DonTsetsi said:
On the other hand, the Monty Hall problem is bullshit. The reason people get the "wrong" answer is because the question is incomplete. It states that the game host knows where the car is, but it does not state how he chooses to use that information.
Consider the two most extreme cases:
The car is behind Door #1. The host likes you, and wants you to win. If you pick Door #1, he opens Door #1 and you win. If you pick either of the other two doors, you would lose, so he lets you repick and even sweetens the deal by removing the other wrong option to try to convince you to pick the remaining right option.
If the host is trying to make you win, switching has a 100% success rate.
The car is behind Door #1. The host hates you, and wants you to lose. If you pick Door #2 or Door #3, he opens that door and you lose. If you pick Door #1, you would win, so he lets you repick and even sweetens the deal by removing one of the two wrong options to try to trick you into picking the other wrong option.
If the host is trying to make you lose, switching has a 0% success rate.
You are misunderstanding something here. In the Monty Hall problem it is a clearly stated rule that once you pick, the host has to open one of remaining the goat doors and then has to offer you to repick your choice. These were the actual rules of the game from which the paradox comes from. He cannot make an arbitrary choice about which door to open whether he likes you or not; in fact you could replace the host with a completely impartial algorithm and you would get the exact same results (except with less gameshow flair).Grumman said:
From Wikipedia:Grumman said:
The entire point of the Monty Hall Problem IS that a losing door is revealed and you are allowed the option to switch. Anything fewer isn't "the Monty Hall Problem."
That first one is killing me... I cannot comprehend it at all... the answer seems to not make sense...
If you actually set it up and recorded the results of the colour of the other coin when this situation happened, then surely 50% of the time the other coin would be gold and the other 50% of the time it would be silver?
If you actually set it up and recorded the results of the colour of the other coin when this situation happened, then surely 50% of the time the other coin would be gold and the other 50% of the time it would be silver?
Dude, I don't owe you any quote because,Grumman said:
a.) I don't have the time to do something you can do just as easily with Google
b.) I actually learned about the Monty Hall problem in a game theory course at the university, and I trust my prof's interpretation better than yours
c.) As I said, this problem is based on an actual game show's rules, and there is no game show in existence where the host's opinion of you can change the rules/outcome, otherwise it would be considered cheating and thus illegal
In other words your interpretation is simply based on a false assumption. End of story.
There are six coins in three pairs, all with an equal chance that you picked that coin: (A, B), (C, D), (E and F).Elementary - Dear Watson said:
You just found out that the coin you picked was A, B or C (one of the three gold coins). That means that the other coin in that pair must be B, A or D, with a 1/3 chance of each. Since there's a 1/3 chance that the other coin is B and a 1/3 chance the other coin is A, there's a 2/3 chance that the other coin is gold.
Much of the trickery in Brain Teasers is obfuscation of the actual question and the information needed to answer that question.Elementary - Dear Watson said:
Teaser 1's actual question: There are 3 Gold coins in 2 boxes: 1 with 2 Gold coins (Ga and Gb) and 1 with only 1 Gold coin (Gc). Someone randomly removes and gives you one of the three Gold coins. What is the chance the Gold coin you were given was Gold coin Ga or Gb?
Most people end up focused on the boxes and not on the Gold coins.
Arg... I kinda get it, but still think it won't hold up to the stats... The probability seem to ignore the fact the coins are in the boxes. That is still a limiting fact. Wouldn't that mean that if you just had the 2 boxes, and you piked a coin at random and it was gold, then the other colour would be gold 2 thirds of the time... even though the choice is really of the 2 boxes, not the 4 coins.Grumman said:
But the boxes are still a factor though, aren't they? If all the coins were in a single box I would understand, but doesn't the rule involving the box (and it being the other coin in the box) affect it in any way?Ultimatecalibur said:
Nope, the boxes are part of the obfuscation along with the Silver coins. The boxes are used to determine pairs, but are not what determines success/failure. Which of the 3 Gold coins you got is what determines success/failure.Elementary - Dear Watson said:
There are 3 possibilities for the Gold coin you pull out of the box:
1. Ga = Success
2. Gb = Success
3. Gc = Failure
So, the problem I've seen quite often when people present brainteasers, especially probability ones, is that they give a flawed presentation and then assume that when people get correct or incorrect answers it's because of their understanding of the problem and not their familiarity with it.
Now, the Monty Hall problem is correctly presented here. Specifically, the often forgotten mention that the host has prior knowledge was included. However, were the other two presented correctly as well?
For the box paradox, I can see the interpretation that would cause the answer given, but I don't see any part of the problem as stated here that would cause that interpretation to be superior to the intuitive one. The issue is in the timing. At the beginning of the problem there were three relevant boxes and six relevant coins, but the question was asked after a coin was removed. According to what I've read of probability, the moment you get more knowledge of a situation the probabilities are subject to change, and in this case the box with the silver coins as well as the silver coins contained therein should stop being relevant.
To summarize, the problem of whether a randomly selected box has a gold coin before checking a coin should not be treated the same as the problem of whether a randomly selected box has a gold coin after checking a coin.
The three prisoners problem presents a similar issue but in reverse. The knowledge that the guard may or may not have flipped a coin has questionable relevance to the central issue. Now, I understand that the problem is presented right before the Monty Hall problem to add the idea that a possibility's chance to happen doesn't necessarily change just because other possibilities did, but the story presented doesn't guarantee that that would be the case. The guard has no reason to actually be honorable.
Now, the Monty Hall problem is correctly presented here. Specifically, the often forgotten mention that the host has prior knowledge was included. However, were the other two presented correctly as well?
For the box paradox, I can see the interpretation that would cause the answer given, but I don't see any part of the problem as stated here that would cause that interpretation to be superior to the intuitive one. The issue is in the timing. At the beginning of the problem there were three relevant boxes and six relevant coins, but the question was asked after a coin was removed. According to what I've read of probability, the moment you get more knowledge of a situation the probabilities are subject to change, and in this case the box with the silver coins as well as the silver coins contained therein should stop being relevant.
To summarize, the problem of whether a randomly selected box has a gold coin before checking a coin should not be treated the same as the problem of whether a randomly selected box has a gold coin after checking a coin.
The three prisoners problem presents a similar issue but in reverse. The knowledge that the guard may or may not have flipped a coin has questionable relevance to the central issue. Now, I understand that the problem is presented right before the Monty Hall problem to add the idea that a possibility's chance to happen doesn't necessarily change just because other possibilities did, but the story presented doesn't guarantee that that would be the case. The guard has no reason to actually be honorable.
No, because you're throwing out all the instances in which you pull out the silver coin.Elementary - Dear Watson said:
Suppose one box has two gold coins (Ga and Gb), and the other box has a gold and a silver coin (Gc and S).
You do this 100 times.
25 times you pull out Ga (so the other coin in the box is gold)
25 times you pull out Gb (so the other coin in the box is gold)
25 times you pull out Gc (so the other coin in the box is silver)
25 times you pull out S (so the other coin in the box is gold)
However, we throw out all the times you pulled out S. We're left with 75 "valid" pulls, and in 50 of them the remaining coin was gold, giving us the proper 2/3 probability.
Ugh. I always hate these questions, because they're always ways for someone that's looked at probability to try and feel smart, when really they misunderstand how things work or are asking the wrong question.
Take the first. The answer is either 75% or 66%, depending on how you interpret the question based on how you interpret the question's English. The question "What are the odds that the other coin in the box is also gold?" revolves around which part of the question I'm now in to determine what the odds are. If we consider that there are 6 boxes that are all relevant, then the answer is 66%, yes. But, I can also take the odds as starting after the coin has been picked. Now the total number of boxes is completely irrelevant and all that matters is that it's one of the two boxes that contain gold coins. There's two valid answers depending on your interpretation of where the question starts. Personally I'd go with 75%, because I find the knowledge that there are additional silver boxes to be completely extraneous information.
Second question: Please in the future, make some mention that the guard has a chance to flip the coin in secret. As it stands, it looks like the guard makes a clear statement that Bart is executed. It's not exactly unknown for questions like these to have a lot of extraneous information to cloud what really happened, and I felt cheated.
The third is just plain bad science. After the door has been opened, I have two choices. To switch, or not switch. Regardless of what I choose, there's a 50% chance that I've now gotten it right, and again you're assuming that irrelevant information is relevant to the question. That there once was a third door that could have been correct is bogus information that doesn't affect my new choice between two doors.
Edit: While I'll certainly stand by my first 2 questions, the third I think ends up being confusing by framing it in the form of a game show. Logically, I should switch every time in such a scenario because the host has decreased the chances of failure. To take it to extremes, if there's 1 million doors, and then all but 2 are closed, I've got a much better chance of switching.
But, by framing it as it does, what you're asking the player to do is in effect to trust or not trust the host.
Take the first. The answer is either 75% or 66%, depending on how you interpret the question based on how you interpret the question's English. The question "What are the odds that the other coin in the box is also gold?" revolves around which part of the question I'm now in to determine what the odds are. If we consider that there are 6 boxes that are all relevant, then the answer is 66%, yes. But, I can also take the odds as starting after the coin has been picked. Now the total number of boxes is completely irrelevant and all that matters is that it's one of the two boxes that contain gold coins. There's two valid answers depending on your interpretation of where the question starts. Personally I'd go with 75%, because I find the knowledge that there are additional silver boxes to be completely extraneous information.
Second question: Please in the future, make some mention that the guard has a chance to flip the coin in secret. As it stands, it looks like the guard makes a clear statement that Bart is executed. It's not exactly unknown for questions like these to have a lot of extraneous information to cloud what really happened, and I felt cheated.
The third is just plain bad science. After the door has been opened, I have two choices. To switch, or not switch. Regardless of what I choose, there's a 50% chance that I've now gotten it right, and again you're assuming that irrelevant information is relevant to the question. That there once was a third door that could have been correct is bogus information that doesn't affect my new choice between two doors.
Edit: While I'll certainly stand by my first 2 questions, the third I think ends up being confusing by framing it in the form of a game show. Logically, I should switch every time in such a scenario because the host has decreased the chances of failure. To take it to extremes, if there's 1 million doors, and then all but 2 are closed, I've got a much better chance of switching.
But, by framing it as it does, what you're asking the player to do is in effect to trust or not trust the host.