Will These Brain Teasers Stump You? Pages 1 2 NEXT | |

Can you solve these probability puzzles, or will they stump you? | |

Aren't these paradoxes just illustrating that our methodology has some flaws? Switching a point of view leads to changes in probability, which should not happen. | |

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. | |

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. 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: | |

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. 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 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, Consider the two most extreme cases: 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 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 | |

Prove it. Quote the part of the Monty Hall problem that states that he | |

From Wikipedia:
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? | |

This is why I hate probability and statistics. The first and third problems are just... | |

Dude, I don't owe you any quote because, 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). 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. 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.
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? | |

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. There are 3 possibilities for the Gold coin you pull out of the box: 1. Ga = Success | |

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. | |

No, because you're throwing out all the instances in which you pull out the silver coin. 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. 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 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. | |

Ah! You guys are stars! You wouldn't beleive how much this hurt my battered, hungover minnd this morning! I can see it now! I had completely not even thought of the 4th discarded option involving the silver coin... it makes a lot more sense now! I think I actually have it! :P | |

The answer is 66%.
First, they describe the process used to determine the warden's answer. If they describe a process in a question like this, you're supposed to solve the problem as if they used the process. And honestly, I find this question easier to understand if you assume that the warden
This is the exact fallacy the Monty Hall Problem is supposed to expose. If you don't believe the answers for any of these, perhaps you should replicate them and run them a few times yourself, and see what results you get; none of these are too hard to reproduce. | |

The problem resides in the English of the question. There are two "beginnings" from which I can calculate the odds. Either from the start of the question, or after I've taken out the gold coin. Which beginning I should start at to then calculate isn't clear.
In tricky questions assuming information not given leads to wrong answers. There's no information given that he ever flipped a coin, and therefore I shouldn't believe that a coin was ever flipped. The only information given about coins was that flipping a coin was involved in one of the answers. | |

The question I'm telling you, reproduce the problem yourself if you don't believe the answer xP
There's no information that he There's no information on whether he flipped a coin or not, and so you should believe that the warden flipped a coin if A is to be pardoned and didn't flip one if C is to be pardoned--just like the question says. | |

I regret coming to this comment section. Try this simplified example. There are 3 events with equal chance of occuring: 1, 2 and 3 Is result A more likely than result B or are they the same? Most of the issues seem to arise with assuming one result always comes from the same event, rather than two similar (but entirely separate) events. | |

And this is how wars are started. Because the problem with these is that a lot of mathematicians don't agree with the way the math is setup for it. But that's none of my business. | |

Could you elaborate? To my knowledge nothing here is a mathematical trick. The monty hall problem is famous and all 3 of these are key to understanding probability theory and why 'bits' of information matter. I don't think it's fair to consider these problems 'tricks'. The second one was a little obfuscating (possibly because of the sheer quantity of words) but they're perfectly logical, the useful information can be separated and you can work from there...in fairness I have a hard time imagining how I thought before learning Bayesian probability...still...it seems obvious. Not all the information is useful, so remove the extraneous data and do the maths... | |

It is stating it: the host always open a door with a goat and then asks you if you want to switch. He never reveals the car at that moment. The main question is: do you switch or not? | |

Actually, there aren't two beginnings. The paradox is in assuming that there the two activities are independent. If the question had been "what is the probability that a box with a gold coin contains another gold coin?", the answer would be 1/2. The participant Let's look at the possible outcomes, labeling each gold and silver coin by number. 1) G1 - G2, pick G1 These are the six possible states. Looking at this, they have a 50% chance of seeing a Gold coin. But, we already KNOW they saw a gold coin. So, only states 1, 2 and 5 are valid. And each of these states are just as likely as the other. So, we are left with: 1) G1 - G2, pick G1 If the question is "what is the probability that the second coin is gold?", this is obviously 2/3 or 66%. Two of the three options have a gold coin. Now posit that the same question was asked prior to seeing the coin in your hand. "Having an unknown coin in your hand, what is the percent chance that the second coin is gold?" In this case, it is 1/2, because 3 of the six possible states have a gold coin. | |

the funny thing is the last one that is supposed to be "difficult" was the easiest for me, the second one got me a bit with not stating if he flipped the coin or not, so I got that wrong, and the first one had me saying 50% initially, but then I remembered my probability class and how I got a similar problem wrong in class so I redid it and got 66% | |

That was really fun! I found the first one easy but the last two took a while. I DID get them all, but in the case of the last one I would have got it wrong if I hadn't just spent 5 minutes working my way through the second one, which actually requires the exact same logical realisation. Thanks for the entertainment! | |

I like your explanation a lot, but it doesn't really satisfy me logically. Or maybe it does, but I want to follow a particular line of inquiry further. Would removing one of the coins from the experiment, selecting it in this case, not remove it from the pool of possible outcomes? By which I mean that there are no longer two potential gold coins to grab, but one, since one of them is now in your hand. Whether you got coin one or coin two from the box is irrelevant, since you know that you either have the S/G box or the G/G box. There can't be a 2/3 chance, because there are only two choices left, right? Wait, hold on, I think I just figured this out another way. If you do count the gold coin as removed from the experiment, by doing so you're also removing two of the silver coins. There is no way the final coin could be one of the ones from the S/S box, so they are effectively irrelevant information. By removing one gold and two silver coins from the original experiment, you're left with two gold and one silver coin in the "new" experiment. Huh, neat. | |

Not true at all, in fact, it's actually the opposite in some ways. Lets look at the 3 door problem. You have a 1/3 chance of getting the car to start. That is, in the end, your exact chance of getting the car no matter what else happens along the way. Why? Well lets look at what is really happening here real quick. You initially choose a door, there is a 33% chance you chose the care. The game announcer knows if you have the car or not, and knows which other door (or doors) do not have the car. Either way, there is a door without a car for him to open. Now, at that exact point in time, you have a 50% chance to have the right door. In no way does the actions of the annoucer opening a door make any difference in your probability of initially choosing the right door. If you do choose again, you now have a 50% chance of being right....but...you have that SAME 50% chance of being right if you don't choose again (it's still the exact same chance). Choosing again, does not change your odds. Saying it does it just plain silly. The annoucer always is going to offer you another pick, and one of the other doors is always doing to have a goat. Your probability of choosing the correct door is not going to change, but....if you choose again at this point, your NEW probability of choosing the correct door will be 50% (instead of the original 33%). That is the same if you choose the door you currently have, or choose the other door. In effect you are starting off with a 33% chance (your initial choice)...which for all intents and purposes is pointless (as long as the announcer always has to give you the option to choose again after getting rid of one door), and then ending up with a 50% chance to win. Your net over all chance to choose the car in this case is 50% (as one goat is always removed). Keep the same door, choose the other door, it doesn't matter, you have a 50% chance to win. | |

As far as the coins one goes. You instantly have knocked out the 2 silver coins boxes (As you grabbed a gold coin). That leaves 2 gold coins left (either 2 in the gold box, or 1 in the gold box and 2 in the silver), and 1 silver coin. That plain and simply is all the coins that are left that it is possible for you to grab. Since 2/3 is 66%...you have a 66% chance to grab a gold coin and a 33% chance to grab a silver coin. Where the coins are, in which box etc, all of it really doesn't matter, it's just a matter of how many gold coins and silver coins you could possible grab....the box with 2 silver coins is out, so you have 2 gold coins and 1 silver coin left....66% chance to get another gold coin. Ps. If you are going to argue the announcer does not need to offer another door (and show a goat), then that makes it even MORE likely you have the right door since if he doesn't always make the offer, that means it's more likely you have the right door and he's trying to get you to second guess yourself and choose a wrong door. Therefore, the announcer would always give the option...meaning you have a 50% chance (or greater) if you stay with your current door (depending on if he always makes the offer or not). Either way, sticking with your first choice at that point is the best (or at least as good) of a bet. 50%...and the fact that the annoucer may know more then you and be making the offer to try and get you to give away the car. | |

As far as the prisoners one, actually I don't agree with it, as the initial random chance to determine who was pardoned was 33% for each of them. That means that each prisoner had a 1/3 chance to be pardoned. When the warden says that bart is being executed (so is not the one pardoned), that means the initial chance has changed from 1/3 to 1/2. At that point in time, when the warden said bart was going to be executed, if the random selection was done then, it would be a 66% chance for charlie to be pardoned. The truth is though, that ship had already sailed and the initial 33% chance for each still meant that both charlie and alried had the same chance of being pardoned (but now there is only 2 of them, so it's a 50% chance). Just because there are more situations where the warden would have said what he did, does not mean it actually changes the initial probability of each of the prisoners being pardoned. What does change the probability is removing bart from the chances to be pardoned, which changes the chance from 33% each, to 50% each for charlie and Alfried.......so I do not agree with that puzzle. | |

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