Will These Brain Teasers Stump You?

[Elementary - Dear Watson]

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
2. Gb = Success
3. Gc = Failure

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.
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.

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!

 

[Bad Player]

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.

The answer is 66%.
If you only have the two boxes, your chance of taking a gold coin (or your chance of taking a coin so that the remaining coin is gold) is 75%, but if you've taken a gold coin your chance of having taken a gold coin with the remaining coin being gold is 66%. (Edited.)

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.

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 always flips a coin (but it only matters when A is being pardoned).
Second, you say that they withheld necessary information, and then complain that these questions usually have too much extraneous information, so I'm not sure what you're actually unhappy about.

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.

This is the exact fallacy the Monty Hall Problem is supposed to expose.
At first, you have a 1/3 chance of choosing a car and a 2/3 chance of choosing a goat. The key is that the host must reveal a door you did not choose and that has a goat behind it. So there's three doors--the chosen door, the revealed door, and the 3rd door. The revealed door is a goat. Therefore, if the chosen door is the car, the 3rd door is a goat, and if the chosen door is a goat, the 3rd door is the car.
In short, it's impossible to choose a goat, and then switch to a goat (or to choose the car and switch to the car). If you initially choose the car and then switch, you get a goat. If you initially choose a goat and then switch, you get the car. Since you initially have a 1/3 chance of getting the car and 2/3 chance of getting a goat, by switching your door you switch these probabilities.


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.

 

[Xeorm]

The answer is 66%.
If you only have the two boxes, your chance of taking a gold coin (or your chance of taking a coin so that the remaining coin is gold) is 75%, but if you've taken a gold coin your chance of having taken a gold coin with the remaining coin being gold is 66%. (Edited.)

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.

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 always flips a coin (but it only matters when A is being pardoned).
Second, you say that they withheld necessary information, and then complain that these questions usually have too much extraneous information, so I'm not sure what you're actually unhappy about.

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.

 

[Bad Player]

The answer is 66%.
If you only have the two boxes, your chance of taking a gold coin (or your chance of taking a coin so that the remaining coin is gold) is 75%, but if you've taken a gold coin your chance of having taken a gold coin with the remaining coin being gold is 66%. (Edited.)

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.

The question is clear.
It isn't asking the probability of drawing a gold coin so that the remaining coin is gold (which is actually 33%), nor is it asking what the probability of drawing a gold coin is, nor is it asking what the probability of drawing a coin so that the remaining coin is gold is.
It's clearly asking what, having already drawn a gold coin, is the probability of the remaining coin being gold.

I'm telling you, reproduce the problem yourself if you don't believe the answer xP

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 always flips a coin (but it only matters when A is being pardoned).
Second, you say that they withheld necessary information, and then complain that these questions usually have too much extraneous information, so I'm not sure what you're actually unhappy about.

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.

There's no information that he didn't flip a coin, though. An assumption he didn't flip a coin is just as much of an assumption as one that he did flip one.

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.

 

[baconsarnie]

I regret coming to this comment section.

Try this simplified example.

There are 3 events with equal chance of occuring: 1, 2 and 3
The outcomes of these events are as follows,
Event 1 causes result A
Event 2 causes result B
Event 3 causes result A

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.

 

[FoolKiller]

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.

 

[Hugga_Bear]

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...

 

[CaitSeith]

It states that the game host knows where the car is, but it does not state how he chooses to use that information.

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?

 

[SurfKansas]

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.

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 perceives themselves to be a part of this independent activity. In reality, they are not. The trick is that you have to identify the probability of them being in a situation that would result in having a gold coin in their hand.

Let's look at the possible outcomes, labeling each gold and silver coin by number.

1) G1 - G2, pick G1
2) G1 - G2, pick G2
3) S1 - S2, pick S1
4) S1 - S2, pick S2
5) G3 - S3, pick G3
6) G3 - S3, pick S3

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
2) G1 - G2, pick G2
5) G3 - S3, pick G3

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.

 

[gmaverick019_v1legacy]

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%

 

[Branindain]

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!

 

[The_State]

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.

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 perceives themselves to be a part of this independent activity. In reality, they are not. The trick is that you have to identify the probability of them being in a situation that would result in having a gold coin in their hand.

Let's look at the possible outcomes, labeling each gold and silver coin by number.

1) G1 - G2, pick G1
2) G1 - G2, pick G2
3) S1 - S2, pick S1
4) S1 - S2, pick S2
5) G3 - S3, pick G3
6) G3 - S3, pick S3

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
2) G1 - G2, pick G2
5) G3 - S3, pick G3

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.

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.

 

[wulfy42]

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.

 

[wulfy42]

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.

 

[wulfy42]

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.

 

[Dimitriov]

This just reveals why probability questions are always stupid. Mathematicians may well understand math, but their ability to frame questions in the English language is always embarrassingly lacking.

Also if you want a pure math problem, stop using real world scenarios in which any non brain damaged person knows that there will be other factors (like human agency) affecting the outcome. Of course I am going to consider whether or not I can trust the warden or the game show host. I'd have to be emotionally stunted not think about that aspect of the scenario.

The prisoner problem is particularly poorly written: there is no opportunity to flip a coin in secret, so no flip can have occurred.

 

[The_Darkness]

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.

Consider a variant on the Monty Hall problem. Lets have 1001 doors. The car is behind a door, 1000 goats are behind the others. You pick a door at random - lets call this door 1. Most likely, door 1 is a goat.

The announcer, who knows where the car is, then proceeds to open 999 of the remaining doors, revealing goats behind all of them. Except he, rather suspiciously, leaves door 752 closed.

Would you switch from door 1 to door 752?
I would. Because either you got the car on your first guess - which is 1 in 1001 odds - or it's behind door 752.

(And, the more important question - who is going to look after our thousand goats?!)

Likewise for the standard Monty Hall problem. Either you got the car on your first guess - which is 1 in 3 odds - or it's behind the other door.

 

[2xDouble]

The "solution" to the "Three Prisoners Problem" is wrong, and why it's wrong creates its own interesting logical/literary paradox. Don't misunderstand, the logic is sound and the math is correct, but there are key details in the text and the nature of the exercise that must be considered.

Because this is a "thought exercise" and math problem, we have to concede that anything which didn't happen in the story didn't happen (damn creativity-stifling thought exercises...). In the story, the warden didn't flip a coin, nor did he walk away from A before revealing the condemned. We cannot make the assumption that the warden flipped an imaginary coin (didn't happen in the text = didn't happen), therefore the coin flip never took place, eliminating the possibility of A's pardon. Similarly, if the story did contain a coin flip, we cannot assume the flip was to obfuscate the "real" answer (because that would be an assumption) and therefore the existence of a coin flip in the story states outwardly that A is being pardoned, else it would not have occurred.

 

[The_Darkness]

The "solution" to the "Three Prisoners Problem" is wrong, and why it's wrong creates its own interesting logical/literary paradox. Don't misunderstand, the logic is sound and the math is correct, but there are key details in the text and the nature of the exercise that must be considered.

Because this is a "thought exercise" and math problem, we have to concede that anything which didn't happen in the story didn't happen (damn creativity-stifling thought exercises...). In the story, the warden didn't flip a coin, nor did he walk away from A before revealing the condemned. We cannot make the assumption that the warden flipped an imaginary coin (didn't happen in the text = didn't happen), therefore the coin flip never took place, eliminating the possibility of A's pardon. Similarly, if the story did contain a coin flip, we cannot assume the flip was to obfuscate the "real" answer (because that would be an assumption) and therefore the existence of a coin flip in the story states outwardly that A is being pardoned, else it would not have occurred.

True. The three prisoners problem would work better if the Warden stepped into a side-room for a moment before giving an answer. Then we'd have no knowledge over whether or not a coin was flipped.

 

[Bad Player]

Because this is a "thought exercise" and math problem, we have to concede that anything which didn't happen in the story didn't happen.

No we don't.

The problem says nothing about whether the warden flipped a coin or not. It never says he didn't flip a coin; assuming he didn't is as much of an assumption as assuming he did.

It's possible to assume that anything not expressly stated in the story doesn't happen, but we don't "have" to, and in fact doing so creates a needless issue in the problem.

The problem lays out a process through which the warden tells A who is going to be executed. The process may or may not involve flipping a coin. The problem is silent on whether the warden actually flips a coin or not. I think the most reasonable explanation is that the warden flipped a coin if necessary. This is explicitly a probability problem, not a logic riddle; if the person who will be executed could be figured out exclusively from whether the warden flipped a coin or not, it defeats the entire problem. So while you "can" assume that he doesn't flip a coin because it doesn't say he did (or you can assume he did flip a coin because it doesn't say he didn't), doing so immediately destroys the issue at question, and so it seems a lot more reasonable to simply read the problem in a way that corresponds to the information provided (that is, the warden follows the agreed-upon procedure) and that properly retains the issue.

tl;dr: assuming he didn't flip the coin because it didn't explicitly say so is just as much of an assumption as anything else, and it's needlessly creating a problem where one doesn't need to exist.
I think people are trying to hard to outsmart the question rather than just solve the probability problem.