A lovely puzzle this week.

I have a box and inside the box are four stones. One stone is white, another is yellow, the third is blue and the fourth stone is also blue. I put my hand into the box and pick up two stones. I bring my hand out in a fist, look inside my fist and remove a blue stone. What are the chances of the other stone in my hand also being blue?

As ever, please do NOT post your answers, but do say if you think you have solved the puzzle and how long it took. Solution on Monday.

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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Think I got it in 2 minutes.

Got it. Tricky one.

One little problem with this puzzle is that it’s incomplete. You can get different answers depending on what assumption you use for the incomplete bit. I feel two answers coming up…

I’ve now done a little bit more work on this. One thing to note is that probability problems based on somebody who has privileged knowledge (as here) with an unknown selection criteria cannot result in just one answer.

I have three answers, and I’m sure one of them will turn out to be the “correct” one, although in one sense there are an infinite number of answers possible.

I agree with you this week Steve. I have two answers. One based on the assumption that everything is completely random, and the other based on the condition that there is at least one blue stone in his hand (which is what the question seems to imply).

Nice to have some agreement, but I fear it isn’t complete for which reasons will have to wait until Monday.

I agree that there is likely to be a trick. The wording implies that you want the probability *given the privileged information*, but that would be too easy.

No trick here. The solution is pretty straight forward. Not sure where there is any confusion here.

What is the probability that he has 2 blue stones in his hand given that he has at least one blue stone in his hand.

@Garret

OK – consider the possibility that the picker only picks a blue stone when he has no choice in his hand. Now consider the case where the picker always chooses a blue stone when there is one available in his hand. Now do the odds. Think those two give the same probability?

Garrett :

…and that he chosed to remove a blue stone.

That’s where the confusion is : in the situation where he has only one blue stone in his hand, how did he chose to remove the blue one and not the other ?

Steve – don’t really understand the scenario you are trying to explain

Yat – Interesting point. Removing 1 at random vs removing a blue stone if a blue stone exists does result in 2 different solutions.

That being said, I think the intent of the puzzle is as I stated above: What is the probability that there are 2 blue stones given that there is at least one blue stone. This solution to this problem is indeed very counter-intuitive.

This question is very easy once you take the problem and re-write it mathematically as the correct question. The correct question asks about choice of pairs at random, as the question above says. I see no ‘implications’ here. I can be wrong.

Don’t make the question more complicated than it is. The question you need to answer is simple,.. what is the probability of drawing two blue stones simultaneously? It’s the only way the first and second stone can both be blue.

Considered that way, the problem becomes a very straightforward probability.

This is not a Monty Hall problem.

OEJ – Not true. The fact that we know at least one stone is blue, gives us a different probability of there being 2 blue stones, than if we didn’t have this information.

Depending on the circumstances surrounding how he chooses which stone to show us, this probability changes.

I agree with Garrett that the wording does create ambiguity but I think the INTENDED question was as One Eyed Jack stated: what is the probability of drawing 2 blue stones from the start.

My interpretation of the question is that knowing that you’ve picked at least 1 blue stone, what is the probability that the 2nd stone is also blue, which changes things, and confirms my suspicion that Richard has sadistic tendencies 🙂

I must admit, I don’t understand where the problem lies. I think Garrett is right. We know that one stone in the fist IS blue. We also know that the picker “looks” inside his fist. This means that the blue stone is not chosen at random if only one of the stones is blue. The question remains what is the probability that the other one is blue too. I may got the answer wrong but the question seems pretty straightforward.

I interpreted it as analogous to the old one about children: I have two children. One is a girl. What is the chance both are girls? If I had two boys, I suppose I’d have had to start with “one is a boy.”

In this one, had he not picked any blue ones, he could have said “I remove a non-blue stone from my hand, and what is the chance both are non-blue?”

To me, it is clear that, in this situation, at least one stone is blue. Had that not been true, he could have rummaged around again a few times, until at least one was blue.

I’m fully with Garrett, in fact, this week the question didn’t strike me as ambiguous at all, like some other times. Pretty straightforward.

@Steve Jones – there is no ambiguity, I think the problem you are having is a commonly misunderstood rule of probabilities and the “trick” in this question.

Steve Jones is correct in that information is missing due to the framing of the question in such abstract terms. Let’s make it more concrete by filling in the missing info:

A smuggler hides contraband in two out of four boxes. The customs officer picks two boxes at random for inspection, however it turns out that there is only time to open one of them, and lets the smuggler choose which of the two to open. The smuggler opens a box that has no contraband. If the officer should then change his mind and open the other box anyway, what are the odds of the smuggler has nothing to worry about.

(In other words, the blue stones represent the boxes that have no contraband.)

We can fill in the missing info differently by changing the story:

An assassin poisons two out of four bottles of wine. A palace servant picks two bottles at random and carries them to the king’s chambers. A guard confronts the assassin, and asks him to choose one of the remaining bottles, and gulp down a mouthful of wine. The assassin dies from his own poison. What are the odds that the king is safe to drink from the two bottles that were sent to his chamber?

(In this case, the blue stones equate with poisoned bottles)

The two scenarios yield different answers because the intent of the chooser differed — the former always preferring to reveal a blue stone, the latter preferring to reveal a non-blue one. The intent is missing from the wording of the puzzle.

A third answer might also arise when we introduce a forgetful smuggler/assassin and a colour blind stone picker.

@Tort

Ah but there is – depending on the protocol taken by the “stone picker”. I’ve written simulator programs for thee three different protocols I’ve come up with and these confirm the mathematics. What people tend not to understand is that the different protocols can eliminate certain possibilities. To take the most trivial example, if the picker only ever chooses a blue stone when he has no choice (that is when the first stage of the process has resulted in two blue stones), then when he reveals the first blue stone there is a 100% certainty the second is blue.

That’s because it has eliminated all the possible solutions where the second stone is white or yellow. We would never see such a scenario. Remember, the question being asked is not what the chance is of both stones being blue at the first stage, but what the chances are after the first reveal.

However, if the protocol is for the “picker” to choose a blue stone if one is present, it leads to a wholly different stage 2 probability. However, if the blue stone is a random choice, that leads to a third probability. The reason for these differences is that the different protocols change what subset of stage 1 (two stone configs) are selected as “blue reveals”.

So to confirm, simulation runs confirm different results for the three scenarios, not because the chances of two blue stones are different, but because of the selections which are eliminated after stage 1.

Okay, Steve, you’ve convinced me. Good argument. My guess is that the puzzle isn’t meant to be ambiguous, but you are right that it is.

I also have more than one answer depending on which interpretation of the question is intended. I don’t think it’s quite right to assume any one simplified version of the question is exchangeable with it as presented. In matters of probability, the wording is often annoyingly important.

I also have two answers depending on the protocol, leading to three possible results.

I think I got it in under a minute but as @Steve_Jones points out above, a part of it does seem incomplete or at least a bit vague.

About 30 seconds for my first answer!

One answer. Easy answer, but tests people’s knowledge of gamblers fallacy. I like it.

Yeah I think I have it in under a minute, but agree there seems to be a bit of incompleteness about it… Might have to keep thinking… Hmmmmm

About 2 minutes. Had to work through it on paper though. (And I don’t think the puzzle is incomplete or that you have to make any assumptions).

I agreed with you initially, but on reflection, I think it is ambiguous.

Option 1: The person chooses one of the two stones at random (50%), reveals it — and the stone is blue.

Option 2: The person observes both stones, chooses a blue stone, and reveals it.

I think the two options lead to different answers…

(the Monty Hall problem has a similar ambiguity — the answer changes, depending on whether Monty knows in advance that he is revealing a goat..)

@repton

Exactly! The problem here is that the wording of “look inside my first” can clearly be understood in two ways. The Monty Hall Problem is a bit different in that while there is no ambiguity in its wording, the solution isn’t quite intuitive.

It’s a common misconception that the Monty Hall problem relies on the hosts knowledge. It only relies on the condition of a goat being revealed – it doesn’t matter if the host knew there was a goat there or if it was revealed by chance.

The same applies here. Whether he chooses a blue stone from his hand or picks a stone at random which happens to be blue does not change the chances that he originally picked up 2 blue stones, which is really all this puzzle is asking. There is no ambiguity.

@thequiet1

You are wrong – about both this and the Monty Hall problem. It’s well known that the Monty Hall problem as originally stated is ambiguous as it doesn’t include the host behaviour based on his privileged knowledge. One of dozens of papers on this :-

http://129.3.20.41/eps/exp/papers/9906/9906001.html

Whilst this is not the same problem, the same issues arise over the motivation and criteria used by the person picking the stone. Statisticians call this the “protocol” and it is absolutely key to understanding these problems.

Steve,

I am not wrong. The paper you link to does not address the issue I raised. Once the host opens a door with a goat the odds do not change whether he opened it by chance or because he knew there was a goat there. Of course the host knew which door to reveal so that the game does not end prematurely, but once the goat is revealed it does not matter how it was chosen, the odds are the same either way.

Option 3: He looks at both stones. If both are blue, he reveals one, otherwise he puts them back in the bag.

Option 4: …..

The possibilities are limitless.

@anonymous

The point you are missing is that in the Monty Hall problems as initially stated, we do not know is the host always shows a goat and gives the option. It may be that he only gives that option when he knows the contestant has chosen the car.

So it’s only true that the contestant double their odds by changing their choice if the host always exposes a goat. If he chooses not to do so when the contestant has chose the car and it’s a means of bluffing, then the contestant should not change his or her mind. There are middling protocols too.

However, I’d certainly agree that if the second door was opened at random to expose a goat, then the improvement in odds would be the same as the host always opening a second door with a goat. However, the overall odds will not be the same as, on one-third of those second door openings, the car will be revealed and that changes everything.

So the host knowing where the car is does matter, as does whether he always follows the procedure of opening a second door.

Steve Jones : “However, I’d certainly agree that if the second door was opened at random to expose a goat, then the improvement in odds would be the same as the host always opening a second door with a goat”

…well, I don’t. Or I misunderstood you, but even if the door exposes a goat, the fact that the door _was_ chosen randomly or not actually is important.

You chose a door, Monty randomly opens one of the two remaining doors, and uncovers a goat, then both your door and the remaining door have a probability of 1/2 to hide the car. In the case where he deliberately opens a door with a goat, this does not give you additionnal information about your door and that why it is still 1/3. If he choses randomly, the fact that he uncovers a goat is more likely to happen if you have chosed the door with the car.

Suppose you always initially choose door 1, and Monty always opens door 2, but the position of prizes is uniformly random. Then there are three possibilities: the car is behind door 1, door 2, or door 3.

Then suppose Monty opens door 2, and reveals a goat. This eliminates the second possibility, so the two equally-likely options are that the car is door 1 or door 3. Your chance of winning is 50%, whether you switch or do not.

(this is sometimes known as the “Monty Fall” problem, because Monty slips on a banana skin and opens a door by accident as he’s walking)

Yep – easy – 20 secs

yeah. It is a lovely puzzle

A couple of minutes just trying to figure out if I’d covered the options.

it’s obvious — i got it immediately. though as jonsie says, different readings of the scenario can led to other results.

Got it in about 1½ minutes. I used the long-winded method of writing out all the possible combinations.

The unwritten assumptions are that the stones have been thoroughly “shuffled”, the hand chooses two stones randomly, the hand cannot feel the colours of the stones according to different shapes or textures, and that the two stones inside the fist are equally likely to be the first one to be observed.

Now I suppose I’ll have to think about what the “intuitive”, “wrong” or “obvious” answer is supposed to be, instead of the mathematically logical one.

There is only one answer!

As has happened many times before, I have AN answer after a couple of minutes. Now I just have to wait until Monday to see if I’m right…

Given the premise set up by the first part of the puzzle, I have an answer, which took around a minute of head thinking. But while I’m confident in my interpretation of the problem, I’ll still be interested to see everyone’s answers on Monday (including, but not limited to, Richard’s “official” one).

Well, I got my approach straight away, then took a few seconds to do the counting and give an actual answer, still under a minute. I’m never confident I have it right with these things though, so can’t say whether I’ve actually solved it or not. I look forward to finding out where I’ve gone wrong on Monday.

I have the answer but I am looking forward to the fall out because many won’t like it. This is very similar to a ‘classic’ problem which I’m sure will be referred to when we get the answer.

@John Loony : I think that, if you consider these assumptions (and more specifically “the two stones inside the fist are equally likely to be the first one to be observed”), then you will get the obvious solution, whereas the problem assumes something like “I open my fist and see at least one blue stone”, leading to a solution like the one in the problem with boys and girls (which, by the way, is also incomplete in all versions I have seen yet).

20 secs!

I don’t know if Richard edited the puzzle since posting, but the current wording seems unambigous to me:

“..bring my hand out in a fist, look inside my fist and remove a blue stone.”

Looking inside the fist ensures that if there is any blue stone at all there, that’s the one that will be picked, i.e. it’s not a random selection from two possibilities.

Which I think leads to only one answer.

(Took me about 60 seconds to get the ‘one’ answer, which I’m fairly sure is correct. Mostly spent trying to figure out if I’d missed something.)

I think the wording implies that one interpretation is more likely to be the intended answer (that he specifically chose a blue stone), but it is not incompatible with him looking inside his hand and choosing a stone randomly (which just happens to be blue).

So there are two possible interpretations with different answers (actually an infinite number depending on exactly how he chose the stone) but I think one is more likely to be the intended answer.

I honestly don’t see how this affects the statistics of the situation. No matter how he selects the stone from his fist, the stones are picked at random from the box and no matter what happens next, the question remains: what are the odds that both are blue.

I can’t see that this has any relation at all to the Monty Hall problem. There is no second selection, no additional knowledge, no psychology, just two stones, selected at random, with the property of being or not being blue.

@Eddy

It seems very much ambigous to me : we need to know what were his intentions when opening his hands. Did he decide beforehand that he would remove a blue stone if he found one, or did he just happen to remove randomly one of the stones in his hands ? If there is, for example, a blue and a red stone in his hands, could he have removed the red one instead ? This is actually what changes the problem and the answer. I can’t go much further without giving both answers…

I get three main answers which provide for two extremes and one neutral (random) version. The best I can say for sure is that what the maximum and minimum odds are.

Those that claim there is just one answer haven’t really understood the wording.

“If there is, for example, a blue and a red stone in his hands”

That would have a probability of 0, since the question doesn’t account for stones changing colour or someone to substitute a different coloured stone 🙂

After all, the three colours that entered the bag were blue, white and yellow.

I think it’s safe to assume the initial selection of two stones was at random, whereas once that selection had been made, the second selection (of a blue from the two) wasn’t at random.

Why would it be safe to assume that ? Nothing is told about the way the second selection is done, I really don’t see why we should consider that this one is not made randomly.

@Steve: same here.

Seems very simple – could be a trick question?! Got it in about 15 seconds if I’m right!

I got an answer, then I read the comments and realized I had assumed that you drew the stones blind, so I resumed my awkward approach and now have 7 correct answers, including one for if you are floating in space.

I got an answer quickly (1 min), double-checked it, then re-read the question and realised I had a wrong answer.

I’m pretty sure I now have the correct answer. About two to three minutes.

15 seconds to solve but some time considering whether there is a trick here

Assuming the two stones were picked at random, there’s a 0.8 chance that at least one blue stone will be in your fist (the possibilities are: B1B2, B1Y, B1W, B2Y, B2W, YW – so the only combination without a blue stone is YW).

As the selections have already been made, this isn’t equivalent to Monty Hall – and as there are only two stones in your fist, I think that regardless of whether you look inside your fist and deliberately pull out a blue stone, or whether you close your eyes and pull out a blue stone, the chances of the remaining stone being blue are the same in each case.

If, instead, we were asked to calculate the odds for the overall process (from bag selection to discovering the colour of the first selected stone to discovering the colour of the second selected stone), the odds may vary depending on whether we’d looked inside our fist to pull out the first blue stone from our selection or chosen it at random from the two stones.

No, the chances of the remaining stone being blue are not the same in each case. You can just try, and you will figure out that combinations with only one blue are eliminated half the time if you draw randomly, but will always be kept if you pull the blue stone deliberately.

You are maybe crossing the grey line over giving too much away. However, I’m quite sure how you work out that 5 out of 6 equals a 0.8 probability…

@Steve – D’oh! I got my head stuck on 5 so somehow created 4 / 5 rather than 5 / 6 (0.8333…)

@Yat – I assumed that the first selection (2 from 6) was made randomly. We’ve now opened our hand and picked out one blue stone (which could have occurred in 5 of the 6 possible combinations). Depending on the method used, I’ve got two possible answers – both of which have a prime denominator.

Ok, now I agree.

Of course, those are the solutions for the two simplests protocols in my point of view (I would say, the ones it could be an honest mistake not to describe), which are “always remove the blue stone if there is one” (probably the expected solution) or “randomly remove one of the two stones” (the one I think it is most reasonnable to assume). As Steve Jones stated several times, it could also be “never remove the blue stone when you have a choice”, or more generally “when there is only one blue stone, remove it with probability P”, with P between 0 and 1.

Got “the” answer in a few seconds, but this is the answer if everything is random (picking stones, and revealing a stone). If there is a conscious thought behind either, the answer will be different, similar to the “boys and girls”-problem a few weeks ago.

About 30 seconds. Still trying to work out the obvious “wrong” answer.

marilyn savant, please pick up the white paging phone…

aside from the inherent ambiguity, it may be helpful to know that the answer is different than if richard had asked the probability of picking the two blue stones out of the box.

About a minute. And I’m 100% sure I’m right. No ambiguity here but a lot of people will get it wrong

So you have the ability to guess with 100% probability the information not actually included in the problem, and necessary to distinguish between the two solutions discussed earlier. I am impressed.

Yat,

If you think there is ambiguity you have not understood the problem. There is one answer.

Ok, we’ll have one on monday, then I will explain the other one, and I challenge you to find an error in it !

@Anonymous

If you think there is no ambiguity YOU have not understood the problem.

Interesting. I “solved” it in a couple of minutes by enumerating the different combinations. I now realise the possibility of a second solution depending on the interpetration of the “look inside my fist” element. Even more interesting would be trying to generalise to larger examples where simple enumeration of outcomes is impractical. Not sure I’m up to that, sadly

A couple of minutes to find an answer, 5 minutes to read the comments and not change my answer! We’ll have to wait until Monday to discuss this…

Well I have an answer but I’m usually crap at probability etc. I understand perfectly when reading the answer but everytime a similar problem comes up I’m scunnered.

Will wait till Monday to see if I’ve been had again.

It is funny how some people try really hard to find something wrong with the puzzle just to give themselves an opportunity to show off some partial knowledge even if it doesn’t apply.

I can see why a psychologist would be interested in this sort of puzzles.

No need to try hard to find that the problem does not contain enough information to be solved. The protocol used is not completely described, so there is no way to chose between the two simple solutions, which both follow correctly the problem wording.

Good for you if you find what you don’t understand funny !

I’m painting my ‘stones’ blue as we speak

IMO the question is not completely unambiguous and, as such, more than one solution can be derived.

Looking forward to the fallout on Monday, when loads of people get a different answer to Richards, and feel the need to justify that they are right anyway! 🙂

Classic!

And they will be ! both solutions follow exactly the problem wording, just interpreting differently the “remove a blue stone” protocol. Both interpretations are equally right because the protocol is not completely defined.

The actual wrong answer is to state that one has _the_ solution, with 100% certainty and see no ambiguity in the question. So of course, when Richard will give the expected solution on monday, it will be with pleasure that I (and probably many others) will explain the other one. Classic !

So… I agree with you completely 😉

I can justify three answers now based on three different policies taken by the “blue stone picker”. Two of those define the maximum and minimum probabilities, but some interim protocols could, theoretically, allow for any in between. However, I’ve stuck to two deterministic and one random protocol.

I think I know the protocol that Richard will have used, and that is that the picker always chooses a blue stone if available rather than only choosing a blue stone if there’s no choice or a purely random choice.

Steve, what makes you think there is a policy at all here? The question simply states “…and remove a blue stone”. It doesn’t say anything about doing this multiple times, or working from any kind of policy.

Never assume that which is not given. The question as stated is unambiguous and has a clear answer. That seems to be a rarity

I’m not assuming that there’s a policy (the absence of which would be equivalent to a random choice). What I’m saying is that we don’t know if there is or is not one and consequently we must either make an assumption or say that it’s impossible to know. We might make a best guess as to the existence (and nature) or non-existence of a policy. However, one thing is for sure, the mathematics and programs to simulate them do give radically different results.

Also, I would say that there is some very direct evidence there is a policy – the question says a choice is made after looking. Why, if it’s a random choice would you look first? Humans aren’t good at random choices – even if it’s unaffected.

So I’m making no assumptions at all – just saying that without knowing if there is any policy (or its nature), we can’t work out the probabilities.

nb. what you call a policy is normally called a “protocol” which would include random.

But it doesn’t say it is at random. All it says is that he removes a blue. Not that he must do this, or that he always does this, just that he does.

I read the problem as follows: take two stones at random. One of them is blue, what are the odds that the other one is as well?

The probabilities here seem rather clear to me, and the question in this form has been in millions of puzzle books already.

30 seconds then read through the comments definitely sticking with my answer.

Got an answer in about a minute.

After reading the comments I got two more.

Now I’ve got a formula that gives the answer for any probability of picking the blue stone.

That’s the super-duper 100% correct completely unambiguous solution 🙂

Solved. Around 15 secs. There is precisely 1 solution. M.

First answer in less than a minute. Then read this thread and reckon it comes down to what exactly is being asked? I think the question is NOT “What are the odds of getting 2 blue stones?” but actually “If I’ve already got one blue stone in my hand what are the odds of having a second one?” Big difference!

Two more minutes to double check the logic and an answer which is quite surprising!

Yes, that’s my line of thinking exactly. There is a fixed number of combinations of two stones that involve at least one blue stone. And there is a subset of those combinations that involves two blue stones. So calculating the probability seems straightforward. Unless I’m overlooking something, which is always possible. 🙂

I think I got the answer, took a couple minutes just to write out the pairs. I just thought about it in terms of this: if you randomly select a pair of stones, what is the chance that one of them is blue? If you KNOW that the pair of stones you have contains AT LEAST one blue, what is the chance that the other is also blue?

I’ll leave this one to more capable mathematical minds.

Come on Paul you can do it. Think of all the possibilities!

I think I solved it, and it took me 10 or 12 minutes.

BTW, I don’t see why some people think there is ambiguity involved. There is no privileged information. This is not the same as the Monty Hall problem.

easiest one ever,,,,,so I’m obviously wrong.

My interpretation of “I look inside my fist” is that the fist (containing two stones) is removed from the box, and then the fist is only slightly open to the extent that the person can only see one of the two stones which are in the fist. The two stones in the fist order themselves randomly as the first/visible one and the second/hidden one.

My second thought is that “I look inside my fist” could be interpreted as meaning that the person opens his fist and can see both stones, and then chooses one which is blue. This puts the person into a similar position as the game show host in the Monty Hall problem, i.e. he has more knowledge (and therefore better than random choice) than the neutral observer / contestant.

Therefore the probability may be (a/c) rather than (a/b).

It’s really a question of maths, do the maths and you’ll probably get it right, have the answer, but still crossing my fingers lol.

For once an unambiguous problem. Took basically no time at all, but only because I’ve seen so many problems of this sort before

20 secs for a first answer. It was easy, really ! If you find the right tought then its easy. But i have to verify my solution.

I got an answer in under a minute. It took me that long to find a pencil and paper. Now I need to wait until Monday.

There is going to be a flame war tomorrow, LOL!

My view is that this is unambiguous and has one solution. It seems to me that the peeking inside the fist is an irrelevant red herring here, designed to confuse. The problem is to be solved from the point of view of the observer, not the person selecting stones, so all that matters is what the observer sees, not the thoughts in the selector’s head. Yes, the selector may have some policy about what to do if there is only one blue stone, but by the time the decision has been made and a blue stone has been removed, the uncertainty has ended: the observer sees the blue stone, and that is enough to solve the problem.

In my opinion some clever guys have already said all what had to be said and have also wasted too much time. What will be sad next Monday, though, will be reading that some naive poster is willing to believe that maybe prof. wiseman was just making some clever experiment by wording the puzzle this way. Oh-Richard-you-are-so-wonderful sort of song. Nothing bad in this, I guess, but be careful…

Haven’t read any of the above. Got it in about 10 minutes and am extremely confident I am right.

Getting ready for the fallout!

I got my answer in under five minutes – most of the time spent looking for a “twist” because the problem seemed too simple and straightforward. That said, I often get these types of problems wrong, so I’m interested to see the answer.

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Got it. A minute. The goat problem helped.

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