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Comment by gametheoretic | Hacker News Reader

This is wrong.

In oversimplifying your table for potential idiot readers by introducing the silent variable of older vs younger child, you have introduced order into a statistics problem wherein order is irrelevant and masked the fact that what matters is which of the children you already know to be a boy. You grant this difference for the BG pairing, thus producing two options, but not for the BB pairing. There is not ONE way of knowing this for BB, as you present in the table, but two: you know the older is the boy, or you know the younger is the boy. 2/4 = 1/2. QED.

What you should have done in the first place, however (unless your aim is to produce a blog post which, apparently, can convince otherwise intelligent people that irrelevant information can magically become relevant) was simply remove the one boy from the equation and reformulate the question. What is the probability that this other child is a boy (and thus that both are boys)? 1/2. Fuck the table. Had you not used a table, this would never have happened. But it feels authoritative, right? Thank you for the psychology lesson.

Downvotes but not refutations, because there aren't any. I'll assume it's my tone. My opinion of the competence of the average reader on this site has plummeted reading the other comments, however, so hey, fuck you guys too. :)

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Your reasoning works if the person says "The older is a boy." You claim that this is correct.

Consider ...

Assuming the sex of a child is 50:50, then consider all couples with exactly two children. Of those, consider only those who can truthfully say "at least one of my children is a boy." Of those couples, 1/3 will have two boys.

Interpreting the initial problem in this manner, the answer of 1/3 is right. What I'd like to know is why you think the initial problem should be interpreted in any different way.

I did an experiment: I kept flipping two coins [1]. If it came up tails, tails I ignored it. If it came up heads, heads I noted it down in one column. If it came up heads, tails I noted it down in another column. After many flips the ratio was 1:3.

As far as I can tell this is an accurate simulation of the problem. We have two things, coins or children, which have a 50:50 chance of being one value or another. We are told one combination of values isn't the case, so we discount them from the simulation. We then use the frequency of the target combination to approximate its probability. We haven't added any artificial ordering into the problem.

I know it's not well argued logical reasoning, but it supports the articles approach. Can you point out where I tripped up in the simulation?

[1] Of course I didn't actually sit there flipping coins. I wrote a program to do it. (let [sample-size 10000000] (/ (get (frequencies (take sample-size (filter #(not= % [:girl :girl]) (repeatedly #(vector (rand-nth [:boy :girl]) (rand-nth [:boy :girl])))))) [:boy :boy]) sample-size))

Instead of actually programming, I flipped 2 coins this morning and left them on my desk. They are still sitting there. Either HT, TH, HH or TT

I just remembered one of them is tails, so is there is a 1/2 chance that they are both tails now? Nope it's 1/3 chance they are both tails. Either HT, TH, or TT.

Of course that's a much simpler way to simulate the problem, but then you get people trying to argue that HT and TH should be collapsed into a single state.

By using a simulation with repetition it's easier to pose the states as 'TT' and 'not TT' and use frequencies to show 'not TT' occurs more frequently than 'TT'. I find this starts to help convince people that the states shouldn't be collapsed, even if it doesn't explicitly explain why.

>We haven't added any artificial ordering into the problem.

Yes you have! To run the experiment, you have to treat TH and HT as different but you "pretend" not to do so by adding them into the same pile. You're pretending to know that the father tells you which child is the "known boy", but by the statement of the problem, you don't know that. Why is this an error? Because the father must follow a rule - he must give you a boy, and there are two ways for him to do that in BB but you only recognize one. If you go the "table-minus" route, which both the author and your experiment do. But you can't do that without affecting the odds! By telling you they aren't both girls, he lowers the chance that either of them might be. Yes, I'm serious.

Correct formulation of table:

a) BG; b) GB; c) BB and he gives you the first; d) BB and he gives you the second.

Isn't the probability of c and d half that of a and b? If BG is the case, there's 100% chance he's talking about the first child. If BB is the case, there's a 50% chance he's talking about the first one, and a 50% chance he's talking about the second one.

    BG, giving the first - 1/3
    GB, giving the second - 1/3
    BB, giving the first - 1/6
    BB, giving the second - 1/6

[Edit] You're right. Tables can be a bad way of thinking about it, because you have to make sure each option has an equal probability for them to work. This is okay when you have a single variable (like in the original post), but gets tricky when you introduce a new variable (as you did: is the father thinking about the first or second). Once you get past one variable, trees are probably a better but still simple representation.

You're oversimplifying.

The problem is the interpretation of the question in precise terms. We are given some background information and then told the observation of a particular random variable. In the given article, the random variable is interpreted as "yes/no: at least one child is a boy born on a Tuesday". Rambling about "order" does not change the fact that the article is correct, given the definition of the random variable that is observed.

The problem is that this interpretation is only correct if you imagine that you are always given the observation of the same random variable no matter what the genders or birthdays of the children actually are. For example, if I had two girls born on Wednesday, I would then tell you "no child is a boy born on a Tuesday". That's an unlikely scenario which is what makes it legitimately surprising.

It's like the Monty Hall problem. What if the host picks the door in advance? What if the host only opens a door with a goat under the condition that you picked the door with the prize? What if the host only opens a door with a goat under the condition that you picked the wrong door?

The article has problems but "wrong" is not one of them.

> so hey, fuck you guys too

Hacker News is a terrible place to visit if you want to take the voting system personally. Don't make this about you. It's about the post.

I'm not, you're overcomplicating! X is a boy. What is the chance that Y is also a boy? 1/2 or 1/3? Non-rhetorical, answer the question as posed.

> answer the question as posed

You can't give a correct answer to an incomplete question. Sit down and define the problem and if you come up with a different answer it's because you made a different assumption about the definition of the observed random variable.

Let's suppose there are four universes.

* Two boys (25% chance)

* Two girls (25% chance)

* Boy and girl (50% chance)

If we assume that the observed random variable has a preimage that contains all possibilities where the statement is true, then we come to the 1/3 conclusion, which is counterintuitive but true. This is the position of the article. I have come to the same conclusion without using order, but I have used the same definition of the observed random variable.

Tell me what you think the random variable is.

We could assume that the random variable has a preimage which only contains the case where both children are boys. Bam, probability 100%, not 1/2 or 1/3. Or we can assome that the random variable has a preimage which only contains the case where one child is of each gender. Bam. Probability 0%. So any answer between 0% and 100% is defensible.

God, I knew this was coming. Since you won't accept my strategy of building up to the complete question (because it is in fact your formulation which is incomplete, and thus I, like you, will not respond to it), let me try this instead:

You misunderstand Monty Hall. I would know; I learned about it by being asked the question and getting it right, not by reading someone else's explanation (which leaves room for misinterpretation). The key to Monty Hall is none of what you stated (or rather, rhetorically asked); it's that the host must pick a door that is both a) bad and b) not yours. If your door is good, then he can leave either, but if your door is bad, then he must leave the good one. So, two possibilities: the door he left was either bad or good, BUT the rule driving his choice wasn't 50/50, so the odds of it being bad or good aren't 50/50, so neither are the odds of yours, even though there are only two left. Take a minute, think about it.

Now ask yourself which "door" (read child) the father told you was a boy. There's the gotcha. Under your/the author's formulation of the question, he doesn't remove a door in the BB universe, but he does in the BG universe. There are two BG's because the table acknowledges that he can tell you which of the two is definitely a boy, and thus he has two options for doing so BUT HE DOESN'T, HE HAS ONE BUT THERE ARE TWO BG UNIVERSES, SO IT LOOKS LIKE TWO.

As mentioned in other replies, it's clear now that you are interpreting the question in a manner other than the usual interpretation, and then being incredibly rude about anyone who disagrees with you.

I have in my hand a device that lets me call up on demand most of the knowledge of the world, and I use it to look at pictures of cats, and to argue with strangers.

Deleting the second option ...

> Since you won't accept my strategy of building up to the complete question (because it is in fact your formulation which is incomplete, and thus I, like you, will not respond to it),

No, his formulation is complete and inevitably leads to the correct answer 1/3. You are constantly avoiding a complete formulation of the problem by diverting to a muddled discussion of an entirely different problem. This is NOT about Monty Hall.

Or you're trolling. In which case you are doing it masterfully.

Wait, explain it to me like I'm 10 years old.

As I see it, there are three options - either it's a boy-boy pair, a girl-girl pair or a boy-girl pair. Now, since OP introduced the age variable, let's give the kids names to make things easier to understand. Let's say that the boys are named John and Jack, the girls Jill and Judy and if it's a boy-girl pair then they're named Jack and Jill.

Now if it's a boy-girl pair, we have exactly two options: either Jack is older than Jill or Jill is older than Jack. This is shown in the first picture.

However, we also have two options for boy-boy and girl-girl pairs: Either Jack is older than John or John is older than Jack. Either Jill is older than Judy or Judy is older than Jill. This is not shown.

Since we know that one of them is a boy, we can dismiss both girl-girl pairs. Now we're left with four options. In 50% of these options the other child is a boy.

What am I missing?

> As I see it, there are three options - either it's a boy-boy pair, a girl-girl pair or a boy-girl pair.

Yes, but those three options are not equally likely.

Let's say the parents decide that their firstborn will be named Jack if male and Jill if female, and the second child will be John if male and Judy if female. Now you see that there are four possibilities: Jack+John, Jack+Judy, Jill+John, Jill+Judy. The last one is eliminated when you know that there is one boy, which gives 1/3 chance that it's the first.

The reason age was introduced was simply was a way of differentiating the two children (giving them identity). You introduced names to differentiate the two children, which is another perfectly good way to do it.

But then you _also_ introduced age. Because you now have two different ways of labelling the children, you've introduced a new variable, and that changes the probability.

A better way to look at it is with two unisex names: Drew and Sam. This completely replaces the age thing as a way of identifying them, so forget all about age.

Our possibilities are:

  Drew: boy, Sam: boy
  Drew: boy, Sam: girl
  Drew: girl, Sam: boy
  Drew: girl, Sam: girl

We know Drew and Sam can't both be girls, because at least one is a boy, so that leaves:

  Drew: boy, Sam: boy
  Drew: boy, Sam: girl
  Drew: girl, Sam: boy

There's a 1/3 chance that both Drew and Sam are boys.

Ok, yes, I get it now :). Thanks for the explanation.

You are an overconfident ignoramus. Thanks for demonstrating the Dunning-Kruger effect so impressively.

The reason for having BG and GB is the fact that it's twice as likely to occur than each of GG and BB.

Hahahaha oh my god what a response. The fact that you think you're pointing anything out to anyone with that second paragraph is just priceless.

In GB and BG, he has only one way of telling you which is the boy. In BB, he has two. Thus, four possibilities a) GB, b) BG, c) BB and he gave you the first, d) BB and he gave you the second.

Your brain will of course refuse to process the above due to your earlier remark and the sweet, sweet motherfucking irony it would mean.

Right, now it's clear that you are answering a different question, and in that question, the answer is 1/2. Many people insist that to be the right question, because the answer matches their intuition, but it is a different question.

Not recognizing the difference is what lets some poker players make money.

Consider this question. A chap flips a coin in the morning and thereby chooses one of his two children. Today he will only talk about that child. Later I meet him and he talks about his son. I thereby deduce that he has at least one boy. What are the chances he has two sons? (corrected in edit - thank you)

Answer: 1/2.

Another version: I meet a man and his son, and in conversation it emerges that he has two children. I can see he has at least one son, what are the chances he has two sons?

Answer: 1/2.

Another version: I meet a man and it emerges in conversation that he has two children. I observe that he is carrying a shopping bag with a child's dress in it, from which I conclude that he has at least one girl. What are the chances he has two girls?

Answer: 1/3. (I had to change gender because I couldn't think of anything that would be specific to a boy.)

Your dismissal of the idea that there are other interpretations than your own is disappointing. The language in which you couch your dismissals even more so. Are you like this in person?

"Consider this question. A chap flips a coin in the morning and thereby chooses one of his two children. Today he will only talk about that child. Later I meet him and he talks about his son. I thereby deduce that he has at least one boy. What are the chances he has two children? Answer: 1/2."

Do you mean "what are the chances he has two sons, not children?

By the way, thanks for your input on this thread, I'm sure I speak for many in saying that I really appreciate it!

  > Do you mean ... "sons," not "children"?

Yes - thank you - edited to reduce confusion.

And thanks for the kind comment.

No. In person I can see what the other party is thinking.

The two are independent variables, man. I don't know how to explain it any more clearly than that. What's tripping people up is the idea that removing GG from the possibilities doesn't affect the local probability of either particular child being girl. It does. You must reformulate. THAT's the counterintuitive part. It's just hella hard to explain, because there are too many 2's and 1/2's floating around the problem, so when you say one cancels another, people think you're not incorporating the first to begin with, so you must be one of the dolts the author is trying to learn some knowledge.

The following, you can take or leave: probability is my thing. I got Monty Hall right when I was in 6th grade. I was an AIME invitee, if you know what that is. I got 800 on the math SAT; 36 on the math ACT in freshman year. None of these things mean anything, logically speaking, I know. I say them only to encourage you (and only you, because you're putting in effort) to consider what I'm saying as fully as you're considering what the crowd is saying.

  > No. In person I can see what the other party is thinking.

So when you have less knowledge about the other person you are more rude. I feel that's logically inconsistent.

  > The two are independent variables, man. I don't know
  > how to explain it any more clearly than that.

"The two" ?? What "two"? Your "explanations" are very unclear.

  > What's tripping people up is the idea that removing GG
  > from the possibilities doesn't affect the local probability
  > of either particular child being girl. It does. You must
  > reformulate.

Again, it's clear that you are talking about a different problem. Yes, the problem you are talking about has an answer of 1/2. No, it's not the problem we're talking about.

  > It's just hella hard to explain, ...

You will find many things difficult if you don't learn how to slow down, take time, be precise, and explain clearly. Be more like Feynman - take time to explain things clearly and precisely.

  > ... you must be one of the dolts ...

Ah.

  > The following, you can take or leave: probability is my
  > thing.  I got Monty Hall right when I was in 6th grade.
  > I was an AIME invitee, if you know what that is.  I got
  > 800 on the math SAT; 36 on the math ACT in freshman year.

Good for you. I speak for and contribute to the training of IOM teams, I've been an invited speaker at the Museum of Mathematics in New York, the Dutch National Mathematics days, Matematikbiennalen in Sweden, and more math masterclasses than I can count. This year I'm speaking in London, Switzerland, Belgium, and giving around 150 talks, masterclasses, and workshops. I have a Bachelors, Masters, and PhD.

  > None of these things mean anything, logically speaking,

Right.

  > I know. I say them only to encourage you (and only you,
  > because you're putting in effort) to consider what I'm
  > saying as fully as you're considering what the crowd is
  > saying.

Hmm. Well, thanks for the encouragement. I was only saying these things to point out that I have considered these things, and have some form in the area. I forgot to mention that I was at the G4G when the Tuesday Boy problem was first mentioned, and spent much of the evening with Tanya Khovanova, Gary Foshee, John Conway, Richard Guy, and others, talking about the various interpretations. I didn't talk about it with Martin Gardner when I spent an afternoon with him, we were talking about my work.

I mention these things only to point out that you might consider that you are wrong.

So, do you answer 1/3 for each of A, B, and C?

Well don't I feel like an idiot for saying any of that. Why were you running a "computer simulation" of the problem? Were you, even?

>"The two" ?? What "two"? Your "explanations" are very unclear.

The only two there are: the two children. The sex of one is independent from the sex of the other.

>You will find many things difficult if you don't learn how to slow down, take time, be precise, and explain clearly. Be more like Feynman - take time to explain things clearly and precisely.

Okay, I'll try, good buddy. Monty Hall is different because it precludes upfront the possibility of there being any ratio other than 1:2. If we were, however, to say that any door could have either a goat or a car, then Monty's elimination of some other door would not inform you on the odds of your door YOUR door. Yea or nay?

The only way for Monty to give you any information at all about whether to switch is if the rule binds him to tell you something about the odds of the unchosen doors. In the original problem, this happens, because the ratio is bound 1:2. In our revised problem, this does not happen - unless he tells you he can't remove a door because neither is a goat. But! Using this as an analogue to our children problem, we've already precluded that possibility upfront by eliminating GG, thus Monty can't provide us any information because his B-G choice is unbound by any rule.

  > Why were you running a "computer simulation" of the
  > problem? Were you, even?

I was reconstructing what happened when Erdős was first told of the Monty Hall problem. He answered 1/2, and was very frustrated for a time. In the end he said something like "But you're not telling me the why!"

Moving on ...

  >> "The two" ?? What "two"? Your "explanations" are
  >> very unclear.

  > The only two there are: the two children. The sex
  > of one is independent from the sex of the other.

So we are assuming that this chap has exactly two children, and that their sexes are independent with probability 1/2.

  > Monty Hall is different because it precludes upfront
  > the possibility of there being any ratio other than
  > 1:2.  If we were, however, to say that any door could
  > have either a goat or a car, then it wouldn't matter
  > whether you stayed or switched.  Monty's elimination
  > of some other door would not inform you on the odds
  > of your door YOUR door. Yea or nay?

Yes, agreed.

  > The only way for Monty to give you any information at
  > all about whether to switch is if the rule binds him
  > to tell you something about the odds of the unchosen
  > doors.

Yes.

  > In the original problem, this happens, because the
  > ratio is bound 1:2.

OK.

  > In our revised problem, this does not happen - unless
  > he tells you he can't remove a door because neither
  > is a goat.

Yes.

  > Using this as an analogue to our children problem,
  > we've already precluded that possibility upfront
  > by eliminating GG, thus Monty can't provide us any
  > information because his B-G choice is unbound by
  > any rule.

I fail to see how that is in any way relevant. There are four equally likely possibilities for the children. If we label them older first then we have BB, BG, GB, GG. Only by labelling them in some consistent order to we get equally likely possibilities. The father says that at least one child is a boy. That eliminates the option of GG. Are you somehow suggesting that the probability distribution of the other options is changed by this?

If so, how? Just as Monty's choice gives no information about the door we have selected (and which therefore remains as chance 1/3 of being the winning door), telling us that it's not the case that both children are girls gives us no further information about the other three equally likely possibilities.

So in https://news.ycombinator.com/item?id=7022615 I asked you this ...

    A: So now let's take another experiment. I tell
       you I'm going to flip two coins until at least
       one of them is heads. I do that. Now I ask you
       to bet on whether or not there is a tail. What
       do you think are fair odds?
    B: ... 
    C: ...

    There are three options.

    1: Your answers to A, B, and C are not all the same;
    2: They are all the same, and not 1/3;
    3: They are all the same, and all 1/3.

In https://news.ycombinator.com/item?id=7022707 you answered:

  > The answer to all 3 is 1/2.

That turns out not to be the case. The answer to A is 1/3. Write a computer program to simulate it and then post it here. Seriously, don't just theorize about it, try it.

>That turns out not to be the case. The answer to A is 1/3. Write a computer program to simulate it and then post it here. Seriously, don't just theorize about it, try it.

I misread A. I agree that it's 1/3 if you flip pairs until at least one is heads. I disagree that living in a world where pairs of tails don't exist and living in a world where pairs of tails do exist but are ignored are the same thing.

>Are you somehow suggesting that the probability distribution of the other options is changed by this?

Yes. I'm saying it's an unwitting double Monty Hall. The fact that he can't give you 3 cars changes the odds. Going (invisibly) from 4 possible worlds (including GG) to 3 changes the prob dist of the remaining 3, by which I mean 2. If Monty must guarantee you at least one boy, then there aren't really 3 remaining possibilities. The unguaranteed can be a boy, or it can be a girl. What is the third?

  >> The father says that at least one child is a boy.
  >> That eliminates the option of GG. Are you somehow
  >> suggesting that the probability distribution of
  >> the other options is changed by this?

  > Yes. ... Going (invisibly) from 4 possible worlds
  > (including GG) to 3 changes the prob dist of the
  > remaining 3, by which I mean 2.  If Monty must
  > guarantee you at least one boy, then there aren't
  > really 3 remaining possibilities. The unguaranteed
  > can be a boy, or it can be a girl.  What is the
  > third?

The standard reasoning is that there are two options, but they are not equally likely. Equally likely options are, labelling oder child first, BB, BG, GB. These were equally probable before we gained the extra information, and the standard interpretation is that they remain equally probable after.

In the Monty Hall case you choose a door, and it has probability 1/3 of being right. It remains 1/3 probable after he chooses a door. The other two doors together have probability 2/3 of concealing the car. It remains 2/3 after he has chosen and opened a door. The probabilities of the branches don't change in any of these cases.

Before you are told anything about the children we have four equally likely possibilities: BB, BG, GB, GG. After one is eliminated, why should the others still not be equi-probable?

  > A: So now let's take another experiment.
  >    I tell you I'm going to flip two coins
  >    until at least one of them is heads. I
  >    do that. Now I ask you to bet on whether
  >    or not there is a tail.  What do you
  >    think are fair odds?

And you say:

  > I misread A. I agree that it's 1/3 if you
  > flip pairs until at least one is heads.

OK. Now B:

  > B: Now I hunt among couples with two children
  >    until I find one that doesn't have two girls,
  >    and I ask: what are the odds they have two
  >    boys?

Do you agree that it's 1/3?

>Before you are told anything about the children we have four equally likely possibilities: BB, BG, GB, GG. After one is eliminated, why should the others still not be equi-probable?

Because in Monty Hall, the effect of changing probability is produced not just via elimination but via process of elimination - which is produced only because sufficient dependencies exist for it: a) the guaranteed 1:2 ratio, b) Monty must eliminate a goat, and c) the goat he eliminates can't be behind your door. These dependencies don't exist in the children problem. Unless Monty's choice occurs under the constraint of avoiding the door we chose, nothing changes. Monty: "You door has a goat behind it" -- so what? Now both of the other doors are 50/50. No similar conflict exists in the other problem. No conflict at all exists in the other problem. Elimination occurs, but it doesn't allow you to do anything further. "No GG" is just one rule, and if one elimination rule isn't sufficient for Monty Hall probabilities to change, why here?

And elimination of what? The elimination not of a possibility, but of a probability? Does that statement even have rational meaning? I'm being dead serious - we can give it mathematical meaning quite easily, but does it even have meaning to begin with?

Yes, I apparently misread that one too. But again, I disagree on the issue of whether the form of the proposed analogue matches the original problem.

  >> OK. Now B:

  >> > B: Now I hunt among couples with two children
  >> >    until I find one that doesn't have two girls,
  >> >    and I ask: what are the odds they have two
  >> >    boys?

  >> Do you agree that it's 1/3?

  > Yes, I apparently misread that one too.

OK, so you agree that this is 1/3. Now on to C:

  >> > C: I hunt among couples with two children until
  >> >    I find one that has at least one boy, and I
  >> >    ask: what are the odds they have two boys?

Do you agree that this is 1/3?

Let's assume for fun that you do agree with that. The original problem was:

  A: I met a man who said that he has two children,
     at least one of which is a boy.  What is the
     probability that both children are boys?

Let's compare that with this one:

  B: Among all the couples who have exactly two
     children, and such that at least one of their
     children is a boy, what is the probability
     that both are boys?

Reading this:

  > But again, I disagree on the issue of whether
  > the form of the proposed analogue matches the
  > original problem.

It seems like you're going to claim that these aren't the same problem.

I've made it very clear exactly what my model is for problem B. If you claim that problem A is different, then you need to provide me with a very clear model for your interpretation.

I think I'm ready to admit being wrong and move on with my life. Thanks for your ridiculous patience, Colin. I will be leaving my comments up as a monument to idiocy. Good god am I frustrated with myself right now.

Again, thanks. I apologize.

It's hard to say this without sounding condescending - please take it at face value ...

I'm pleased we got to the heart of it. I remember many times being in a similar situation as yours and I owe much to those too numerous to mention here who were patient with me. All I would say now is that in the future when you find yourself in disagreement, ask if you can learn something, don't just assume you are right. I found that easy, because I had more determination than talent, and was wrong often enough to be reminded constantly that there was still much to learn.

Tone matters on HN. That is why you're being downvoted.

> The fact that you think you're pointing anything out to anyone with that second paragraph is just priceless.

Posturing does not make your arguments more convicing.

> In GB and BG, he has only one way of telling you which is the boy. In BB, he has two. Thus, four possibilities a) GB, b) BG, c) BB and he gave you the first, d) BB and he gave you the second.

How is "ways of telling" you possibly relevant to the puzzle?

You're still apparently committing the same basic fallacy of ennumerating possibilities without assigning them probabilities and basing your argument on the wrong assumption that they must be equally probable.

> Your brain will of course refuse to process the above due to your earlier remark and the sweet, sweet motherfucking irony it would mean.

That describes exactly your behaviour in every single comment here.

> In GB and BG, he has only one way of telling you which is the boy. In BB, he has two. Thus, four possibilities a) GB, b) BG, c) BB and he gave you the first, d) BB and he gave you the second.

With respect to only the sentence above, why is the following table incorrect?

BG 1/3 GB 1/3 BB 1/6 BB 1/6

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