Probability Problem: Find P(Product=6 | Die1=2)

[Samurai44]

Homework Statement

Two fair dices are thrown and the product of the numbers on the dice is recorded, given that one die lands on 2, find the probability that the product on the dice is exactly 6

Homework Equations

P(A|B)=P(AnB)/P(B)

The Attempt at a Solution

I drew the table of the products

The intersection is 2/36
But how can I find the probability of landing on 2?

 

[Orodruin]

But how can I find the probability of landing on 2?

Looking at your outcome table: How many combination of dice rolls end up with at least one 2?

 

[Samurai44]

Looking at your outcome table: How many combination of dice rolls end up with at least one 2?

2 combinations? 2 and 1, and 1 and 2

 

[Orodruin]

What about 2 and 3 or 2 and 4? Those have at least one 2.

Edit: Let me put it this way. How many combinations have a 2 on Die 1? How many combinations have a 2 on Die 2?

 

[Samurai44]

What about 2 and 3 or 2 and 4? Those have at least one 2.

Edit: Let me put it this way. How many combinations have a 2 on Die 1? How many combinations have a 2 on Die 2?

So total of 10 combinations right?

 

[Orodruin]

So total of 10 combinations right?

This depends on how you interpret the problem formulation (which is not clear). If the formulation is that only one die should show a 2, then 10 is correct. If the formulation also allows both dice showing 2, then you get an additional possible outcome.

 

[Samurai44]

This depends on how you interpret the problem formulation (which is not clear). If the formulation is that only one die should show a 2, then 10 is correct. If the formulation also allows both dice showing 2, then you get an additional possible outcome.

How do I know if I count for 2 dices or just one? Or in other way why I can't count for only 1 die landing on 2

 

[Orodruin]

How do I know if I count for 2 dices or just one? Or in other way why I can't count for only 1 die landing on 2

This would depend on problem formulation. As you have formulated the problem above, I would interpret it as at least one die landing on 2. Other possibilities would include:

• Dice 1 landing on 2 and Dice 2 arbitrary.
• Only one die landing on 2 and the other on a different number.

I would say it is up to the problem maker to formulate the problem such that it is unambiguous what is intended. For the case of at least one die landing on 2, the possibilities are the 10 possibilities I suspect you counted (2x and x2 where x is not equal to 2) plus the possibility of both dice landing on 2.

 

[Samurai44]

This would depend on problem formulation. As you have formulated the problem above, I would interpret it as at least one die landing on 2. Other possibilities would include:

• Dice 1 landing on 2 and Dice 2 arbitrary.
• Only one die landing on 2 and the other on a different number.

I would say it is up to the problem maker to formulate the problem such that it is unambiguous what is intended. For the case of at least one die landing on 2, the possibilities are the 10 possibilities I suspect you counted (2x and x2 where x is not equal to 2) plus the possibility of both dice landing on 2.

That explains a lot thanks
The problem is there is a similar question where it says that there're two spinners(1 to 4), given that at least one spinner lands on 3, and they counted the outcome of one spinner only! :\

 

[haruspex]

I would say it is up to the problem maker to formulate the problem such that it is unambiguous what is intended. For the case of at least one die landing on 2, the possibilities are the 10 possibilities I suspect you counted (2x and x2 where x is not equal to 2) plus the possibility of both dice landing on 2.

I have a slightly different view. It says

given that one die lands on 2​

I agree that, almost surely, the poser intended to convey

given that at least one die lands on 2​

but that is not what it says. There are two other interpretations of the actual words that are at least as reasonable:
- that exactly one die landed on 2, producing 10 cases
- that a specific die landed on 2... maybe the red one, or the first rolled, or the first the poser looked at... producing only 6 cases.

 

[haruspex]

The problem is there is a similar question where it says that there're two spinners(1 to 4), given that at least one spinner lands on 3, and they counted the outcome of one spinner only! :\

I couldn't understand that description. Would you post that question exactly as worded please?

 

[vela]

Homework Statement

Two fair dices are thrown and the product of the numbers on the dice is recorded.

How do I know if I count for 2 dices or just one? Or in other way why I can't count for only 1 die landing on 2

Just thought you might want to know that dice is the plural of die, so it's "two dice" not "two dices". :)

 

[Orodruin]

but that is not what it says. There are two other interpretations of the actual words that are at least as reasonable:
- that exactly one die landed on 2, producing 10 cases
- that a specific die landed on 2... maybe the red one, or the first rolled, or the first the poser looked at... producing only 6 cases.

But those two cases are exactly the cases included in my post #8 as other possibilities. This is why I have been pointing out that the formulation is ambiguous.

 

[haruspex]

But those two cases are exactly the cases included in my post #8 as other possibilities. This is why I have been pointing out that the formulation is ambiguous.

I'm very sorry - for some reason I thought you'd only only covered the exactly one alternative. I must have read it too quickly.

 

[statdad]

" given that one die lands on 2, find the probability that the product on the dice is exactly 6"

would typically be taken to mean "at least one die" - if a single die were intended then it would read "exactly one"
For the OP: at least one means or or the other or possibly both: would it be possible for the sum to be 6 if both of the dice end up with 2 dots showing?

 

[haruspex]

" given that one die lands on 2, find the probability that the product on the dice is exactly 6"

would typically be taken to mean "at least one die" - if a single die were intended then it would read "exactly one"
For the OP: at least one means or or the other or possibly both: would it be possible for the sum to be 6 if both of the dice end up with 2 dots showing?

The only reason it typically means that in probability questions is that it makes it an interesting question. In everyday usage, that is about the least likely of the three possibilities.
If a friend were to say to you "I have two children. One is a girl", you would assume the other is a boy.
Alternatively, suppose he only told you that he had two children, and happened to have a girl with him at the time, you might report back to your partner that "Albert has two children. One of them is a girl". You might even say "at least one is a girl", yet the probability that the other is a girl remains 1/2, because you only know about one child.
So while the required answer here is almost surely 2/11, it is a badly worded question. The English language is not good at conveying logical relationships, so a conversational style is fraught with traps for the unwary problem setter.

 

[statdad]

"The only reason it typically means that in probability questions is that it makes it an interesting question."
That is the context - when we (I and others) write problems for tests, or text, the specific case (exactly 1) is the one that must be stated.

"So while the required answer here is almost surely 2/11, it is a badly worded question."
In everyday discussion you would be correct. In the general academic setting - no: it is the way things are. You are mixing two cultures. (You can make a very good case that academics in general, and textbook authors in particular, need to be more diligent in their writing and I would not disagree with you. That's not how the current situation is framed.)

 

[haruspex]

"The only reason it typically means that in probability questions is that it makes it an interesting question."
That is the context

The context is that it must be an interesting question? How many students are taught that, and how would they know when to apply it?
No, of course you mean that the context is probability questions set in courses, but if that's your defence then the syllabus had better include learning this arcane usage.

when we (I and others) write problems for tests, or text, the specific case (exactly 1) is the one that must be stated.
In everyday discussion you would be correct. In the general academic setting - no: it is the way things are.

I don't accept any of that as excusing the problem setter. It is a practice that has become accepted through custom, but it creates an unnecessary hurdle for every generation of students that has nothing to do with testing their understanding of probability.

An Australian HSC question about 15 years ago set the classic question about two balls, one of which was red. It framed it as Joe having the two balls behind his back, drawn from four red and four black say, but dropped one, and that was seen to be red. Clearly this framing doesn't work. Identifying a specific ball as red means the other ball has a probability 3/7 of being red, not the 3/11 that was marked correct. This customary usage is so pernicious it even catches out exam setters.

You are mixing two cultures.

No, that's my complaint: the problem setter mixes two cultures by adopting a conversational style.
For a mathematical context there's mathematical style, something like "the set of numbers shown by the dice includes 2". A conversational style is OK, but it is up to the question setter ensure that no ambiguity arises.

Interestingly, in writing my previous post I came to realize that even saying "at least one" does not really solve it. Consider the example I gave about meeting Albert and his little girl:

you might report back to your partner that "Albert has two children. At least one of them is a girl".​

Your partner cannot estimate the probability that the other is also a girl without knowing how you knew that - did you see both children or only one? Or did Albert say something that implied one was a girl, and if so what?

 

[Ray Vickson]

The context is that it must be an interesting question? How many students are taught that, and how would they know when to apply it?
No, of course you mean that the context is probability questions set in courses, but if that's your defence then the syllabus had better include learning this arcane usage.

I don't accept any of that as excusing the problem setter. It is a practice that has become accepted through custom, but it creates an unnecessary hurdle for every generation of students that has nothing to do with testing their understanding of probability.

An Australian HSC question about 15 years ago set the classic question about two balls, one of which was red. It framed it as Joe having the two balls behind his back, drawn from four red and four black say, but dropped one, and that was seen to be red. Clearly this framing doesn't work. Identifying a specific ball as red means the other ball has a probability 3/7 of being red, not the 3/11 that was marked correct. This customary usage is so pernicious it even catches out exam setters.

No, that's my complaint: the problem setter mixes two cultures by adopting a conversational style.
For a mathematical context there's mathematical style, something like "the set of numbers shown by the dice includes 2". A conversational style is OK, but it is up to the question setter ensure that no ambiguity arises.

Interestingly, in writing my previous post I came to realize that even saying "at least one" does not really solve it. Consider the example I gave about meeting Albert and his little girl:

you might report back to your partner that "Albert has two children. At least one of them is a girl".​

Your partner cannot estimate the probability that the other is also a girl without knowing how you knew that - did you see both children or only one? Or did Albert say something that implied one was a girl, and if so what?

The context is that it must be an interesting question? How many students are taught that, and how would they know when to apply it?
No, of course you mean that the context is probability questions set in courses, but if that's your defence then the syllabus had better include learning this arcane usage.

I don't accept any of that as excusing the problem setter. It is a practice that has become accepted through custom, but it creates an unnecessary hurdle for every generation of students that has nothing to do with testing their understanding of probability.

An Australian HSC question about 15 years ago set the classic question about two balls, one of which was red. It framed it as Joe having the two balls behind his back, drawn from four red and four black say, but dropped one, and that was seen to be red. Clearly this framing doesn't work. Identifying a specific ball as red means the other ball has a probability 3/7 of being red, not the 3/11 that was marked correct. This customary usage is so pernicious it even catches out exam setters.

No, that's my complaint: the problem setter mixes two cultures by adopting a conversational style.
For a mathematical context there's mathematical style, something like "the set of numbers shown by the dice includes 2". A conversational style is OK, but it is up to the question setter ensure that no ambiguity arises.

Interestingly, in writing my previous post I came to realize that even saying "at least one" does not really solve it. Consider the example I gave about meeting Albert and his little girl:

you might report back to your partner that "Albert has two children. At least one of them is a girl".​

Your partner cannot estimate the probability that the other is also a girl without knowing how you knew that - did you see both children or only one? Or did Albert say something that implied one was a girl, and if so what?

If you just report what you saw then one can make a good case that the probability the other child is a girl is 1/3 (because if you see a girl it eliminates the case BB, leaving three equally-likely cases GB, BG, GG--in order youngest-oldest). However, if you know, for example, that the child you saw is the oldest (or youngest) that changes the probability of the other being a girl to 1/2 (assuming the usual independence and equal birth rates, etc.)

 

[haruspex]

If you just report what you saw then one can make a good case that the probability the other child is a girl is 1/3 (because if you see a girl it eliminates the case BB, leaving three equally-likely cases GB, BG, GG--in order youngest-oldest). However, if you know, for example, that the child you saw is the oldest (or youngest) that changes the probability of the other being a girl to 1/2 (assuming the usual independence and equal birth rates, etc.)

Well, Ray, I did not expect you to fall into that trap. You've illustrated the issue nicely.
If you see one child and that child is a girl, you have eliminated two equally likely cases:

  case; child you see; other child
    1        g           g
    2        g           b
    3        b           g
    4        b           b

Seeing a girl eliminates 3 and 4.
Any arbitrary reason for selecting a particular child of whom to state the gender brings the probability back to 1/2.

 

[Orodruin]

Well, Ray, I did not expect you to fall into that trap. You've illustrated the issue nicely.
If you see one child and that child is a girl, you have eliminated two equally likely cases:

  case; child you see; other child
    1        g           g
    2        g           b
    3        b           g
    4        b           b

Seeing a girl eliminates 3 and 4.
Any arbitrary reason for selecting a particular child of whom to state the gender brings the probability back to 1/2.

Or, without the table, using Bayes' theorem:
P(2girls | see girl) = P( see girl | 2 girls) P(2 girls) / P(see girl) = 1*(1/4)/(1/2) = 1/2.

 

[Ray Vickson]

Well, Ray, I did not expect you to fall into that trap. You've illustrated the issue nicely.
If you see one child and that child is a girl, you have eliminated two equally likely cases:

  case; child you see; other child
    1        g           g
    2        g           b
    3        b           g
    4        b           b

Seeing a girl eliminates 3 and 4.
Any arbitrary reason for selecting a particular child of whom to state the gender brings the probability back to 1/2.

Look at it differently. The three relevant outcomes for what the children could be are BG, GB, GG (where these are in the order oldest-youngest). However, we need to look also at the "seeing" aspect, so this produces a 6-element sample space (the three where we see the youngest and the three where we see the oldest). Among these 6 sample points, two of them have two girls, giving P(2 girls) = 1/3, as I suggested before. Of course, as soon as you know you are seeing the oldest (or the youngest) that collapses the resulting sample space into two elements, giving you the 1/2 probability. And, as you say, we need to assume that the announcement of gender was random, which in reality might be difficult to guarantee. While I am not completely sure, I suspect that we could devise some types of mixed announcement strategies that produce a 2G probability between 1/3 and 1/2.

I don't think this problem is substantially different from the following coin-tossing example: "Your friend tossed a (fair) coin twice and announced that he got at least one head. What is the probability that he got two heads?" If you have no information about the 'position' of the announced head (first vs. second) the probability is 1/3, but if you are told the position it reverts to 1/2. If you have some possibly imperfect information about the position, perhaps a result between 1/3 and 1/2 can be obtained.

 

[Orodruin]

Look at it differently. The three relevant outcomes for what the children could be are BG, GB, GG (where these are in the order oldest-youngest). However, we need to look also at the "seeing" aspect, so this produces a 6-element sample space (the three where we see the youngest and the three where we see the oldest). Among these 6 sample points, two of them have two girls, giving P(2 girls) = 1/3, as I suggested before. Of course, as soon as you know you are seeing the oldest (or the youngest) that collapses the resulting sample space into two elements, giving you the 1/2 probability. And, as you say, we need to assume that the announcement of gender was random, which in reality might be difficult to guarantee. While I am not completely sure, I suspect that we could devise some types of mixed announcement strategies that produce a 2G probability between 1/3 and 1/2.

I don't think this problem is substantially different from the following coin-tossing example: "Your friend tossed a (fair) coin twice and announced that he got at least one head. What is the probability that he got two heads?" If you have no information about the 'position' of the announced head (first vs. second) the probability is 1/3, but if you are told the position it reverts to 1/2. If you have some possibly imperfect information about the position, perhaps a result between 1/3 and 1/2 can be obtained.

Even if you randomly chose which child to announce the gender of (without selection bias), the probability becomes 1/2. This becomes evident from the argument using Bayes theorem. In order to get 1/3, you need to introduce a selection bias towards girls so that P(see girl) = 3/4. One possibility being that your friend was told to bring a girl if he has one.

 

[Ray Vickson]

Even if you randomly chose which child to announce the gender of (without selection bias), the probability becomes 1/2. This becomes evident from the argument using Bayes theorem. In order to get 1/3, you need to introduce a selection bias towards girls so that P(see girl) = 3/4. One possibility being that your friend was told to bring a girl if he has one.

No selection bias was used in getting the 1/3---in fact, just the opposite: the selection was random, as I tried to explain rather carefully. Again, the Bayes results need a sample space to start with, and one must be careful not to force the result just by the way the data is presented. If my two out of six argument is wrong, I would like to know how and why.

 

[Orodruin]

No selection bias was used in getting the 1/3---in fact, just the opposite: the selection was random, as I tried to explain rather carefully. Again, the Bayes results need a sample space to start with, and one must be careful not to force the result just by the way the data is presented. If my two out of six argument is wrong, I would like to know how and why.

Assuming we quote the kids in age order, you have four outcomes, not six.
See older kid: GG, GB (BG is not allowed as you would see a boy)
See younger kid: GG, BG (GB is not allowed, you would see a boy)

 

[Ray Vickson]

Assuming we quote the kids in age order, you have four outcomes, not six.
See older kid: GG, GB (BG is not allowed as you would see a boy)
See younger kid: GG, BG (GB is not allowed, you would see a boy)

The sample space is all the things that could have been seen, but some were not. IF the gender announcement is truly random it is just a matter chance that we did not see a boy if one was actually part of the family.

I have seen this problem and the associated issues discussed in some introductory probability textbooks, but I sold off a lot of my library and cannot find the sources anymore---and I do not remember exactly where I saw it.

 

[Orodruin]

The sample space is all the things that could have been seen, but some were not. IF the gender announcement is truly random it is just a matter chance that we did not see a boy if one was actually part of the family.

Given that you saw a girl (which is what the probability is conditioned to), you could not have seen the oldest if the children were BG and therefore it is not a possible outcome. This leaves GG and GB for the case where you saw the oldest child.

As for the application of Bayes' theorem, the starting point is when I only know the guy has 2 kids and we assume that their genders are independent and equally probable. The probability that I will see a girl is then:
P(see girl) = P(see girl | see old kid) P(see old kid) + P(see girl | see young kid) P(see young kid) = P(old kid girl) P(see old kid) + P(young kid girl) P(see young kid)
= (1/2)(P(see old kid) + P(see young kid)) = 1/2.
Note that we do not even need to know the probabilities for seeing each kid. The assumption is that the selection of old/young is independent of the kids being girls or boys.

It should be pretty obvious that P(see girl | GG) = 1 and that P(GG) = 1/4.

See also http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
In particular, this is related to the second question and the interpretation relevant for this case is the second. We are not choosing a family, we are choosing a child of a family that we know has two kids. If our friend has two kids and is out walking with a random kid, which turns out to be a girl, the probability of the other kid being a girl is still 1/2. It would be different if we were in a conversation with our friend and asked "Do you have any daughters?" and he answered "Yes".

 

[statdad]

"No, that's my complaint: the problem setter mixes two cultures by adopting a conversational style.
For a mathematical context there's mathematical style, something like "the set of numbers shown by the dice includes 2". A conversational style is OK, but it is up to the question setter ensure that no ambiguity arises."
You are making a complaint where none should exist. Part of the intent of introductory material (which this problem clearly is) is to avoid becoming bogged down in mindless pedantry. The discussions about wordings of problems, in this manner, is done in texts and in class.
"Interestingly, in writing my previous post I came to realize that even saying "at least one" does not really solve it. Consider the example I gave about meeting Albert and his little girl:
you might report back to your partner that "Albert has two children. At least one of them is a girl".
Your partner cannot estimate the probability that the other is also a girl without knowing how you knew that - did you see both children or only one? Or did Albert say something that implied one was a girl, and if so what?"
Certainly he can. Every probability is based on assumptions, using information in the posing of the problem. So, if there are at least two children there are four possibilities: if at least one is a girl you know BB is not an option: For the other to be a girl the children have to be GG so, as Ray correctly points out, the probability would be 1/3. Adding the other items into the calculation is introducing items into the problem that were not included in the problem.

If you wish to make other assumptions (the "which was seen" stuff) then you could make this arbitrarily difficult - but again, you are missing the point: introductory material is introductory material because the point is to stress the basic ideas: complicating factors are introduced later. The complicating ideas discussed above seem to be presented simply because they can be.

 

[Orodruin]

you might report back to your partner that "Albert has two children. At least one of them is a girl".
Your partner cannot estimate the probability that the other is also a girl without knowing how you knew that - did you see both children or only one? Or did Albert say something that implied one was a girl, and if so what?"

If you tell your partner that at least one of them is a little girl you are misrepresenting the information that you have or simply not giving all of the information you have. Based on this you should not be surprised that this leads to different probabilities (in fact this is the entire premise of misrepresentation when playing poker for example, you try to feed your opponents misinformation or incomplete information so that they will make decisions based on the resulting probabilities). If you tell your partner that you have met one of them, which is a girl, then you are giving a completely accurate account of your available information, and your partner will also arrive to the conclusion of 1/2 probability of Albert having two girls (i.e., do not do this when playing poker).

Please also see the Boy or Girl paradox link in my previous post. It is well known that the formulation is ambiguous and Gardner even acknowledged this. There is also the possibility of Albert preferring to walk with girls, if you have this information, the outcomes BG and GB would be equally probable to GG and you would get back to 1/3. If there is no such selection bias, GG is twice as probable as BG or GB to result in Albert walking with a girl. Thus GG and (BG union GB) are equally probable to have resulted in Albert walking with a girl and thus the probability of two girls is 1/2.

I believe the bottom line is this one:

The English language is not good at conveying logical relationships, so a conversational style is fraught with traps for the unwary problem setter.

While saying "at least one is a girl" would be generally acceptable English language (and true), it is not conveying the full information you have if you make the mathematical interpretation of "at least one".

On a side note I suggest using the Reply feature for quoting posts. It will be much more easy to understand who and what you are quoting and when you are quoting. (If you mark part of the text, you will get a Reply button right next to it, clicking it includes only that part of the text in the quote.)

Edit: See also http://dx.doi.org/10.1016/0010-0277(82)90021-X

 

[statdad]

"If you tell your partner that at least one of them is a little girl you are misrepresenting the information that you have ..."
That is blatantly false. If you see a girl you know that there at least one - what else could it be, none? What you don't have is evidence there is exactly one.

 

[haruspex]

If you just report what you saw then one can make a good case that the probability the other child is a girl is 1/3 (because if you see a girl it eliminates the case BB, leaving three equally-likely cases GB, BG, GG

To be clear, this is the formulation I am challenging: if you see Albert with a daughter, and know by some means there is another child, that other child is equally likely boy or girl, not 2:1 a boy.

No selection bias was used in getting the 1/3

Yes there was. If you had seen a boy and the other child was a girl, you would not have reported that at least one was a girl. In the standard (valid) formulation you will report that at least one is a girl whenever that is true. In the above formulation you will not achieve that.

So, if there are at least two children there are four possibilities: if at least one is a girl you know BB is not an option

Yes, but it's not that simple. In order to deduce the 1/3 answer you have to make the assumption that if either is a girl, you would have the information that at least one is a girl. This is where Ray's argument above breaks down. Having seen a girl, he knows at least one is a girl, but there is a case where at least one is a girl yet he does not have that information.

you are missing the point: introductory material is introductory material because the point is to stress the basic ideas: complicating factors are introduced later.

That seems to me a total misrepresentation of the matter.
There is much evidence that this problem is rather subtle. It was not properly understood by some HSC examiners, and appears to have fooled the eminent Ray Vickson. (You have not posted a view on Ray's argument, so I cannot tell whether it fooled you too. )
To make it 'introductory level', it will be essential to minimise confusion. The advanced level could then look into the epistemological niceties.

I view your argument as "this problem is always poorly worded in tests, so we have to teach students to interpret it in a certain way". Well, that's nothing to do with teaching mathematics; that's teaching interpretation of sacred texts.

 

[haruspex]

"If you tell your partner that at least one of them is a little girl you are misrepresenting the information that you have ..."
That is blatantly false. If you see a girl you know that there at least one - what else could it be, none? What you don't have is evidence there is exactly one.

First, perhaps Orodruin should have written that you are understating the information that you have, i.e. you have thrown some information away. On that he is correct, and is likely to be the case whenever someone fails to state how they know the information.
Secondly, your argument supports my case exactly! If you believe that relaying the information as "at least one of them is a little girl" correctly conveys the information that you have (that you saw a girl and have no idea of the gender of the other child) then you are forced to the view that the statement "at least one of them is a little girl" does not support a 2:1 likelihood that the other is a boy.