[bonanova]

A magician blindfolded seven subjects, then for each of them he rolled a

pair of fair dice and asked them the probability of their dice totaling 7.

He said to the first, here's a hint: truthfully, at least one of your dice shows a 6.

The subject counted 11 cases of at least one 6, two of which, 1-6 and 6-1, total 7.

So he answered, my chances of having 7 are 2/11. Very good, said the magician.

He then said to the second, I've looked at your dice, and at least one of them is a 5.

This subject counted 11 cases of at least one 5, of which 2-5 and 5-2 made 7.

So he answered 2/11. Very good, said the magician.

He told the third subject, I see at least one 4 on your dice.

That subject also found 11 cases of 4, of which 3-4 and 4-3 made 7.

So he answered 2/11. Very good, said the magician.

The next subject was told at least one of his dice was a 3.

Like the others, he found 11 cases, and of them only 4-3 and 3-4 were favorable.

So he answered, my chances of having 7 are 2/11. Very good said the magician.

The next two were told their dice showed at least one 2 and one 1, respectively.

They found 5-2, 2-5 for one, and 6-1, 1-6 for the other, among 11 cases gave 7.

They both answered their chances of 7 were 2/11. Very good said the magician.

The seventh subject had been listening to all of this. And before the magician

could speak, he said, I don't need a hint. I know that you're going to tell me

some number appears on at least one of my dice. And you've already confirmed

what the right answer is in each case. So whatever you were going to say,

I know most certainly what the odds probability of my dice totaling 7 are is.

My answer is 2/11.

But we know his odds are probability is 1/6, right?

Edited December 29, 2008 by bonanova
Scraff's observation

[Guest]

Oddly enough, as I reread through the post, I came across Rossbeemer's initial answer, which is correct. I don't know why he recanted on this since it is logically infallible. The only thing I'd do is remove the line about A and B BOTH being 6, because he already covered that possibility in his previous 2 scenarios and it results in a 0 anyhow. Regardless, that was a slick, simple, complete, CORRECT proof. Good job Rossbeemer. Now all you have to do is believe yourself instead of being convinced by someone else.

Thank you, I appreciate that.

There's no paradox here. If 1/3 of all of the couples have two girls, and we select any of the couples at random, guess what the probability is that the couple has two girls? That's right- 1/3! Claiming it's 1/2 is nonsensical!

There does seem to be a paradox here, and I think the paradox is in how you look at the problem, or as BeastMaster would say, how far you break down the subset.

These have both been said before, obviously, but I'm trying to simplify it into one quick post:

Suppose you consider the groups of families that would have at least one girl. The groups would be GG, BG, or GB. That considered, the probability of having GG is 1/3.

But there's another way to look at it!

Suppose the family has at least one girl. The probability of the first child being a girl is 1/2, in which case the probability of the second child being a girl is 1/2. So (1/2)(1/2)=1/4

Now the probability of the second child being a girl is 1/2, in which case the probability of the first child being a girl is 1/2. So (1/2)(1/2)=1/4.

And then 1/4 + 1/4 = 1/2.

So it seems that your probability depends on how far you break down the subset, or whether you specifically consider the individual cases or not.

Hopefully I explained that clearly and sensibly, if I didn't I apologize.

Something is confusing me though...I'll think on it and post again later.

[Guest]

Bolding mine

There does seem to be a paradox here, and I think the paradox is in how you look at the problem, or as BeastMaster would say, how far you break down the subset.

These have both been said before, obviously, but I'm trying to simplify it into one quick post:

Suppose you consider the groups of families that would have at least one girl. The groups would be GG, BG, or GB. That considered, the probability of having GG is 1/3.

But there's another way to look at it!

Suppose the family has at least one girl. The probability of the first child being a girl is 1/2, in which case the probability of the second child being a girl is 1/2. So (1/2)(1/2)=1/4

Now the probability of the second child being a girl is 1/2, in which case the probability of the first child being a girl is 1/2. So (1/2)(1/2)=1/4.

And then 1/4 + 1/4 = 1/2.

There is no "the family".

Look again at what I wrote and how he answered. I asked:

Out of all couples with exactly two children that have at least one girl, how many have two girls?

He answered:

1/2

He then later claimed the answer is 1/3 and that he never answered 1/2.

Then he claimed the following:

I haven't disagreed with this statement. Of all couples with exaclty two children that have at least one girl, 1/3 WILL have 2 girls. However, if you want to know the PROBABILITY of a any particular family in that population having 2 girls, you MUST select one (any one) and then continue along with the rest of my proof, which leads you to the answer of 1/2. This is the incredible paradox of statistics.

See what's wrong with this? He said if you pick one family out of the 75% we have of the total (25%GG, 25%GB, 25%BG), the probability of having a couple with two girls is 1/2. This is incorrect.

[Prof. Templeton]

1/2

Here's the equation for conditional probability (used by ALL statistics books)

P(A|B)=P(A "intersect" B)/P(B)

Where P(A|B) means, the probability of A given B

(here's the link for the source if you'd like it - http://mathworld.wolfram.com/ConditionalProbability.html line 1)

This would read, for the child problem as:

Probability of A(child is a girl) given B(child is a girl) = probability of A(child is a girl) AND B(child is a girl) divided by the probability of B(child is a girl)

When P(A) and P(B) are independent events (which is true of die rolling and childbirth-the die rolls and births are separate) P(A "intersect" B)=P(A)*P(B)

(here's the link for the source if you'd like it - http://mathworld.wolfram.com/Probability.html line 18)

So the equation becomes

P(A|B) = {P(A)P(B)}/P(B) = P(A)*{P(B)/P(B)} = P(A)*1 = P(A)

In the case of the child problem P(A) = 1/2 (a single child being a certain gender)

Given this stated question

Out of all couples with exactly two children that have at least one girl, how many have two girls?

I don't believe you have your equation worded correctly. It should be P(A|B) = probability of A (both children are girls) given probability of B (at least one of them is a girl), which when the correct numbers are plugged in becomes

P(A|B) = P(A ∩ B)/ P(B) = ¼ / ¾ = 1/3

Same equation, but the set-up is key. I'm sure you would agree that there a 1/4 chance that both children are girls (GG out of GG,BG,GB,BB) and a 3/4 chance that there is at least one girl (GG,BG,GB). So the correct and only answer must be 1/3. QED

I hope my html character entities show up properly. Prime would be proud.

[bonanova]

With these I believe everyone agrees:

1. Of all the couples with two children, at least one of which is a girl,
   equal numbers have, naming the older child first,
   {girl boy}, {boy girl}, {girl girl}.
   
   
   
2. Of all the people who have rolled two fair dice, at least one of which is a six,
   equal numbers obtained, naming the dice in the order they came to rest say,
   {1 6}, {2 6}, {3 6}, {4 6}, {5 6}, {6 6}, {6 5}, {6 4}, {6 3}, {6 2}, {6 1}.
   

If not, identify an unequal class.

If so, treat them as axioms.

Note the statements deal with populations, and they do not mention

how we know [at least one girl] or [at least one six]

[Guest]

What opened my eyes was an attempt to sketch a simulation program and "remove doubt."

I love simulations, because you have to program them [i.e. specify your premises and argument precisely everywhere]

and because they do distinguish between sufficiently different outcomes.

Easy. Repeat this a million times:

1. Roll the dice.
   
2. Say to your subject, one of your dice shows a ...........long pause for thought...........
   
3. Ouch! the dice came up 3-5. What do I say - 3, or 5?
   
4. Eyes begin to squint from the bright light that all of a sudden was shining.
   
5. End of simulation algorithm.
   

See why I love simulations?

Finally, finally:

OP is incomplete - it doesn't give guidance on which die to identify to our subject.

If we go one way it creates p[7] = 2/11.

If we go another way it creates p[7] = 1/6.

What do we do?

Well, we could just say the puzzle is flawed, and walk away disgusted.

Better, take the usual and most reasonable choice:

Assume that for areas where OP is silent, no bias is introduced.

Under the assumption of random choice between A and B, p[7] = 1/6 in all cases.

*head explodes*

It seems my first inclination was correct after all, that the probability should be 1/6 in all cases. There was something tiny nagging in the back of my mind about how the magician somehow magically rolled the dice in such a way that "at least one of them is a 1" for the first subject, etc. without introducing some sort of influence (looking for specific numbers on the dice, rerolling them if the number he was looking for didn't show up, or lying to the subject about what one of his dice shows) and removing the pure randomness of it all. However, the 11 possible combinations in each case, knowing the value of at least one of the dice, does paint quite the vivid picture of a scenario that is actually a red herring.

I took just one statistics course in college, but I don't remember problems like this coming up in class. Good thing too, for I would have surely failed. Good puzzle, bonanova, although it left questions unanswered that I didn't even realize needed to be asked: How did the magician determine which die to reveal to each subject, and how lucky must he really be to roll them in such a way that the first subject had at least a 1 showing, the second a 2, and so on if each roll was completely random and unbiased? Anyone want to calculate THAT probability?

1771561/2176782336 or (11/36)⁶, which is approximately 0.0814%. He must truly be a magician to have been successful with those kinds of chances.

I really hope I did that right; with all the second guessing I've done already, I may have overlooked something else. My head exploding didn't help matters much either.

[Guest]

Out of all couples with exactly two children that have at least one girl, how many have two girls?

1/2

Here's the equation for conditional probability (used by ALL statistics books)

P(A|B)=P(A "intersect" B)/P(B)

Where P(A|B) means, the probability of A given B

(here's the link for the source if you'd like it - http://mathworld.wolfram.com/ConditionalProbability.html line 1)

This would read, for the child problem as:

Probability of A(child is a girl) given B(child is a girl) = probability of A(child is a girl) AND B(child is a girl) divided by the probability of B(child is a girl)

When P(A) and P(B) are independent events (which is true of die rolling and childbirth-the die rolls and births are separate) P(A "intersect" B)=P(A)*P(B)

(here's the link for the source if you'd like it - http://mathworld.wolfram.com/Probability.html line 18)

So the equation becomes

P(A|B) = {P(A)P(B)}/P(B) = P(A)*{P(B)/P(B)} = P(A)*1 = P(A)

In the case of the child problem P(A) = 1/2 (a single child being a certain gender)

Ho hum... just while we're establishing our credentials, I'm also a maths graduate, having spent three years at Oxford University. (Just out of interest what's the signifigance of the time you mentioned after each course? Presumably its not the amount of time spent studying it... cos that would mean a degree after 45 hours of learning) Is it the length of the exam(s)?)

So...

we should be calculating:

P(there being two girls GIVEN there is at least one girl) = P(there are two girls AND there is at least one girl)/P(there is at least one girl)

= 1/4 / 3/4 *

= 1/3

*P(there are two girls AND there is at least one girl) = P(there are two girls) = 1/4

P(there is at least one girl) = P(GB)+P(BG)+P(GG)=1/4 + 1/4 + 1/4 = 3/4

I think that covers everything on the Girl Boy problem. I believe your error crops in because you are talking about specific children (children A and B) rather than children in general. You've fallen into the common trap of mistating the question in your calculations.

editadded scraffs original quote

Edited January 2, 2009 by armcie

[Guest]

Given this stated question

I don't believe you have your equation worded correctly. It should be P(A|B) = probability of A (both children are girls) given probability of B (at least one of them is a girl), which when the correct numbers are plugged in becomes

P(A|B) = P(A ∩ B)/ P(B) = ¼ / ¾ = 1/3

Same equation, but the set-up is key. I'm sure you would agree that there a 1/4 chance that both children are girls (GG out of GG,BG,GB,BB) and a 3/4 chance that there is at least one girl (GG,BG,GB). So the correct and only answer must be 1/3. QED

I hope my html character entities show up properly. Prime would be proud.

Well said Professor Templeton. I think that finally cleared it up.

Incidentally (if anyone is interested, which probably nobody is ) I found the error in my own analysis:

Suppose the family has at least one girl. The probability of the first child being a girl is 1/2, in which case the probability of the second child being a girl is 1/2. So (1/2)(1/2)=1/4

Now the probability of the second child being a girl is 1/2, in which case the probability of the first child being a girl is 1/2. So (1/2)(1/2)=1/4.

And then 1/4 + 1/4 = 1/2.

My entire problem was that I was not considering all the cases. I was thinking "the probability of the first child being a girl is 1/2, it could be a boy or a girl." But really, because of the choices GB, GG, BG, if the first child is a girl (2/3 of the time), the second child will be a girl 1/2 of the time. (2/3)(1/2)=1/3

Same if the second child is a girl.

So as Prof. Templeton, Scraff, and others have already said, the probability is 1/3. I was just glad that I found my own error.

[Guest]

Alright, I concede.

I haven't read any posts since my last because I was busy for new years, but I had emailed my old college professor and he helped me see my error.

First, my apology, then the proof of my error.

I am sorry for being arrogant. I have eaten the biggest piece of humble pie in the world just now. I am obviously NOT better than everyone at brainden at this stuff and the moderators and Scraff have been correct all along. In fact, I should probably return my math degree because of my idiocy.

However, here's where I went wrong. I hope that this will help any others who may still be doubting (like I was).

I was looking at the events as

A: You have a boy or girl

B: You have a boy or girl

IF this were the case, then these would be independent events and all my previous proofs would be correct. HOWEVER, the real probability events we are looking for are

A: You have at least one girl

B: You have two girls

Now, in this case, we already know P(A) = 3/4 because there are 3 scenarios (out of 4) where this is true. These are BG, GB, GG.

Also, we know P(B) = 1/4 because there is only 1 scenario (again, out of the original 4) where this is true. This is GG.

So the conditional probability equation is...

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

BUT, this can't be simplified by replacing P(B interstect A) with P(B)*P(A) because you can only do that when P(B) = P(A) and we just established that that isn't true. So, what is P(B interstect A)? Well, B intersect A is the condition when BOTH B and A are true. That means, when B: "both are girls" and A: "you have at least one girl" are both true. This obviously happens only when both are girls, so P(B intersect A) = P(B) = 1/4

So now, P(B|A) which read "the probability of B given A", or in this case "the probability of both children being girls given at least one child is a girl" can be calculated as follows

P(B|A) = (1/4)/(3/4) = 1/3

This is just as Scraff and others have stated all along.

This same thing would extend to the die problem (and coins too).

If you know you have a 3 and want to know the probability of totaling 7, you calculate

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

where

A: you have at least one 3

B: you have a total of 7

So you can read the equation as

P(totaling 7|you have at least one 3)=P(Totaling a 7 with a 3)/P(you have at least one 3)

And this calculates to..

P(B|A) = (1/18)/(11/36) = 2/11

Like I said, I was totally confusing the probability events and screwing up the whole thing. This is the kind of proof I was looking for and since I didn't get it, I felt validated in my solution. I was obviously dead wrong.

Again, I concede and feel like an absolute idiot (as I'm sure I should). All of Scraff's previous proofs are valid and true, even if they didn't look like I wanted them to.

Scraff, feel free to commence the beating.

[Guest]

Alright, I concede.

I haven't read any posts since my last because I was busy for new years, but I had emailed my old college professor and he helped me see my error.

I just saw that Prof. Templeton explained just this same thing.

Like I said, I was way way way wrong.

This is when arrogance really sucks, when you're wrong. Being arrogant is fun when you're right, but it sure tastes like crap when you have to eat your words.

[Guest]

Scraff, feel free to commence the beating.

No way, BeastMaster. It takes a big man to do that. Just glad it's over.

[Guest]

No way, BeastMaster. It takes a big man to do that. Just glad it's over.

Well, I appreciate that. I probably should've been a bigger man and listened to reason earlier, but oh well. I'm with you, I'm glad it's over, even if it did end with me being wrong.

Now that I've seen the error of my way, I'd like to tackle the actual question bonanova asked, but I don't have time for that right now. I'll do it in a later post.

[bonanova]

Well, I appreciate that. I probably should've been a bigger man and listened to reason earlier, but oh well. I'm with you, I'm glad it's over, even if it did end with me being wrong.

Now that I've seen the error of my way, I'd like to tackle the actual question bonanova asked, but I don't have time for that right now. I'll do it in a later post.

The moderator wipes a real tear from his eye as the denizens shake hands.

Observation: formulas are conclusions without context - easy to use, impossible to refute.

It's the context that is appropriate or refutable.

Use with skill and care.

Happy New Year guys.

[Guest]

I would like to thank all of you who made this such an interesting thread. I was entirely out of my depth, but the passion of the arguments on both sides was admirable and highly entertaining. And I ended up learning some things, since you stretched it out so long. I ended up having to figure some of it out on my own and come up with an opinion (which changed twice as I learned more).

You guys ran me out of popcorn!

Thanks BN for yet another entertaining topic.

[bonanova]

Let's see if we can tidy up here.

Make a table of outcomes [dice total] when at least one die is 1.

There is one outcome [1-1] that makes 2; two outcomes [1-2 and 2-1] that make 3; and two more outcomes each for totals of 4, 5, and 6 -- a total of eleven.

Complete the exercise for die values of 2, 3, 4, 5, 6.

Note there are eleven outcomes for each - a total of 66 in all; not 36, because we count occurrences of two numbers twice; doubles are counted only once.

At this point, we are not choosing a die value to talk about. Just sorting out the occurrences for each value.

Note some entries are blank. There is no way we can have a total of 2, for example, if one of the dice is 2 [or higher].

We can answer some questions immediately from this table.

Look at the circled value of 2 in the "3" column and the "5" row.

There are 11 outcomes in the "3" column. The circled 2 tells us when at least one of the dice is 2, the probability of the total of the two dice being 5 is 2/11 -- 2 occurrences out of 11.

There are 8 outcomes in the "5" row. Now the circled 2 tells us when the total of the two dice is 5, the probability of at least one die being a 3 is 1/4 -- 2 occurrences out of 8.

Finally, look at the gray circled 2 in the "6" column. That's our old friend telling us that if one or both dice is 6, the probability of the total being 7 is 2/11.

It will be instructive to notice how the probabilities for the red circle [3 and 5] and the gray circle [6 and 7] change throughout this post.

Now let's choose one of the dice by some means.

To help matters, let's color them. Let's dye die A blue and dye die B green. Now we can distinguish them.

Let Method 1 for choosing one of the dice be to choose die A - the blue one.

We suspect the results will be the same for Method 2 - choosing die B - the green one.

If they are the same, then Method 3 - a random choice say the average of A and B - will be the same, as well.

Filling in the columns, as in the first case we ask how many ways can the first die be 1 and get 2 total - just one [1-1].

How many ways can we have a total of 3 if the first die [or 2nd die] is 1? Now the answer is just one: [1-2] and [2-1], respectively.

And for a random selection of these cases, just one of them will be picked. Similarly for all the other entries.

By choosing just one number from each roll, we count each of the 36 outcomes only once.

Using one of the three methods mentioned, one of the dice is seen to be say, a three.

Notice the probability of a 3 if the total is 5 is still 1/4, but the probability of rolling 5 given at least one 3 is now 1/6.

Note also the probability of rolling 7 given at least one 6 is also 1/6.

It's tempting to think that whenever we pick just one die, all the probabilities are such as they are here.

This turns out not to be the case.

Suppose we adopt Method 4 - picking the smaller of the two numbers. Or Method 5 - choosing the higher.

Still choosing just one of the dice, the probabilities are changed.

[edit]

Note that since these are exclusive and exhaustive cases, they average to the random Method [3] - Add them up and divide by two.

[/edit]

In particular, note that if we choose the higher ranking die, the probability of rolling 7 of at least one of the dice shows 6, is our old nemesis 2/11.

In the OP, if the magician was using Method 5, and if the first subject suspected as much [how, we could only guess], then his answer of 2/11 is the correct one.

But that doesn't help much. The table shows that Method 5 tells Subject #2 ["one of your dice is 5"] to say 2/9, not 2/11.

Well then.

We ask, is there a consistent algorithm the magician might have used so that all six of the 2/11 answers were correct?

It turns out there is, although the Subjects would have to have been psychic to find it.

It's Method 6: Predetermine a number, and call it out if it's showing on one of the dice.

And if not, then choose randomly from the two numbers that are showing.

The final table shows Method 6 when the predetermined number is 3:

If a 3 is showing, and if the magician predetermined he would call it if so, then the probability of rolling a 7 is in fact 2/11.

You can read that from the table, where the 7 and 3 intersect, even though that 2 is not circled.

What is the likelihood of this happening?

1. The magician would have to have predetermined 6, 5, 4, 3, 2, 1 in that order.
   
2. The subjects would have to have guessed that was his strategy.
   
3. Each of the predetermined numbers would have to have shown in the dice that were rolled.
   

No one would make money betting all of that would happen.

But let's suppose it did: to accept the OP, where the magician complimented each for divining a correct response, it must have happened. The relevant probabilities are not fixed. They're neither 1/6 in all cases nor 2/11 in all cases. They depend on the choice Method; so we must assume a Method that makes OP consistent. And, unlikely as it is, we've found it.

So finally we're left with accepting or disproving the argument by Subject #7 that ... for reasons noted in OP ... his unconstrained probability of rolling 7 must be 2/11.

Well we've seen the [isolated and improbable] conditions for a 2/11 result, and Subject 7 didn't mention any of them, so we don't believe the argument. But to disprove it, we have to point out a fallacy. And there's one that is fairly obvious. For the first six subjects' answers to have been correct, as we've just seen, the choosing method [i.e. the value of the predetermined number] would have to be different in each case. Thus the six outcomes are not exclusive and exhaustive cases: they are "apples and oranges." Therefore they can't be combined, as Subject 7 reasoned they could.

Thus the paradox is explained.

Edited January 6, 2009 by bonanova
Added a comment [in dark red] after Mehods 4 and 5

[Guest]

Tis truely the season of miracles: someone on Brainden admits they were mistaken... especially miraculous after such a heated debate. Much kudos all round.

[Guest]

I've just read all the posts on this and on the two kids (boy/girl) problem and want to make sure I have properly understood it all. Please bear with me on this, as this is quite a long post, but I'd really appreciate people's thoughts, especially those who have so passionately argued this case.

I believe from Bonanova's last post that the probabilities for each of people 1 to 6 is NOT definitively 2/11, but is actually indeterminate. The fact that the magician tells them that they are correct is actually the only useful information we can take from the puzzle which leads us to Bonanova's final conclusion that the magician was using a very odd (and lucky) way of choosing what number they had at least one of and proving that person 7 was wrong.

I'm hoping that I'm right in what I'm saying, so not going to give my reasoning, but would be grateful if someone could confirm it! However, I will explain my reasoning in relation to the two kids problem (which I believe uses the exact same logic so should be able to be extended to this problem). I'm also going to try to explain using reason, as I'm not happy with a math formula that proves one thing but common sense that says something else!

So, with the original kids problem (posted here) I again believe that the answer is indeterminate based on the wording of the question.

Now, before anyone starts shouting at me, I agree that the answer to the question "Out of all couples with exactly two children that have at least one girl, how many have two girls?" is 1/3. However, that's not what the original kids problem states. Instead it states "They have two kids, one of them is a girl, what is the probability that the other kid is also a girl."

As was stated in that thread several times, we do not have any information as to why we were told one of them is a girl. However, in order to arrive at the answer of 1/3, we have to make the assumption that if there is at least one girl then we will be told about it.

I think this is best explained by looking at the reduced sample sets in a similar way to how Bonanova has done it in this thread. People generally take "one of them is a girl" to mean that the reduced set is BG, GB or GG with equal probability. However, this is only the case if we know that we are always told about a girl if one exists.

The reason I need to get this clear in my own head is the paradox that this produces by changing the question slightly (the same kind of paradox that this thread has posed), ie.

A couple has two kids, at least one of them is a girl, what is the probability they have a boy and a girl? A. 2/3

A couple has two kids, at least one of them is a boy, what is the probability they have a boy and a girl? A. 2/3

A couple has two kids, at least one of them is a boy or at least one of them is a girl, what is the probability they have a boy and a girl? A. 1/2

Changing boys and girls to heads and tails, this would mean that if I were to toss two coins whilst blindfolded and guess at the chance of a head and a tail (in any order) I would have 1/2 chance of getting it right. However, if I asked a friend to tell me that at least one of them was either a head or a tail (it wouldn't matter which as long as the person was being honest) then my odds of getting it right would have increased to 2/3. This obviously makes absolutely no sense!

Looking at the 100 coin experiment that I think Martini originally suggested and has been referred to numerous times, I think there is a fallacy in this experiment. If we are told "at least one is a tail" then sure we can discount all the heads/heads, but that doesn't mean we can include all other 75 instances with the same probability. Let's look at two cases:

1) We are told "at least one is a tail AND I can confirm that I would always tell you that provided I see at least one tail"

In this case we can discount the H/H only and we are left with a probability of head and tail (in any order) of 2/3.

2) We are told "at least one is a tail AND if faced with a situation where I could tell you either at least one is a head or at least one is a tail I will pick one at random"

In this case we can discount the H/H but we must also discount half of the H/T and half of the T/H (taking "random" to mean a probability of 1/2 for each case). This now leaves us with a probability of head and tail (in any order) of 1/2.

Now, it seems to be widely agreed that, in the absence of specific information, the "unwritten rule" is that we must take the random event, i.e. case 2 above. This would mean that the kids problem's technically correct answer would be 1/2 although, if interpreted as in case 1 above, would be 1/3. (It thus seems odd that people have interpreted the original kids problem as "Out of all couples with exactly two children that have at least one girl, how many have two girls?" as this means that people are going with case 1 which is against the "unwritten rule", but that's another story)

This also solves my tossing coins paradox. If my friend randomly chooses which to tell me, then my odds haven't increased at all as we are simply following case 2 above. However, if he always tells me if he sees a head for example (and says nothing if he doesn't) then I know that if he says something my chances have increased but if he says nothing I have no chance at all. Putting this together, my overall chance of having a head/tail combo in any order are:

P(him saying something)xP(being head/tail in any order) + P(him not saying something)xP(being head/tail in any order) = 3/4x2/3 + 1/4x0 = 1/2.

So, in summary, at the risk of opening a whole new can of worms, I think technically (using the "unwritten rule") the original kids problem should be answered as a 1/2 or alternatively must be left as indeterminate. Using this same logic I believe that the 2/11 probability posted in this thread is also incorrect as the problem is also indeterminate and it is only the magician's answers that apply any certaintly (as stated above).

Put another way, 2/11 would be correct if I were to say "Out of all the people who have rolled at least one 6, what is the probability they have rolled a total of 7" or "You have rolled at least one 6 AND I confirm I would always tell you that provided I could see at least one 6, what is the probability you have rolled a total of 7." But that's not actually what he asked.

A final comment - I am not trying to be obtuse or wind people up. I have read the entirety of both threads and I also have a degree in Mathematics from Cambridge in the UK, so hopefully feel qualified to input on this!

[Guest]

2) We are told "at least one is a tail AND if faced with a situation where I could tell you either at least one is a head or at least one is a tail I will pick one at random"

In this case we can discount the H/H but we must also discount half of the H/T and half of the T/H (taking "random" to mean a probability of 1/2 for each case). This now leaves us with a probability of head and tail (in any order) of 1/2.

Sorry for my courage, as my profession is away from math. But I have an objection:

I suppose you're able to make software simulations on pc. Write a software and give this order to pc: Randomly get millions of H or T letters, randomly make couples with them, then randomly get a couple, and finally randomly get one letter in that couple, and tell me.

If your pc says H, it is the same thing as saying at least one of letters (or coins) is H. By this information, you understand that both of the letters are not E, so they are eather HE or HH. So make your software tell you the number of couples and the ratio of HH/HE, it will say:1/3. You see that in your words '2) We are told "at least one is a tail AND if faced with a situation where I could tell you either at least one is a head or at least one is a tail I will pick one at random" ', the pc did everything as in your second item, and give 1/3????

[Guest]

One final addition - I realised I argued by reason, but now having thought through all that I also believe I know why the mathematical arguments given in this thread are also wrong.

I haven't got time to go into the detail, but I think the problem has been that people have been basing their mathematical arguments on two events happening, namely die 1 is rolled and die 2 is rolled. (It doesn't matter which order, or which is which, but the two dice are independent of each other and so are two separate events.)

However, there is a third event happening - the person is being told what at least one of their dice are. This event must also be taken into consideration when doing the maths and so must the probability of this event.

In order to arrive at the answer 2/11, the probability of (for example person 1) being told of a 6 is being ignored and thus assumed to be 1 in all the formulas given. This is the same as saying that we know we will be told if there is a 6 present for person 1.

However, the probability of being told of a 5 for person 2, a 4 for person 3 and so on is also assumed to be 1. This is the same as saying "we know we will be told if there is a 6 present for person 1 AND we know we will be told if there is a 5 present for person 2 AND we know we will be told if there is a 4 present for person 3 AND so on" which is possibly one of the biggest assumptions I have ever seen (as it can only be true in a very obscure case as Bonanova has pointed out).

It is this false assumption that leads to the maths being wrong and that provides the paradox.

One final observation, is that this doesn't contradict conditional probability. You could increase your odds of guessing a 7 by getting someone to tell you if you have (for example) at least one 6 present. As long as they always talked about the same number you would know that you had a 2/11 chance of having a 7 if they said something and a 4/25 chance of a 7 if they said nothing. Hence you are better off only betting if they say something. (And this is why casinos won't let you do this!!)

[Guest]

Sorry for my courage, as my profession is away from math. But I have an objection:

I suppose you're able to make software simulations on pc. Write a software and give this order to pc: Randomly get millions of H or T letters, randomly make couples with them, then randomly get a couple, and finally randomly get one letter in that couple, and tell me.

If your pc says H, it is the same thing as saying at least one of letters (or coins) is H. By this information, you understand that both of the letters are not E, so they are eather HE or HH. So make your software tell you the number of couples and the ratio of HH/HE, it will say:1/3. You see that in your words '2) We are told "at least one is a tail AND if faced with a situation where I could tell you either at least one is a head or at least one is a tail I will pick one at random" ', the pc did everything as in your second item, and give 1/3????

The problem with your logic here is where you say "randomly get a couple". At that point you now have a couple, so why do you later return to the entire set to figure out what the probability is of the other coin in that one couple.

Let's take your simulation one step further - do everything that you have said up to it randomly telling you one letter. Now get it to tell you what the other letter is and make a note of it. Now get it to randomly select another couple, randomly give you one letter from that couple and tell you what the other letter is. Do this hundreds (or thousands) of times.

I haven't done this myself yet, but will do as I want to confirm it to myself (and still perfectly willing to be proven wrong!), but my thought is that in all your results you will see that when it randomly told you H, 50% of the time the other letter was a H and 50% of the time the other letter was a T. This means the probability is still 1/2.

[Guest]

The problem with your logic here is where you say "randomly get a couple". At that point you now have a couple, so why do you later return to the entire set to figure out what the probability is of the other coin in that one couple.

Let's take your simulation one step further - do everything that you have said up to it randomly telling you one letter. Now get it to tell you what the other letter is and make a note of it. Now get it to randomly select another couple, randomly give you one letter from that couple and tell you what the other letter is. Do this hundreds (or thousands) of times.

I haven't done this myself yet, but will do as I want to confirm it to myself (and still perfectly willing to be proven wrong!), but my thought is that in all your results you will see that when it randomly told you H, 50% of the time the other letter was a H and 50% of the time the other letter was a T. This means the probability is still 1/2.

Your simulation is inside the entire coin pool, thus it is obvious that you get 1/2.

Let me ask in that way: If the magician, or pc, randomly looks at only one coin, and tells you that it is tail, what is the probability for the other coin. This is like your second statement, but that was not clear for me. But my last question's answer must be 1/2, and I'm convinced myself now. Thus the dice questions answer for subjects 1-6 can be both 1/6 and 2/11. Thanks.

[Guest]

Your simulation is inside the entire coin pool, thus it is obvious that you get 1/2.

That's really the whole point. In both questions it would be normal for us to have to consider the entire "pool". The only way we can get anything other than 1/2 is if we unnaturally limit the pool. E.g. for the kids question, we would have to say "of all people with 2 kids and at least one girl, what is the probability of a second girl", whereas really the question is "of all people with 2 kids, what is the probability of 2 girls if I tell you that they have at least 1 girl but don't tell you why I am telling you that."

[bonanova]

The conclusion reached here is that there are two cases that apply to the added information:

something is true; and we are told something is true.

Case [1] facts are present that depict everything that can be said about a situation [for want of a better way to say it]

We construct the population of equally likely cases, remove the impossible ones, and divide the favorables by the total.

Thus,

Of the population of 2-children families from which those with two boys have been removed, girl-girl families make up 1/3

Two fair dice have been rolled, repeatedly, until it is true that the number six is absent from the pair of numbers obtained.

Removing from 36 outcomes the 25 where no six is seen, we note that [only] 1-6 and 6-1 -- 2 cases out of 11 -- total 7.

But see how [any why] that changes in the 2nd case.

Case [2] an honest observer looks at a particular outcome and gives us a partial picture of it.

It may take an irritating paradox to force an intuitive thinker to see that [and how] this case differs from Case [1].

And it's not because we're not certain the observer has told the truth.

It's rather that the truth he has told is only a partial picture, and we have to wonder about the part that was not said.

What we've been told has been filtered, in a manner unknown to us, by the reporter.

Thus we cannot be certain of the result [analyze the complete picture] unless

[1] we are told what the observer's filter is or

[2] we make the [unwritten rule] assumption the statement has been chosen at random from all of the similar statements that could also be true.

A family with two children is said to have at least one girl.

Certainly the boy-boy cases have been eliminated.

But if what's left gives a 1/3 probability of girl-girl, we have to wonder whether the observer could have told us there was at least one boy.

In 2/3 of the cases, we reason, that statement is also true. And if he had, we would have 1/3 boy-boy.

Combined, both girl-girl and boy-boy cases would be eliminated, dictating a mixed-gender outcome.

Separately, the statements leave us partially in the dark.

Constructing tables [like table 1 in my previous post] leads to a paradox.

If you are told one of two dice you rolled is a 6, how many outcomes are possible? 11; two of which total 7.

But the symmetry of rolling a 7 [all numbers participate on an equally likely basis] demands six [equal] answers that total unity - i.e. 1/6.

Either being told at least one die is a 6 is meaningless, or the probability must be calculated in a different way.

Exploring the matter as I did in previous post, the problem is seen to arise from the overlap of cases permitted by the six statements.

There is no overlap when the first die, or the second die, or the lower die, or the greater die is described.

And in each of those cases p[7] = 1/6.

And because {1st die, 2nd die} and {greater die, lesser die} are exclusive, exhaustive and equally likely,

their averages must give the result for {randomly chosen die} = 1/6.

If we are told the {older, younger} child is a girl, then p[girl-girl] = p[mixed] = 1/2

These cases also are exclusive, exhaustive and equally likely.

So here it is: if the child that is described in the statement "One of the children is a girl" is randomly chosen from {older, younger},

then p[girl-girl] and p[mixed] must be the statistical average [add them up and divide by 2] of those two cases.

Similarly,

If we are told the {older, younger} - child is a boy, then p[boy-boy] = p[mixed] = 1/2

If we are told about a child who is a random choice of the two, the probabilities must be the same.

But if our intuition still screams out that p[mixed] = 1/2 of the general population,

so it therefore must be greater than 1/2 of any reduced population,

Then we need further help to see how p[mixed] is affected by statements like these.

In the unconstrained case, p[mixed] = p[girl-boy] + p[boy-girl] = 1/2

Why does it remain 1/2 when we are told something about

[A] the older child: because p[mixed] is now calculated given the older child is girl.

(B) the younger child: because p[mixed] is now calculated given the younger child is girl.

[C] a random selection of {older child, younger child}: what is p[mixed] in this case?

Here's the answer:

[A] p[mixed] = p[girl-boy] = p[girl-girl] = 1/2 - these are the only two cases when older child is girl

(B) p[mixed] = p[boy-girl] = p[girl-girl] = 1/2 - these are the only two cases when younger child is girl

[C] p[mixed] = average of [A] and (B) = 1/2

As with the dice - which is the topic of this thread - only when the observer is biased in his choice of which die [or child]

to talk about, does the "modified general population" analysis apply:

"This statement is true [about a child or die] AND it's all that I will tell you.

If it weren't true, you can be assured I would have said nothing."

[Guest]

As with the dice - which is the topic of this thread - only when the observer is biased in his choice of which die [or child]

to talk about, does the "modified general population" analysis apply

Great - although we've said it in different ways, I think we are in agreement.

Can you confirm then with this way of thinking that, with the wording of the original boy/girl question, the "best" answer (based on the unwritten rule) is actually 1/2 and not 1/3 as most had said it was?

[bonanova]

Great - although we've said it in different ways, I think we are in agreement.

Can you confirm then with this way of thinking that, with the wording of the original boy/girl question, the "best" answer (based on the unwritten rule) is actually 1/2 and not 1/3 as most had said it was?

The OP for that thread says,

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

If the OP is taken as the "honest reporter" then we assume the unwritten rule and come up with 1/2.

But there has to be a way, of laying out facts, that implicitly says "this is a complete specification - no other assumptions are warranted - answer, based just on what is stated."

Otherwise you could never describe a general population from which one segment has been removed while leaving the others intact.

I mean, saying there are three families - none of which has two boys but two [and only two] of which have the two cases of mixed gender. What is the probability of a family with two girls?

There you have, unequivocally, 1/3. But it wouldn't be much of a riddle.

My approach to solving puzzles is to take the OP as a complete specification of fact, from which a unique answer [if it's well worded] will arise.

So my personal answer for that puzzle [since it lacks an honest reporter clause] is 1/3 - general population minus the boy-boy cases.

Note that in the present thread [dice] I include a statement by the magician. "I have looked at your dice [after the fact] and I can tell you one of them is a six" [or paraphrase of that]

Not to disagree with the analytical method just discussed - just to say that's how I view what a puzzle specification [rightly] is.

For me, the other thread's OP would have to have included a statement like: "we are told by the father that one of his children is a girl" in order for me to invoke the rule.

I would guess, from the volume of debate in that thread, that's where the majority of differences arose.

Thanks for the light you've shed on this discussion.

- bn

[Guest]

My approach to solving puzzles is to take the OP as a complete specification of fact, from which a unique answer [if it's well worded] will arise.

So my personal answer for that puzzle [since it lacks an honest reporter clause] is 1/3 - general population minus the boy-boy cases.

I see where you are coming from. I actually agree with your approach to how puzzles should be taken, but probably still disagree with the 1/3 answer simply because of the way the question is worded and that I think we have to make a bigger assumption to get to the 1/3 answer than we do for the 1/2 answer.

If the question was something like "A family has two children and at least one is a girl. What is the probability that the other is a girl?" then I would say 1/3. This is because of how the family is selected.

In any probability question you should be able to say something like "If I were to repeat that question a thousand times and find out the correct answer, how many times would the requested answer appear?"

In my reworded version, repeating the question would mean that each time the question is asked the family has to be chosen as a family with two children and at least one girl, i.e. it must be chosen from the sub set of all two children families with no boys.

With the original wording, we are first told that we have a family with two children. We are then told that one of the children is a girl. To me, if we were to repeat the question then the family is simply chosen from all families with two children. We are then simply told something about that family - i.e. what we are told about them is NOT a criteria for how the family was selected in the first place. It then follows that, if the family selected has 2 boys, then for the question to be posed in a similar form we must be told that there is at least one boy. If this happens then we must assume that for boy/girl combinations we will be told of either, otherwise we are introducing bias into the problem. This then leads us to the 1/2 answer.

Basically I think the wording is critical because, depending on how we are told "at least one is a girl" it can either be taken as a condition for the family or just an observation of it.

In the real world, this situation is most likely to occur when we or someone else only knows about one child (but not knowing which one). Obviously, in the real world, we are not likely to only ever meet two children families from the subset of all families without two boys. So, in the real world, we will be talking about a family that has been selected from the entire population and then has an observation made about them.

So I now know that in the real world (for example, you hear a couple has two kids and know one is a girl and want to know the probability of the other being a girl), the answer is definitely 1/2. (So I am no longer tempted to make a bit of money on the side by placing bets with people! )

However, I do also acknowledge that there is a mathematical interpretation of the question that could lead to the answer being 1/3.

And then to get to an answer that requires absolutely no assumptions, I would have to say "somewhere between 0 and 1"!

I suppose this is just an example where the english language and interpretations thereof can really lead you to screwing up your sums and that you need to be really careful before you go making assumptions of any kind!

Hopefully this also means that all those that dissed the answer of 1/2 previously can now see that it is a perfectly reasonable answer.

Thanks Bonanova - I've really enjoyed this topic! I had read the kids problem previously and was convinced the answer was 1/3, but this thread got me thinking again. It's great to be able to challenge your own thoughts and, as ever, this forum continues to make you do just that!!