[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]

Sure. We can stop whenever.

But did we debunk his argument?

I'm not just trolling for more posts: I find it persuasive. I don't have a tightly-worded rebuttal.

I paraphrase your answer as [he was given no clue] therefore [1/6 cannot change.]

That ignores his argument. It says, since my argument is right and we differ, you must be wrong.

If [he had heard nothing] then clearly [he was given no clue.] But [he heard the first six conversations].

I am trying to prove, to my satisfaction, that [hearing them] does [not lead to a clue].

I haven't succeeded yet.

It's interesting, probably pivotal, that had he [not heard anything] he could nevertheless reason exactly the same!

Suppose you have a coin, one side of which is marked "1" and the other side "2".

There are 4 equally probable ways two coins can land:

1 - 1, 1 - 2,

2 - 1, 2 - 2

A magician blindfolds three people, then for each of them he flips a pair of fair coins and asked them the probability of their coins totaling 3.

He said to the first, here's a hint: truthfully, at least one of your coins shows a 2. The subject counted 3 cases of at least one 2, two of which, 1-2 and 2-1, total 3. So he answered, my chances of having 3 are 2/3. Very good, said the magician.

He then said to the second, I've looked at your coins, and at least one of them is a 1. This subject counted 3 cases of at least one 1, of which 2-1 and 1-2 made 3. So he answered 2/3. Very good, said the magician.

The third 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 coins. 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 coins totaling 3 are is. My answer is 2/3.

The third subject is wrong.

********

This now, is exactly the same as the Boy-Girl Problem. And because of that, it's (hopefully) easier to see that the two cases covered by "person one" and "person 2" are not mutually exclusive.

The magician essentially tells person 1, "the two coins are NOT 1 - 1." Then he tells person 2, "the two coins are NOT 2 - 2." Person 3's error here is to assume that these two statements fully describe all possibilities.

If that makes sense, your original problem is just an extrapolation of this simplified one.

Someone who understands probability more than I do explained it in a way bonanova would be able to appreciate more than I can (over my head).:

The problem is with the reasoning of the seventh person. He is forgetting that the magician's statements are not all mutually exclusive. In standard probability language, the seventh person is (mis)using the conditional-probability formula

P(A) = sum_B P(A|B)P(B) .

To be used correctly, this sum must be taken over a complete set of mutually-exclusive events B. The seventh person takes as event A "the dice sum to 7" and as the six events B "at least one 1 is showing", ..., "at least one 6 is showing", so that all P(A|1)=...=P(A|6)=2/11. Now he implicitly assumes (wrongly) that these six events are mutually exclusive and (also falsely) that all six P(B) are equal to 1/6, and therefore gets P(A)=2/11.

If instead I define my events B as "the smallest number showing is a 1", ..., "the smallest number showing is a 1" then these events are mutually exclusive and complete, and so the formula above can be used to get the 1/6 value we expect. However, in this case the six values P(A|B) are not all equal to 2/11 (and nor are the values of P(B) equal to 1/6). Alternately, I may use an inclusion-exclusion argument to compensate for my overcounting.

And from someone else:

The truth is, if we were to analyze the case a little differently, the probabilities might be different. There are the following 66 possibilities:

Roll: 1, 1. Told: At least one is a 1

Roll: 1, 2. Told: At least one is a 1

Roll: 1, 2. Told: At least one is a 2.

Roll: 1, 3. Told: At least one is a 1

Roll: 1, 3. Told: At least one is a 3.

...

Roll: 5, 6. Told: At least one is a 5.

Roll: 5, 6. Told: At least one is a 6.

Roll: 6, 6. Told: At least one is a 6.

Now, just because there are 66 possibilities, we shouldn't assume that each has probability 1/66 (to assume "N possibilities" always means "N equally probable possibillities" is a common probabilistic mistake). Indeed, by implicit convention, we instead have that each possible die roll outcome is equally probable; thus, the first row above has probability 1/36, the sum of the second and third rows' probabilities is 1/36, etc.

However, beyond this, we haven't been given any information about, and have no clear convention regarding, the particular probabilities of row 2 and row 3. They could be 1/36 and 0, or 1/72 and 1/72, or all sorts of other things. Much depends on this.

If we go with the most natural "1/72 and 1/72" setup (e.g., after rolling the pair of dice, the magician flips a coin to decide which one to tell you about), then everyone who announces "2/11" is actually wrong. For example, if the magician says "You have at least one 1", then this encompasses "Roll: 1, 1; Told 1" (prob. 1/36) as well as "Roll: 1, n; Told 1" (prob. 1/72) and "Roll: n, 1; Told 1" (prob. 1/72) for each of n = 2, 3, 4, 5, 6. The total probability of this is (1/36 + 5/72 + 5/72) = 12/72, out of which only 2 possibilities of probability 1/72 each have the dice adding up to 7; this gives a final conditional probability of (2/72)/(12/72) = 2/12 = 1/6. NOT 2/11. In this setup, everyone who announces "2/11" is wrong.

The point is, when the magician announces "At least one of your dice is a 4" or whatever, he is not merely giving you the information that at least one of your dice is a 4; you are also being provided with the further information that, also, out of the two dice, the one he chose to tell you about was a 4. This further information can cause the correct response to not be "2/11", but something else (in the above setup, "1/6"; in other setups, perhaps something else). Taking this into account, there will be no paradox.

Now, remember... the problem given is not fully specified. After rolling the two dice, we have no idea what process the magician uses to decide what to tell you; perhaps he always tells you about the first die rolled, perhaps he flips a fair coin to decide, perhaps he flips an unfair coin, perhaps he generally prefers to announce the minimum number but doesn't always, perhaps he has a much more complicated system.

So, "He flips a coin to decide what to tell you" is just one possible way to finish off your problem, not the only one. That warning having been emphasized, let's look at that particular case another way, in case the above discussion was confusing. We can most naturally construe it as containing 72 possibilities (6 die faces * 6 die faces * 2 coin faces), and, particularly nicely, all of these 72 possibilities will be equally probable.

After being told "At least one of your coins is a 4", there will be 2 possible ways to have a dice-total of 7 ("3, 4, tails" and "4, 3, heads"), while there will be 12 total possibilities still in play ("n, 4, tails" and "4, n, heads" for each of n = 1, 2, 3, 4, 5, 6); thus, in this setup, the conditional probability of having a dice-total of 7 will be 2/12, and NOT 2/11.

[Guest]

This blows my mind...I now understand that the probability is 2/11 (thank you Bonanova, Scraff, and others for convincing me)

But I find that complete amazing- learn something new every day I guess.

But so now, as for the 7th person...

It reminds me sort of Schrodinger's Cat (I'm sure you're all familiar with that, where the cat is both alive and dead at the same time?). It seems that the 7th person's probability is both 1/6 and 2/11 at the same time. There is uncertainty about the roll, so his probability is 1/6, but as soon as he is told "at least one of your dice is a..." his probability changes to 2/11.

Hopefully that makes sense, it probably doesn't.

Here's another interesting thing to ponder (as though this thread weren't crazy enough):

Suppose the magician says "At least one of your dice is a 3." But it isn't! The magician is lying! Is his probability of having a seven 1/6 or 2/11? Because the subject believes it to be 2/11, but he has received invalid information...

Somebody will probably answer that in like 5 minutes, but I thought it was an interesting point.

[bonanova]

rossbeemer: We may assume the magician was not lying.

Scraff: With your last quote, I think all the cards are now on the table.

Now for a longish post that I think will be my last, here.

The following assertions are mine, and I have reasoning for each that at least satisfies me.

1. Subject 7's argument basically says [can be reworded to say] that his answer should be the same as the other six.
   
2. 7's argument can be rigorously formalized. [i don't disagree with Scraff's objection to it; I recognize it, and assert it can be accommodated.]
   
3. The six 2/11 answers *could* be right. But they needn't be. The assumptions that lead to 2/11 may be correct. But they needn't be.
   
4. I arrived at the preceding assertion after mentally sketching a simulation of the problem. As stated, I do not believe the puzzle can be simulated.
   
5. With modification the puzzle can be simulated; information would be added to the OP.
   
6. Clearly "luck" involved - that all 6 magician statements could be true.
   

Here's what seems to fall out of all this.

If the OP stopped after conversation #1, one could reasonably go either way on 2/11.

The strong preference, however, is for an assumption that makes p[7] = 1/6 for all seven subjects.

Luck still is needed for all 6 magician statements to be true.

The "proofs" for and against 2/11, mine included, could have been worded more convincingly

[completeness and relevance of supporting facts], with the result of seeing the conditions

[assumed, or un-compellingly explained IMHO by all] under which the result [for and against 2/11] is true.

In the following spoiler I'll formalize Subject 7's argument in a way that I believe is rigorous,

show why I could not simulate the puzzle, and finally support my statement in red, above. ]By symmetry - none of the six conversations 7 heard has more import than the others - each of the conditions [outcome probabilities] that validly follow from the magician's true statements applies to Subject 7 with equal likelihood. That is, 7's situation might fit Subject 1's or Subject 5's or ... with equal likelihood. By symmetry that probability is 1/6. [Note, we're not talking about p[6]=1/6 here]

[spoiler=Formalization of 7's argument, and its final consequence

Now, probability is changed by knowledge. We start with 6x6 = 36 EL [equally likely] outcomes. Each of Magician's statements eliminates 25 of those cases, leaving 11. Both sides agree with that statement. I'll say more about their comparative likelihood later. We can construct 7's outcome table by adding together the probabilities of the 11 outcomes for each of the six, multiplied by 1/6. The manner in which the 66 = 6x11 outcomes distribute themselves on 7's outcome table is telling: one major diagonal term 1-1, 2-2, etc from each, and two off-diagonal terms 3-5, 1-2, 5-6, etc collectively from the six. The 36 outcomes for 7 occur equally, except for the "doubles" outcomes. That is, the off-diagonal outcomes happen twice, while the "doubles" cases happen only once. They sum to 66 [6x10 + 6x1]. I think this touches on Scraff's objection to 7's reasoning - as if there were an assumption they are all on equal footing. They are not. But that's OK, because we've counted [accounted for] them all.

But now look at the NE to SW diagonal - the outcomes where 7 is the total. There are 2's in each of those boxes. That is, 12 of the 66 outcomes have 7 as a total. So if all 66 outcomes are equally likely [recall so far we've only counted them as said there are 66] then p[7] = 12/66 or 2/11 for Subject 7. Notice the if. It's the same if that leads to p[7] = 2/11 for the first 6 subjects. OK that's a long [rigorous] path to saying p[7] for Subject 7 is the average of the p[7]s for the first six. Note that whatever the probability distribution of the 11 outcomes for each of the first 6 subjects [the distributions are all the same by symmetry] the the same probability distribution will come out in the average. That is, all p[7] values will be the same. Subject 7's argument is thus valid: If p[7] = 2/11 for the first six, p[7] = 2/11 for Subject 7; if p[7] = 1/6, 0, 0.0001, ... whatever, for the first six, 7's is the same.

I said that each of the magician's statements eliminates 25 outcomes. That's not in dispute. Differences arose in assertions of the remaining 11 outcomes being equally likely. More to the point: if not, why not. I assert now that they are, maybe, or they're not, maybe, equal. That is, the OP does not tell us enough to be sure. In the "one of your dice is 4" case, if the magician looked for a 4 and called it out if it was there [see how luck enters - what if the dice were 3-5?] then all 11 outcomes [14, 24, 34, 44, 54, 64, 41, 42, 43, 45, 46] have the same likelihood, and p[7] = 2/11. If the magician glanced at the dice and chose between the dice at random, then the special case 44 is twice as likely as the others: he's compelled to say 4, where in the other 10 cases he could have equally likely have called 1, 2, 3, 5 or 6. And then p[7] = 1/6. Thus,

1. if Magician looked for a 6, 5, 4, 3, 2 and 1 in sequence as he talked to the first six subjects AND FOUND HIS NUMBER IN EACH CASE [more and more luck] then p[7]=2/11.
   
2. if Magician chose at random from the two dice AND IT JUST HAPPENED THAT WAS ABLE TO AND DID CHOOSE 6, 5, 4, 3, 2 and 1 in sequence [how much luck is involved there???] then p[7] = 1/6 in each case. But we don't know for sure, because the OP does not tell us the Magician's algorithm.
   

Now the final point in formalizing 7's argument: what do we do with case 1? We've said that 7's p[7] must be the average of the first six. Yet we can't live with an unconstrained p[7] different from 1/6. The answer is that in case 1 the argument is invalid, while in case 2 it's valid. How convenient, you say - when the answer is wrong, you say it doesn't apply, but if you like the outcome you say it's OK.

Well, maybe I've done that once, in the course of my life. But not here.

There's a valid objection in case 1. Here it is. It's the apples and oranges thing. To justify taking the average of 6 cases, they must be of the same type. In our case this means the SAME ASSUMPTIONS must hold in each case. They don't: first we looked for 6's then we looked for 5's and so on. That removes the justification for averaging. In case 2, we're doing precisely the same thing each time, and averaging is permitted. The fact that luck is necessary does not introduce bias: we can re-roll the dice until luck shines on us. Certainly each roll of fair dice has the same p[7]. Finally, note that you can compute the average p[7] for both cases. It's just not valid to do so in case 1. Bottom line is no valid logic compels us to deduce p[7] = 2/11 for Subject 7, even though there is a valid set of circumstances for which p[7] = 2/11 for the first six. That is the conditions of case 1. That, in essence is the solution to the puzzle: why and how we can escape unconstrained p[7] different from 1/6.

Now it's usually implied in logic puzzles that when a process is not specified, we assume the absence of mechanisms that produce bias. That is, if there is a mechanism that produces bias, we expect it to be identified in the puzzle. Thus, the preferred algorithm for magician is that he choose at random from the two dice when he talks with each subject. Now we come to p[7] becoming 1/6 for the first six subjects in case 1. In both cases, there are are absolutely 11 outcomes. A=n or B=n absolutely removes 25 outcomes. But let's consider the likelihood, for the n=6 case, of the outcomes 5-6 and 6-6. In the 5-6 case, under the conditions of case 1, 50% of the time Magician will say "one of your dice is 5". But in the 6-6 case, 100% of the time Magician will say "one of your dice is 6".

Notice, this is not the same as saying A=5, B=6 is the SAME CASE as A=6, B=5. Arguing that way says there are ONLY 6 OUTCOMES, with the A and B values separately not mattering. That statement is false. There are two distinguishable cases, just as they were distinguishable in the unconstrained 36-outcome case. What LEADS TO THE SAME RESULT but, as desired IS ALSO TRUE is this: one ascribes, rightly, twice the likelihood that "one of your dice is 6" suggests a 6-6 configuration than you, rightly, do to it suggesting a 5-6 configuration. Thus, the 5-6 and 6-5 cases carry the same "logical weight" TOGETHER as the 6-6 case does BY ITSELF.

Several solvers [most likely] saw this, and invented ways to make it happen:

1. 5-6 and 6-5 are the SAME CASE, said some, but just saying so is (a) wrong and (b) not surprisingly, not justified.
   
2. 6-6 should be COUNTED TWICE, said others, but with no justification given, other than p[7]=1/6 if you do so.
   Approaches like these might get you partial credit on a test. I know they did for me, on occasion.
   

Both assertions give the desired answer; but neither stands up to scrutiny.

Finally,

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.

p.s. [those initials stand for mea culpa]:

My contribution to the heat-but-little-light aspect of the discussion was to insist

the 11 remaining cases HAD TO HAVE equal probability.

They started out that way, and "Nothing the magician said changed that." Is what I think I said.

It still has sort of a compelling ring.

But I would rather think of it this way.

Start with "One of your dice is a 6" and put allowable cases into the outcome table.

Clearly I could not put any of the forbidden 25 in, but as I put the others in, I would

have the opportunity of asking what probability they have for being there.

The 6-6 case certainly must be there. Whatever the magician was up to, he'd have to name a 6.

But the 5-6 case is there only if magician happened to choose the 6 to call out - 50% of the time if magician was random in his choice.

Thinking this way, it's clear that 5-6 [AND 6-5] BOTH belong in the table, but their likelihoods of producing the magician's statement are half of the 6-6 case.

So again, in case 1, the 11 cases do not have equal probability: p[7] = 1/6;

in case 2 they do: p[7] = 2/11.

[Guest]

Here's another interesting thing to ponder (as though this thread weren't crazy enough):

Suppose the magician says "At least one of your dice is a 3." But it isn't! The magician is lying! Is his probability of having a seven 1/6 or 2/11? Because the subject believes it to be 2/11, but he has received invalid information...

Having a wrong clue will not raise his chance from 1/6 to 2/11, but it will also not reduce it from 1/6 to any lower value. My answer = 1/6.

I want to recall my old post for somebody who find it diffucult to believe that subjects 1-6 have 2/11 odds.

Indeed it comes me unlogical why these subjects odds raise when they heard at least one of their dice is any number.

In all condition, a number of course will be on their dice. It doesn't seem to be a clue.

The real clue they have is not that. The real clue is: Both of his dice are not among the group of numbers those are 1,2..6 minus given number.

Of course these two clues are same things, only wordings are different. But the second wording seems to give a definite information. Because in all conditions, both of his dice must not be among that group.

This wording comes me very logical and convinces me about 2/11.

[Guest]

You ask why I don't say whether 2/11 is correct for subjects 1-6. I did make a case; but it's in a spoiler, since it's part of the puzzle.

The puzzle is solved either of two ways:

[1] Prove that p[7] = 1/6 for all seven subjects.

[2] Either

(a) Give up the notion that Subject 7's p[7] has to be 1/6, or

(b) Debunk S7's argument.

You took approach [1], giving a proof [one that satisfies you] that p[7] = 1/6 for all seven subjects.

You've finished the puzzle.

If I was perceived to put you [or anyone else] down, that's not ever my intent; I apologize.

I believe it's within the bounds of respect for a person to point out what might be

a weak point in an argument, but certainly I will not impugn anyone's intelligence.

I reserve for myself the right to make a mistake; and, frankly, I exercise that right fairly regularly.

So I this give that same latitude to others.

Bonanova, I fully respect that and apologize for any negative connotations my comments had toward you.

As for Scraff, I don't care if 1,000 people with 1,000 "math" websites agree with you. If you're wrong, you're still wrong, regardless of how many people agree (and are also wrong). Maybe I am being arrogant, but I have a degree in Mathematics. I shouldn't have to do this, but here's a listing of the courses I took.

STA 2023 Statistical Methods 3 hrs

MHF 3302 Logic and Proof in Mathematics 3 hrs

MAP 2302 Differential Equations 3 hrs

MAS 3105 Matrix & Linear Algebra 4 hrs

MAS 3106 Linear Algebra 4 hrs

MAP 4363 Applied Boundary Value Prob I 3 hrs

STA 4321 Statistical Theory I 3 hrs

MAS 4301 Abstract Algebra I 3 hrs

STA 4322 Statistical Theory II 3 hrs

COP 3502C Computer Science I GEP

MAA 4226 Advanced Calculus I 4 hrs

MAA 4227 Advanced Calculus II 3 hrs

MAD 4203 Applied Combinatorics 3 hrs

MTG 4302 Introduction to Topology 3 hrs

MAP 4307 Appl of Complex Variables 3 hrs

Notice the 2 called Statistical Theory. These courses are focused on all the theory behind where those handy statistics equations come from. It's a logic and proof course on statistics. Do you know any of this? With all due respect to Bonanova and the other moderators, do they know any of this (maybe they do)?

Also, notice the class called Combinatorics. Do you know what that is? That is exactly this thread. It was a whole semester of questions like this one. I aced that class by the way, without EVER studying. The whole semester I sat in class listening to other students using false logic and coming to false conclusions, which is why I get so frustrated by this thread. I started out wanting to shed some light and wisdom on the subject but you have refused to see what I've plainly laid out for you. So, yes, I'm getting arrogant because I have a right to. I know this stuff like the back of my hand. I can't speak for the 3 math sites you posted, because I don't know the credentials of anyone on those sites, but I know that I'm better at this stuff than anyone on BrainDen. Again, with all due respect to the moderators, because they do a wonderful job and I do respect them, this thread isn't about a brainteaser, it's about statistics, which I am definitely qualified to be arrogant about.

[Guest]

I started out wanting to shed some light and wisdom on the subject but you have refused to see what I've plainly laid out for you.

I saw it; you're wrong.

So, yes, I'm getting arrogant because I have a right to.

I didn't claim you didnn't have a right; having the right to be rude doesn't make it any less rude. And you didn't answer my questions:

What agenda are you accusing bonanova of having? Do you really believe he's dismissing your answer for any reason other than he knows it's wrong due to his knowledge of mathematics? Why would he not like the truth regarding the answer to this riddle?

but I know that I'm better at this stuff than anyone on BrainDen.

Oh brother.

Answer this simplified question regarding the Boy/Girl problem.

First, assume all children born have a 1/2 probability of being a boy and a 1/2 probability of being a girl.

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

[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)

In the case of the die problem P(A) = 1/6 (a single die roll resulting in a desired number)

By the way, this thread is proof I know this stuff better than anyone on BD because I'm the only one left who hasn't bought into the erroneous logic you've put forth.

This is a simple statistics problem gone way way way wrong. I don't blame anyone or judge anyone on here for getting it wrong because it is very common. The commonality of this mistake is why there aren't more statisticians and mathematicians around. I don't fault people for getting it wrong but you've told me I'm wrong about something I know I'm 100% correct on. I have given you 4 or 5 solid proofs (some a little better than others). This one is the best yet as I've simply gone back to the basics of the equations used in statistics books, as opposed to trying to show the logic behind them. That was my point in showing my course history at college. Anyone can look into a statistics book and see the above equation and presumably come to the correct solution. I, however, spent over a year proving WHY that is true.

[Guest]

1/2

Great. Then answer what would happen if we tried this real world experiment.

We'll simulate the question I asked with coins.

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

Let's lay out 100 pairs of coins. Heads we'll call boys and tails we'll call girls. Since in the problem above we know there are no Boy/Boy pairs (there's at least one girl), we'll remove all Head/Head pairs.

Out of those coins left, you believe about 1/2 will be Tail/Tail, correct?

[Guest]

Great. Then answer what would happen if we tried this real world experiment.

We'll simulate the question I asked with coins.

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

Let's lay out 100 pairs of coins. Heads we'll call boys and tails we'll call girls. Since in the problem above we know there are no Boy/Boy pairs (there's at least one girl), we'll remove all Head/Head pairs.

Out of those coins left, you believe about 1/2 will be Tail/Tail, correct?

It's true that in that case, 1/3 would be Tail/Tail...

[Guest]

It's true that in that case, 1/3 would be Tail/Tail...

That's right. And it's the same for the question "Out of all couples with exactly two children that have at least one girl, how many have two girls?" because that question is essentially the same as "Out of all pairs of coins that are flipped that have at least one Head showing, how many have two Heads showing?"

Let's say there are one million couples in the world that have exactly two children. We can flip a pair of coins for each of these pair of children and the coins should roughly reflect what the sex of the children are.

Out of the children there should be about 250,000 BB pairs, about 250,000 GG pairs, about 250,000 GB pairs and about 250,000 BG pairs.

Likewise, the coins should be about 250,000 HH pairs, about 250,000 TT pairs, about 250,000 TH pairs and about 250,000 HT pairs.

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

To simulate with the coins we have, we'll remove all the BB pairs. That leaves us with about 250,000 HH pairs, about 250,000 HT pairs, about 250,000 TH pairs. About 1/3 of the pairs of coins will have two heads and about 1/3 of the children pairs will be GG. BeastMaster said the answer would be 1/2. Maybe he'll concede that he in fact isn't "better at this stuff than anyone on BrainDen."

[Guest]

Great. Then answer what would happen if we tried this real world experiment.

We'll simulate the question I asked with coins.

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

Let's lay out 100 pairs of coins. Heads we'll call boys and tails we'll call girls. Since in the problem above we know there are no Boy/Boy pairs (there's at least one girl), we'll remove all Head/Head pairs.

Out of those coins left, you believe about 1/2 will be Tail/Tail, correct?

In this case, you would be left with 1/3 of your new population as Tail/Tail pairs, BUT, that's not probability. You've changed the data set, so you have to think about it differently.

That's what I keep trying to tell you, you can't look at probability that way. Once you change your data set (you say you have at least one girl), you can no longer use the original data set to calculate probability. The whole problem changes once you assert a known value. Look at it this way.

You tell me you have 2 children. You then tell me that at least one of them is a girl. That means one of two things MUST be true. Either child A is a girl or child B is a girl. They could both be girls, but at least one or the other MUST be true. Now, simple if-then statements will lead you to conclude that...

IF child A is a girl, then child B COULD BE a boy or a girl. There are 2 options, so the probability for girl/girl, in this scenario, is 1/2.

IF child B is a girl, then child A COULD BE a boy or a girl. There are 2 options, so the probability for girl/girl, in this scenario, is 1/2.

Now, I know the argument is that I repeated the situation with child A being a girl and child B being a girl, but in reality this was not a repeat because it all comes down to which was known and which was a variable.

KNOWING the child A is a girl and child B COULD BE a girl is very different from KNOWING child B is a girl and child A COULD BE a girl. In this case these 2 seemingly repeat scenarios of A=girl/B=girl are not repeats at all because they belong in different data sets.

By the way, the same thing is true for the coins, and for the dice.

The problem here is that you, and others, have fully convinced yourself of your solution, over however long you've been doing these problems. Now, it's almost impossible for you to see the reality because you've been believing your own falsehoods so long. I don't mean this derrogatorily, it's just the way it is. It is incredibly difficult to change your point of view when you've convinced yourself your right. I admit that there is potential here for me to have the same problem, except that my view is supported by all major statistics books, formulas, proofs, and logic. Yours is supported by wikipedia (which is user submitted) and Dr. Math, who was questioned on his site but refused to acknowledge the infallible math that was placed before him.

You've convinced yourself that your method is the right method to arrive at the solution, but you have yet to show why the logic of that method is correct. All your proofs revolve around doing the same problem with the same method just using a different named variable (coins, children, dice, etc.). Multiple samples of the same incorrect logic do not create a proof. A proof is a mathematical procedure of showing truth through logic and sound mathematical principles. You have not done this. I have done this 4 times in 4 different ways. I have explained the logic and I have shown you the tried and true, proven equations of statistics. Maybe you'll never believe me and the solution I've proven, but I sure hope you do. In the long run, people getting this wrong doesn't affect my life, but I have a hard time sitting idly by while people make mistakes. It's OK that people make mistakes, I just see it as my responsibility to correct them when I see it.

Scraff, I truly do apologize for any antagonistic statements I've made. I apologize for being arrogant. It truly hasn't been my intention to be rude, but I admit that I was rude. You didn't deserve that and neither did the forum. Like I said, I don't blame people for getting this stuff wrong, I just get frustrated when I show people how they're wrong and they don't see it. I shouldn't have taken my frustration out as anger, it was inappropriate.

Please do go back to my previous post and look at the equation I showed, and the site that shows it as well. If you can prove why that statistical equation is incorrect, I will concede. I doubt this will happen though, because proving that equation to be incorrect will mean turning hundreds or thousands of years of mathematical study on it's head. It's been tried and proven.

I do understand where you're coming from in believe the 1/3 child probability and the 2/11 dice probability. Especially when you explain it, it seems to make sense, which is why people have believed you. The problem is, that no matter how good your solution sounds or SEEMS to make sense, the method you use to arrive at it is not based on sound statistical theory. I'm not trying to be rude or mean by saying that, it's just the way it is.

[Guest]

...

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

To simulate with the coins we have, we'll remove all the BB pairs. That leaves us with about 250,000 HH pairs, about 250,000 HT pairs, about 250,000 TH pairs. About 1/3 of the pairs of coins will have two heads and about 1/3 of the children pairs will be GG. BeastMaster said the answer would be 1/2. Maybe he'll concede that he in fact isn't "better at this stuff than anyone on BrainDen."

Like you'll see in my last post, I do concede that in the scenario you lay out, that 1/3 of the pairs will in fact end up being GG, but that isn't the same thing as probability. Your question was "Out of all couples, how many will have...". That's not the same thing that the magician asked. The magician asked, "You have at least one "x", what is the PROBABILITY your die total 7". That's a very different question. I know it's not intuitive, but read my last 2 posts for the full explaination.

[Guest]

In this case, you would be left with 1/3 of your new population as Tail/Tail pairs, BUT, that's not probability.

That IS probability. It seems you are conceding that 1/3 of the population will be Tail/Tail pairs, but then you go on to say this:

You tell me you have 2 children. You then tell me that at least one of them is a girl. That means one of two things MUST be true. Either child A is a girl or child B is a girl. They could both be girls, but at least one or the other MUST be true. Now, simple if-then statements will lead you to conclude that...

IF child A is a girl, then child B COULD BE a boy or a girl. There are 2 options, so the probability for girl/girl, in this scenario, is 1/2.

IF child B is a girl, then child A COULD BE a boy or a girl. There are 2 options, so the probability for girl/girl, in this scenario, is 1/2.

The two examples are the SAME! If you get 1/3 Tail/Tail pairs, then you get 1/3 GG pairs. You DON'T split probabilities the way you did above. If it is right to do so, then you would have to do the same when figuring probability for coin pairs like this:

IF coin A is showing tails, then coin B COULD BE showing heads or tails. There are 2 options, so the probability for tail/tail, in this scenario, is 1/2.

IF coin B is showing tails, then coin A COULD BE showing heads or tails. There are 2 options, so the probability for tail/tail, in this scenario, is 1/2.

But as you already admitted, there is a 1/3 chance of the coins showing tail/tail, and the above is not how we figure the probability for the coin problem. The coin problem and the Boy/Girl problem as I re-worded it are identical!

Scraff, I truly do apologize for any antagonistic statements I've made. I apologize for being arrogant. It truly hasn't been my intention to be rude, but I admit that I was rude. You didn't deserve that and neither did the forum. Like I said, I don't blame people for getting this stuff wrong, I just get frustrated when I show people how they're wrong and they don't see it. I shouldn't have taken my frustration out as anger, it was inappropriate.

I appreciate that. Let's move on.

Now, I know the argument is that I repeated the situation with child A being a girl and child B being a girl, but in reality this was not a repeat because it all comes down to which was known and which was a variable.

Contradicting yourself a little? You know the situation is a repeat but in reality it's not a repeat? Sorry, that's a complete contradiction. The coin problem is the same as the Boy/Girl problem. You can't say the probability of having a Tail/Tail pair remaining is 1/3 and the probability of having a Girl/Girl pair is 1/2 without also contradicting yourself.

KNOWING the child A is a girl and child B COULD BE a girl is very different from KNOWING child B is a girl and child A COULD BE a girl. In this case these 2 seemingly repeat scenarios of A=girl/B=girl are not repeats at all because they belong in different data sets.

You need to do the same thing for the coin problem then, and come to the same conclusion.

[Guest]

BeastMaster, I showed in post #61 how the coin and Boy/Girl problem are identical. Can you show me at which step in that particular post that my analogy between the two situations is incorrect?

[Guest]

In this case, you would be left with 1/3 of your new population as Tail/Tail pairs, BUT, that's not probability. You've changed the data set, so you have to think about it differently.

BeastMaster, can you explain what you mean by this? I'm not challenging you, just asking a question so I can try to understand All these posts have been so confusing I'm probably going to go crazy- but it's definitely fun to discuss this.

If you have 250000 BB, 250000 BG, 250000 GB, and 250000 GG, and you remove all the BB (because we are considering couples with "at least one" girl) then isn't your new sample space 750,000? Which would mean that if you randomly selected a couple, the probability of getting two girls would be 1/3?

Edited December 31, 2008 by rossbeemer

[Guest]

But as you already admitted, there is a 1/3 chance of the coins showing tail/tail, and the above is not how we figure the probability for the coin problem. The coin problem and the Boy/Girl problem as I re-worded it are identical!

No, I didn't admit that there is a 1/3 CHANCE. I admit that when you take the original data set and remove the BB or HH pairs, you end up with 1/3 of the resultant population being GG or TT pairs. That is NOT the same as saying the CHANCE of having a GG or TT pair is 1/3 if you know that at least one of the elements of the pair is a G (or T).

Like I said, your "logic" is convincing, but based on erroneous methods. Now, I've answered your question, although you may not be satisfied with my response. What I would like you to do is prove that the following statistical equation is NOT true.

This is a very basic and common equation in statistics, used for conditional probability problems (of which this is one).

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

Where

P(A|B) means probability of A given B (order doesn't matter, it just means you know the outcome of B)

P(A interstect B) means the probability of A AND B BOTH happening when nothing is known or given

P(B) means the probability of B happening in an isolated case

Now, we all established that P(Girl intersect Girl) when nothing is known or given is 1/4

We also established that P(Girl) when you are only considering a single child is 1/2

So P(Girl|Girl) or the probability of both children being girls when you know at least one (doesn't matter which one) is a girl = (1/4)/(1/2) = 1/2

Die rolling, child gender, and coin flipping are all known to be independent events (ie. if you flip a coin 99 times and happen to get heads each time, the probability of your 100th flip being a head is still 1/2 : No flip depends on any other flip for its probability)

When 2 events are statistically independent, P(A intersect B) = P(A)*P(B) where P(A) and P(B) are the probability of attaining a desired result independent of any other factors.

So we could also solve the problem by simplifying the equation. Here's how.

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

but we just said that P(A intersect B) = P(A)*P(B) for independent events, so

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

or more simply

P(A|B) = P(A)

and I think we all agree that the probability that a single random child is a girl is 1/2.

The reason your admitadly convincing logic fails is that in statistics you can't just reduce your original data set and then consider all the remaining events as equally probable, because they aren't equal anymore. You have to rewrite your statistical events to match your new problem with your new knowns and variables.

I have already written this, but seeing as how you didn't see this the first time, I'll say it again.

There are actually 2 ways to solve the child problem (and by extension, the coin and dice problems). One way where you don't care about order, and one where you do.

If you don't care about order and you know at least one is a girl, then the other child could be a brother or sister. A hermaphordite would be a third option, but I think we can reasonably ignore that. So, 2 options means P(G with a sister) = 1/2

If you do care about order and you know at least one is a girl, then the other child could be an older brother, younger brother, older sister, or younger sister. 4 options means P(GG) = P(G with older sister) + P(Girl with younger sister) = 1/4 + 1/4 = 1/2.

Either way you get 1/2.

Again, as I said before, once you have a known and a variable, instead of 2 variables, you have to rewrite your events. You can't just use your old events and remove the ones that don't fit the known or given value. So saying that you know at least one is a girl means you remove BB from the data set of BB, BG, GB, and GG is where the logic is flawed and leads you to the incorrect conclusion of P(G|G) = 1/3. However, it is still true, that if you take a discrete population of any size you will end up with 1/3 of the families being GG if you eliminate all the BB families. That's just a paradox of statistics.

I just gave you 3 solid infallible proofs.

What about this is confusing?

[Guest]

Ahhh, I've figured out a way to explain the paradox that Scraff has pointed out that while I assert P(G|G) = 1/2 it is also true that only 1/3 of the total population of a data set would actually be GG pairs.

Here's where the population vs. probability issues come from.

For the sake of identification, let's label all the indiviuals in the pairs as either child 1 or child 2 (this doesn't indicate birth order, just allows us to select one or the other for clarification)

so child 1 could be a girl or boy and child 2 could be a girl or boy

So our possible combinations are

1=B; 2=B

1=B; 2=G

1=G; 2=B

1=G; 2=G

now if we say that there are 100 families in our data set then we all agree we'd have

25 1=B;2=B pairs

25 1=B;2=G pairs

25 1=G;2=B pairs

25 1=G;2=G pairs

Now lets take a random pair and reveal the gender of one of the children to be a girl and we want to know the probability of her having a sister.

Well our population has been reduced to

25 1=B;2=G pairs

25 1=G;2=B pairs

25 1=G;2=G pairs

so this family must belong to one of these. Now it's been stated we can now easily say that the probability of this girl having a sister is 1/3 because there are 25 GG pairs and 50 alternative pairs. However, we don't know if this girl is child 1 or child 2, so our data set is not yet fully reduced. You must further reduce your data set by assuming that she is either child 1 OR child 2, since she can't reasonably be both or neither. When you do this you end up with one of 2 scenarios. You either have

25 1=G;2=B pairs

25 1=G;2=G pairs

if you assume that the girl is child 1

or

25 1=B;2=G pairs

25 1=G;2=G pairs

if you assume that the girl is child 2

now you can clearly see that the probability of this girl having a sister is 1/2. It doesn't matter if she is child 1 or child 2 so we can assume either, but we must assume one or the other in order to properly reduce our data set. If you don't reduce your data set, you are saying that the single child you know the gender of can simultaneously be both child 1 and child 2, which she can't. She must be one or the other, although both children can still be girls. You can't say that the probability is 1/3 because you don't know which child you know the gender of (child 1 or child 2), so you have to further reduce your subset.

That is where the paradox comes from.

Scraff, I hope I have sufficiently explained where the discrepancy between your solution and my solution came from and why your solution, although seemingly accurate, was not complete enough.

Again, I've explained this using proven statistical calculations (on my previous post), and now using Scraff's own logic. I hope this resolves the issue.

Edited December 31, 2008 by BeastMaster

[Guest]

No, I didn't admit that there is a 1/3 CHANCE. I admit that when you take the original data set and remove the BB or HH pairs, you end up with 1/3 of the resultant population being GG or TT pairs.

That's admitting there is a 1/3 chance.

That is NOT the same as saying the CHANCE of having a GG or TT pair is 1/3 if you know that at least one of the elements of the pair is a G (or T).

Yes, it is.

Let's start over:

Great. Then answer what would happen if we tried this real world experiment.

We'll simulate the question I asked with coins.

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

Let's lay out 100 pairs of coins. Heads we'll call boys and tails we'll call girls. Since in the problem above we know there are no Boy/Boy pairs (there's at least one girl), we'll remove all Head/Head pairs.

Out of those coins left, you believe about 1/2 will be Tail/Tail, correct?

You don't believe the answer is 1/2, but 1/3, correct?

Then how do you explain saying this?:

You tell me you have 2 children. You then tell me that at least one of them is a girl. That means one of two things MUST be true. Either child A is a girl or child B is a girl. They could both be girls, but at least one or the other MUST be true. Now, simple if-then statements will lead you to conclude that...

IF child A is a girl, then child B COULD BE a boy or a girl. There are 2 options, so the probability for girl/girl, in this scenario, is 1/2.

IF child B is a girl, then child A COULD BE a boy or a girl. There are 2 options, so the probability for girl/girl, in this scenario, is 1/2.

Do the coin problem and the Boy/Girl problem receive the same answer of 1/3, yes or no? If no, explain why not.

I asked you this:

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

And you answered this:

"1/2"

Explain why the coin problem gets a different answer when all I did was replace "boy" with "heads" and "girl" with "tails".

[bonanova]

The OP admits to either answer, 2/11 or 1/6, or even 0, in some cases.

The algorithm used by magician [on which OP is silent]

determines whether the eleven outcomes actually have equal probability.

My previous post was too long to read, being written late at night,

so I'll excerpt here how [i see] what the magician was doing could change things.

Consider my simulation problem where magician sees 5-6. He has the

liberty to say "one of your dice shows 5" or "one of your dice shows 6"

My next line of code was going to be a random choice function.

But it could have been pick the first value, or pick the higher value,

or pick a 6 is one is present, else pick a .... whatever!

That's when the light came on.

What you logically can infer from hearing a chosen number depends on how the number was chosen.

So, leave the magician now, and consider the Subject.

By premise, he hears "one of your dice shows 6".

What is he to think?

1. If the subject reasons: if the magician sees that either of my dice is a 6, then he'll tell me about it, then all eleven cases consistent with A=6 or B=6 have equal footing. Because in each of the eleven cases, necessary and sufficient conditions are present for the magician to make his "six" statement -- including the 6-6 case, where magician's algorithm doesn't matter: he must say "six". Thus p[7] = 2/11
   
   
   
2. If the subject reasons: the magician will tell me about the higher/highest of the dice, then, again, all eleven cases have equal footing. Because if either die is 6 it will be the higher/highest [highest applies to 6-6 where neither technically is higher]. Again p[7] = 2/11
   
   
   
3. If the subject reasons: the magician, having not told me how he chooses the number to call out, can be assumed to have made a random selection, then the eleven cases no longer are on equal footing. The subject considers the 6-6 case. If magician saw that, he must call out the six. But if magician saw any of the other cases, say A=5 B=6 or A=6 B=5 then half of the time he will call out the 5. Similarly with 4, 3, 2 and 1. So it is only with 50% confidence that I can place the other 10 possibilities into my outcome table.
   
   That fact changes p[7] into [.5 + .5]/[.5 + .5 + .5 + .5 +.5 +.5 +.5 +.5 +.5 +.5 +1] = 1/6.
   
   
   
4. If the subject reasons: the magician will tell me about the lower/lowest of the dice, then, only the 6-6 case will be mentioned, and since it totals 12, p[7] = 0!
   

Because the OP is silent on the magician's algorithm,

the subject cannot be certain of how to calculate p[7], and neither can we.

Now, I admit did not believe the assertions that the 11 cases are not always on equal footing, given by others. To me they were conclusions with words wrapped around them, rather than clear expositions of the problem from start to finish. Once I came to the line of code in the simulation that addresses it, I had to put it into my own words, and of now course it is perfectly clear.

[Guest]

Ahhh, I've figured out a way to explain the paradox that Scraff has pointed out that while I assert P(G|G) = 1/2 it is also true that only 1/3 of the total population of a data set would actually be GG pairs.

That's a ridiculous contradictory statement! You can't logically say that the probability for GG pairs is 1/2 and at the same time say "1/3 of the total population of a data set would actually be GG pairs."

That's a clear contradiction!

Here's where the population vs. probability issues come from.

For the sake of identification, let's label all the indiviuals in the pairs as either child 1 or child 2 (this doesn't indicate birth order, just allows us to select one or the other for clarification)

so child 1 could be a girl or boy and child 2 could be a girl or boy

So our possible combinations are

1=B; 2=B

1=B; 2=G

1=G; 2=B

1=G; 2=G

now if we say that there are 100 families in our data set then we all agree we'd have

25 1=B;2=B pairs

25 1=B;2=G pairs

25 1=G;2=B pairs

25 1=G;2=G pairs

Now lets take a random pair and reveal the gender of one of the children to be a girl and we want to know the probability of her having a sister.

Read that last sentence again. Is that what I asked? No!

What I asked was:

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

We don't randomly do anything.

Of all couples with exactly two children that have at least one girl, 1/3 will have two girls. You have disagreed and answered 1/2. That's the wrong answer.

[Guest]

That's a ridiculous contradictory statement! You can't logically say that the probability for GG pairs is 1/2 and at the same time say "1/3 of the total population of a data set would actually be GG pairs."

That's a clear contradiction!

It isn't contradictory, look at the proof. If you take the population of all families that have at least one girl, 1/3 of them have 2 girls. However to answer the question of this riddle and the child riddle, if you have a single particular family, and you know that at least one of their children is a girl, then the probability that family has 2 girls is 1/2, which is clear from my proofs.

Read that last sentence again. Is that what I asked? No!

What I asked was:

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

We don't randomly do anything.

Of all couples with exactly two children that have at least one girl, 1/3 will have two girls. You have disagreed and answered 1/2. That's the wrong answer.

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.

I would still like you to refute the statistical equation I posted. All you've done is try and confuse the issue with semantics and changing the question from what was originally asked in the riddle. The question was one of probability of a single particular pairing based on a known value of one of the items in the pair.

I am not being rude, but if you're going to prove a mathematical riddle, please use proofs (as in logic and proof from basic mathematics classes), not incomplete population reductions. You have yet to put forth a true proof and I have put forth 4 different proofs, all resulting in the same result for the dice riddle, coin riddle, and children riddle.

Edited December 31, 2008 by BeastMaster

[Guest]

Of course, another way to look at it is...

Consider the two dice A and B. And at least one of them is a 6.

The probability of A being a 6 is 1/2, and then the probability of making a 7 (B being a 1) is 1/6. So (1/2)(1/6)=1/12

Now the probability of B being a 6 is 1/2, and then the probability of making a 7 (A being a 1) is 1/6. So (1/2)(1/6)=1/12

Now the probability of A and B BOTH being a 6 is probably 1/6 (since one of them must be a 6), but regardless of the initial odds, the probability of that making a 7 is 0. So (1/6)(0)=0.

Now 1/12+1/12+0=1/6, which is the actual probability of having a 7.

EDIT: Clarified "probability"

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.

[bonanova]

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.

An argument given in my previous post shows that you can't be sure p[A=6] = 1/2.

Depending on how magician chooses his number, it could be 1/2 [random choice].

But,

It could be 6/11 [pick higher number or pick 6 if it's there].

It could be 1/36 [pick lower number].

[Guest]

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.

That's not what you said!:

I asked you:

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

You responded:

"1/2"

(See post #58)

Wanna take that back now?

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.

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!

Again, this can be proved with the penny experiment. If we have 75 pennies that are now 25% TT, 25% TH, and 25% HT and number them 1-75, we can ask someone to randomly pick a number from 1 to 75. What are the chances that th number they pick will match up with that numbered penny pair and be TT? That's right- 1/3. According to you it will be 1/2.

I would still like you to refute the statistical equation I posted. All you've done is try and confuse the issue with semantics and changing the question from what was originally asked in the riddle.

I've attempted to confuse nothing and I resent that charge! There you go being rude again! I have responded using plain English that should help anyone understand the correct answer.

[bonanova]

Let me take the moment to wish you all a very Happy New Year.

I'm going out for a walk in some freshly fallen snow.

My planned visit to NYC for celebration there is off,

so I'm making the best of it.

Talk to you all next year.