[bonanova]

The storied probability question of the two-child family with at least one son was

a while ago in original form, and again lately with a few twists.

In the latter post, I suggested /page__view__findpost__p__247003 about the

information given in the puzzle, and in some cases the probability asked for might be 1/2.

In the first topic I had argued most vehemently for 1/3, although I never stated the

assumptions that are needed to support that answer.

Assumptions were also debated in where I asked the probability of

rolling a 7 with a pair of fair dice, making a specious argument that it is 2/11, not 1/6.

I invited a refutation. It comprised more than one hundred posts that demonstrated

assumptions affect the answer.

If assumptions do affect answers, then either (a) puzzles of this type are hopelessly

indeterminate, or (b) there is a best, unwritten perhaps, set of assumptions that we

should invoke, when none are stated, that will give these puzzles well defined answers.

We don't like (a), so let's assume (b).

This topic asks: what is that best set of assumptions?

Consider:

An trusted observer looks at a situation [children, dice, whatever] and makes a statement:

e.g. "At least one child is a girl," or "at least one die is a six." What might we now assume?

Here are some possibilities.

[1] It is the only statement the observer could make; that is, it's a complete description.

[2] There are other relevant statements that are true, and the observer chose from them at random when he spoke.

[3] There are other relevant statements that are true, and the observer chose from them with some type of bias.

Alternatively, there is a position that says "We should make no assumption at all."

But does that position serve to remove ambiguity?

A starting point could be /page__view__findpost__p__247110.

Enjoy.

[Guest]

I didn't even notice this offshoot until just now.

That rolling 7 problem is giving me a lot to think about.

I'll get back to this when I feel satisfied with my understanding of it.

I don't really want to wade through 106 pages, so I'll just think about it on my own .

[Guest]

Just a thought for the two children, at least one is a boy, what is prob that both are boys:

Rule 1: If the written question does not make any distinction between children, then the solver should assume that the informant's method makes no distinction between children either.

Of your methods (included here): only #2 and #3 make no distinction between children.

Monkey wrench, anyone?

This question was

with yours truly and others insisting the answer is 1/3.

However, consider the following argument that we don't have enough information!

That is, there is a built-in, unspecified probability linking the information, given to us

by our informant, to the gender distribution, that is known by the informant. That is,

Our informant followed some [unspecified] algorithm when s/he gave us the information.

Possibilities include, but are not limited to the following:

1. I will pick a child at random. If it's a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter.
   Probability of two sons is 1/2.
   
   
2. If either child [in the inclusive sense] is a boy, I will say "at least one child is a son."
   Otherwise I will [have to] say "at least one child is a daughter."
   Probability of two sons is 1/3.
   
   
3. If both children are boys, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1.
   
   
4. If the older child is a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1/2.
   
   
5. If the taller child is a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1/2. Maybe. See note following case 7.
   
   
6. If the child whose name comes before the other's in the alphabet is a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is 1/2.
   
   
7. I will ask the children to raise their right hand. If the hand that is raised first belongs to a boy, I will say "at least one child is a son."
   Otherwise I will say "at least one child is a daughter."
   Probability of two sons is unknown.

Why is case 7 unknown? The speed with which a boy raises his hand might differ from a girl's speed.

We can't assume that in the mixed cases of BG and GB, that a boy's hand will be raised first 50% of the time.

This is a case where the elements of the sample space don't necessarily have equal a priori probability.

This might apply to case 5 as well: if all boys are taller than all girls, then case 5 becomes 1/3, equivalent to case 2.

If some girls are taller than some boys, but boys tend to be taller than girls, then case 5 lies somewhere between 1/2 and 1/3.

Why method 3 should not be assumed:

Rule 2: If the solver must decide between two informant methods where, given all available information, the set of possible outcomes of one method is a subset of the set of possible outcomes of the other method, the solver should assume the method corresponding to the superset.

method 3 has the following possible values {MM} given "at least one son" was communicated

method 2 has the following possible values {MF,FM,MM} given "at least one son" was communicated

method 2's possible values are a superset of method 3's, so method 2 should be assumed over method 3.

There is no reason for the solver to assume that the informant used a method in which the outcomes {MF, FM} can never be realized given the information.

This rule is slippery, however.

if we modified the question to say:

a man has two children, at least one is a daughter, what is the probability that the man has two daughters?

According to my rule 1, the solver should assume method 2 or method 3.

According to my rule 2, the solver should in this case assume method 3, not method 2.

[Guest]

Bonanova,

I'm sorry I'm not going to take the time to read all the posts in that last discussion topic (rolling a 7), but I don't see what's wrong with any of the logic. The question you had posed simply asks what's the conditional probability that die 1 plus die 2 is 7 given that die 1 or die 2 is 6--that is, P(A+B=7|A=6 or B=6). The math works out to 2/11 because of the condition. (And of course, this holds for all other possibilities in addition to 6). Math below:

P(A+B=7|A=6 or B=6) = P(A+B=7 and (A=6 or B=6))/P(A=6 or B=6)

P(A+B=7 and (A=6 or B=6)) = P(A=6 or B=6|A+B=7)*P(A+B=7) = 2/66 * 6/36 =2/36

P(A=6 or B=6) = 11/36 (just draw a table and count if you don't believe me)

P(A+B=7|A=6 or B=6) = P(A+B=7 and (A=6 or B=6))/P(A=6 or B=6) = (2/36)/(11/36) = 2/11

And that calculation includes the fact that P(A+B=7)=6/36=1/6.

I don't see how this involves any sort of assumption on the part of the reader or anything like you're talking about here.

[Guest]

Bonanova,

I'm sorry I'm not going to take the time to read all the posts in that last discussion topic (rolling a 7), but I don't see what's wrong with any of the logic. The question you had posed simply asks what's the conditional probability that die 1 plus die 2 is 7 given that die 1 or die 2 is 6--that is, P(A+B=7|A=6 or B=6). The math works out to 2/11 because of the condition. (And of course, this holds for all other possibilities in addition to 6). Math below:

P(A+B=7|A=6 or B=6) = P(A+B=7 and (A=6 or B=6))/P(A=6 or B=6)

P(A+B=7 and (A=6 or B=6)) = P(A=6 or B=6|A+B=7)*P(A+B=7) = 2/66 * 6/36 =2/36

P(A=6 or B=6) = 11/36 (just draw a table and count if you don't believe me)

P(A+B=7|A=6 or B=6) = P(A+B=7 and (A=6 or B=6))/P(A=6 or B=6) = (2/36)/(11/36) = 2/11

And that calculation includes the fact that P(A+B=7)=6/36=1/6.

I don't see how this involves any sort of assumption on the part of the reader or anything like you're talking about here.

Let's say that one die is red, and the other is blue.

Let x be a number between 1 and 6 inclusive.

Case A:

What if you are told that the red die is x.

What then is the probability of having 7 as a sum?

The answer is 1/6.

Because....

This translates to: we have a blue die, what is the probability that the blue die has a value of 7-x? This probability is 1/6.

Case B:

What if you are told that the blue die is x.

What then is the probability of having 7 as a sum?

The answer is 1/6.

Case C:

What if you are told that at least one of the dice is x.

As you just showed, the probability of having 7 as a sum is 2/11.

Where is the fundamental difference here?

What if the magician/informant looked only at the value of the red die without looking at or considering the blue die at all, but then told the solver that at least one of the dice is x (not mentioning the color red).

Would this be case A or case C?

If the solver believed that this is what the magician/informant did, then the solver would give an answer of 1/6 as this would be equivalent to case A.

I can't think of a good scenario equivalent to case C, in which the magician acts on his own without the solver asking a question.

He cannot randomly pick numbers until he gets one that matches one of the dice and then tell the player that at least one of the dice has that value. If he does this, then the solver would actually give an answer of 1/6 still.

Here is why: The informant will eventually say: at least one of your dice has the value v.

The probability of the two dice having the same value is 1/6.

The probability of the two dice having different values is 5/6.

If the dice have different values than the informant must pick one of the two values. Therefore the 5/6 probability for each such event is reduced to 5/12.

If ! = not

Notation: (Dice 1's value, Dice 2's value)

P[(v,v)] = 1/6

P[(v,!v)] = P[(!v,v)] = 5/12

(v,!v) consists of 5 outcomes, one for each value of !v

So an individual outcome has probability 1/5*5/12 = 1/12

Similar reasoning goes for (!v, v).

As far as sums are concerned, (x,y) is equivalent to (y,x).

Therefore if x != y, then the probability that the sum is x+y is the sum of the probabilities of the two outcomes: (x,y) and (y,x).

If x=y, then the probability is the sum of the single outcome.

So if the magician says that at least one die is v,

then the solver assumes that the sum can be 1+v to 6+v.

1+v has two outcomes of 1/12 each, with net probability 1/6

2+v has two outcomes of 1/12 each, with net probability 1/6

...

v+v has one outcome of 1/6, with net probability of 1/6

...

6+v has two outcomes of 1/12 each, with net probability 1/6

For any value of v, 7 will be one of the possible sums, and as shown above, it will have a value of 1/6.

--------

Anyway, I assume there must be some method the magician can do to transfer the information such that the problem is equivalent to case C.

If the solver believed that this is what the magician/informant did, then the solver would give an answer of 2/11.

The solver's belief in how the magician obtained and transferred the information thus has an effect on the solver's probability calculations.

Edited September 26, 2010 by mmiguel1

[bonanova]

Bonanova,

P(A+B=7|A=6 or B=6) = P(A+B=7 and (A=6 or B=6))/P(A=6 or B=6) = (2/36)/(11/36) = 2/11

And that calculation includes the fact that P(A+B=7)=6/36=1/6.

I don't see how this involves any sort of assumption on the part of the reader or anything like you're talking about here.

It doesn't, and we don't disagree. We agree that p[7|"one of your dice is 6"] = 2/11.

It's rather the 100^th contestant that makes a statement we disagree with.

by specious argument, that p[7|No Statement] = 2/11.

That's clearly wrong: p[7] unconstrained is 1/6.

So, what's wrong with his argument? Where's the hole?

The hole is that C₁₀₀ makes an assumption. One that is not warranted.

So let's find it.

Without loss of generality, say the dice are 3 5.

"Whatever number you say, I already know the calculated probability is 2/11."

That statement is flawed. It's true only if the magician chooses 3 or 5.

If so, then he gets p[7|"One die is 3" or "One die is 5"] = 2/11.

But to assume it's from {3,5} is not warranted.

If we make no assumptions, the magician will choose from {1,2,3,4,5,6} with equal likelihood.

Note that it's less likely to be from {3,5} than from {1,2,4,6}.

So let the unconstrained choice be x. We know that for any x

. . . . . 11/36 of the time it will be on one of the dice. Call this the Yes case. p[Yes] = 11/36

. . . . . 25/36 of the time it will be on neither of the dice. Call this the No case. p[No] = 25/36

2 of the 11 Yes cases total 7. If he chose 3 [one of the dice] they are 3-4 and 4-3. p[7|Yes] = 2/11

4 of the 25 No cases total 7. If he chooses 4 [neither of the dice] they are 1-6, 6-1, 2-5 and 5-2. P[7|No] = 4/25

To get p[7|No Statement] C₁₀₀ must take a weighted sum of these probabilities.

p[7|No Statement] = p[7|Yes] x p[Yes] + p[7|No] x p[No].

. . . . . . . . . . . . . . . .= 2/11 x 11/36 + 4/25 x 25/36 = 2/36 + 4/36 = 6/36 = 1/6.

That's what C₁₀₀ should expect.