With both the Fortran simulation example and the Bayes Theorem quotation, the event space is skewed because again, it assumes that BG & GB are separate events, but that GG is only 1 event, even though the 2 Gs are separate entities (i.e. G1G2 & G2G1 are separate events in the event space).

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If

and

are seperate,

then you could model the situation as follows.

Of the 4 children Helen, Mary, Roger and Sean,

in how many ways can you form 2-child families with these children ?

This will be the "Sample Space"

The first child will be the younger one

**Helen Mary**

Mary Helen

Helen Roger

Roger

**Helen** **Helen** Sean

Sean

**Helen** **Mary** Roger

Roger

**Mary** **Mary** Sean

Sean

**Mary**
Roger Sean

Sean Roger

If Helen is the younger child, there is a 1/3 chance she is the younger child of 2 sisters.

If Mary is the younger child, there is a 1/3 chance she is the younger child of 2 children.

If the younger child is a girl, there is a 2/6 chance she is the younger of 2 sisters.

If Helen is the older child, there is a 1/3 chance she is the older child of 2 sisters

etc etc.

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Therefore I stick with my original proof:

I define 2 sets c1 & c2, each containing the set of possible states each of the coins can take after each is thrown:

c1 = {H1, T1}

c2 = {H2, T2}

Throwing both coins at the same time gives the following set of possibilities:

{H1, H2}

{H1, T2}

{H2, T1}

{T1, T2}

So I throw the 2 coins so that you can see them when they land, but I cannot. You tell me that one of the coins is in a state that is a member of the set {T1, T2}. You then ask me the probability that the other coin is in a state that is also a member of this same set. The set {H1, H2} is therefore eliminated, leaving the following possibilities:

{H1, T2}

{H2, T1}

{T1, T2}

**3 equally likely scenarios with a 1/3 chance of TT**
But if a throw of c2 gives T2, then {H2, T1} is eliminated, so there remains a 50% chance that c1 = H1 and a 50% chance that c1 = T1

**One way we can model this with children is to tie a girl and boy back to back using rope.**

With both boys tied to both girls facing in opposite directions, we can turn them around and around

and stop them in an unbiased way.

Now, suppose Helen is tied to Roger and you are told that after spinning, Helen is facing the front.

In this way, we can get Roger out of the equation.

After spinning the other pair of children around, we can say that Mary and Sean

have equal likelihood of facing the front also.

Then the probability of Mary facing the front is unrelated to the probability of Helen facing front and is 0.5

Conversely, if a throw of c1 gives T1, then {H1, T2} is eliminated, so there remains a 50% chance that c2 = H2 and 50% chance that c2 = T2.