>>Let's make that a statement:
>>If I pick a pancake at random, look at one of its sides and
>>see the colour X, then there is a two third possibility that
>>the other side has the same colour.
>>
>>Now, if we substitute "brown" for X - just substituting,
>>nothing else - then why should the probability change? And
>>likewise for "gold"?
>
>You can't do that. Say, you are picking a fruit out of a bag
>that contains a peach and a pear. The probability of getting
>a fruit is 1. The probability of getting a peach is 1/2 as
>is the probability of getting a pear.

The comparison is not completely equal to the pancake problem. If you were to quote it correctly, it would be something like: "you grab a fruit out of the bag and see that it is fruit X. Now that you have made this observation, there is a 100% chance that you grabbed fruit X. Now let's substitute "peach" for X."

I got to my approach in the following way:
If you see a brown side, there is 2/3 chance that you have the BB pancake, a 1/3 chance of having the GB pancake and a 0 chance of having the GG pancake.
If you see a golden side, there is a 0 chance that you have the BB pancake, a 1/3 chance of having the GB pancake and a 2/3 chance of having grabbed the GG pancake.

In diagram:

B/B B/G G/G
2/3 1/3 0/3 if we see a brown side
0/3 1/3 2/3 if we see a golden side
-----------
2/6 2/6 2/6 when we add the two situations up and then divide by 2.

I think that this adding up is valid. And if we've done that, we can say that there is a 2/3 chance that we've drawn a monochromatic pancake. If we then substitute a colour for "colour X", we fall back to one of the two situations that made up the addition.

OK, maybe I am cheating, but I don't think that I'm doing it to such an extent as you are suggesting with the fruit example.

>
>>You can test the thing with the help of a pack of cards and
>>a friend. Take three red and three black cards out of the
>>pack. Ask your friend to make three stacks of them; one with
>>two red cards, one with two black cards and one with one red
>>card and one black one. Then draw one card out of one stack
>>at random and look at its colour. Then check the other card
>>out of that stack and see whether the colour is different or
>>equal. Repeat this a large number of times and count how
>>often the colour is the same.
>
>This is exactly the case when a wrong argument leads to the
>right answer. For a multitude of other examples see
>Mathematical Fallacies, Flaws and Flimflam by
>Edward Barbeau.

Oh, Alex, you do have to explain this. If we glue the backs of the two cards in each stack together, we have the same situation as with the two-sided pancakes, but if we leave the cards unglued, it is a different situation? Sorry, but I don't buy that.
In the experiment above I did not suggest gluing the cards together because otherwise the diamond 7 would always be paired with the 9 of hearts (substitute other cards at will).

Whymme