>>>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.

If you assertion is based on the fact that there are two monochromatic pancakes against one that is not, then the reasoning is misleading. For, in that case X stands not for an arbitrary but fixed color, but for any color. The difference is the same as in the common algebraic usage between constants a, b, c, ... and variables x, y, z, ...

>>>
>>>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."

This is the hardest part making interpretations of somebody else's words. You may take to heart the fact that two people (I and Bob) might have misunderstood your meaning.

>
>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.

Why?

>And if we've done
>that, we can say that there is a 2/3 chance that we've drawn
>a monochromatic pancake.

There's indeed an a priori 2/3 chance of drawing a monochromatic pancake simply because there are two of them among three. Your addition does not help to comprehend this simple fact.

Let's follow Bob's idea of having say three colors and all possible combinations by two. There are 9 in all, of which 3 are monochromatic. Then the a priori probability of having a monochromatic pancake is 3/9. But, once an A is drawn, the probability that it's a side of AA is 1/2, because now you draw from AA, AA, AB, and AC.

>Oh, Alex, you do have to explain this.

I've no problem with your stack experiment. It should lead to a correct estimate.