Thus, there is a 2/3 chance that you drew the B/b pancake and are looking at one of its sides.

Whymme

I apparently was, but no longer. Thank you for correcting it. At the time I was probably too drowsy to answer intelligently.

the answer is 1/2

your can exclude pancake with two golden sides (g/g)

now i will use different notation for the pancakes: b/b is the pancake with two sides brown

you are either seeing the b side of the b/g pancake
or one of the b sides of a b/b pancake

however the b sides are identical (they are just described as brown)
picking up the b/b pancake either side up is still picking the same pancake
so in reality there are 2 pancakes you can choose

one of them gives the wanted result

therefore the answer is 1/2

Why are the possibilities equal?

It was not my intention to bring more confusion, but it turned out to be the outcome. We have once and for all to come to some agreement,(but judging by Monty Hall problem, I doubt that we will).

It is clear that the colors of the pancakes are symmetrical.
So, if we ask the same question for the other color, we have to get the same probability. If we ask a "colorless" question: i see a side and now what is the probability that the other side is the same color?
then the probability is 2/3, for 2 out of 3 have the same color on both sides.
This probability is an "or" of two color specific probabilities, and
it makes each of them equal 1/3.

So which one is correct? 1/3 ? 1/2 ? 2/3 ?

The Brown/Gold cake can be positioned with either color up (the first two)
The Brown/Brown cake can be positioned with either Brown side up (the second two) and
The Gold/Gold cake likewise can be positioned with either Gold side up (the third two)

One out of three initial choices will be Brown, and if so, of the three possibilities. two of them
are Brown on the bottom side..

So, the 2/3 answer is correct.

We have the three differently coloured pancakes. You see one side of one pancake, and it has colour X (let's leave in the middle whether this is brown or gold). As you say, there is a 2/3 probability that the other side of that pancake is also colour X; two of the three pancakes have the same colour on both sides, after all.

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 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.

Whymme

Thank you for the response. I'm still not convinced. You slightly
misinterpreted what I said: prob. 2/3 is total prob. for both colors,
(i.e. prob. that the color on the other side is the same as the one you see, and it is 2/3 just because there are 2 out of 3 cakes colored this way. In case of 3 colors there would be AA,BB,CC,AB,AC,BC
6 cakes and the prob. of the other side being the same 1/2 or 3 out of 6 for all the colors combined). So for a specific color the total prob. should be devided by number of colors, for they "OR"ed into the
total prob. because of the symmetry.

If you look at the Bayes probabilities for the 3 cakes

P(BB)=1/3 prob to get Brown/Brown cake
P( not BB)=2/3 prob of not BB cake

P(side B/BB)=1 you are looking at Brown side
P(side B/not BB)=1 you are looking at Brown side

Then

P(BB/side B)=P(side B/BB)* P(BB) / ( P(side B/BB)*P(BB)+P(side B/not BB)*P( not BB) )= 1*1/3 / (1*1/3+1*2/3) = 1/3

It'seems the trick is in the fact that once you picked the cake and
see a side there is no flexibility on choosing the other: it is already chosen and the prob. is 1 out of 3 for 2 colors or 1 out of 6 for 3 colors.

Please correct me if I'm still wrong.

Thanks.

BG

Why? There are two pancakes GG and GB with four sides, so that

P(side B/not BB) = 1/4.

If you substitute that into your Bayes formula

> P(BB/side B)=P(side B/BB)* P(BB) / ( P(side
>B/BB)*P(BB)+P(side B/not BB)*P( not BB) )

you get

P(BB/side B)	= 1·1/3 / (1·1/3 + 1/4·2/3)
	= 1/3 / (1/3 + 1/6)
	= 1/3 / (3/6)
	= 6/9
	= 2/3

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.

>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.

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

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.

You take one pancake and look at one side of it. You can have two situations: either that side is brown or it is gold. Those situations are mutually exclusive and together they add up to 100%. I thought that in those situations those chances can be added up. I may be mistaken, however, it's been years since I studied maths or statistics other than just as a hobby.

>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.

The addition was how I got to the idea of getting the general colour X, for which we can then substitute brown or gold. IMHO my addition shows how the general situation of drawing colour X can be split up in the specific situations of either drawing brown or gold.

The idea of this general situation is also in a way a validation of the method of the repeated experiment, with the three stacks of cards, that I described.

>
>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.

Hehe. There is a mistake in your example here. Do you want to find it yourself or do you want me to tell you? I wrote a hint at the bottom of my post, but I guess that you can find out the mistake without it.
If you correct the mistake, you'll find that my addition still works

>
>>Oh, Alex, you do have to explain this.
>
>I've no problem with your stack experiment. It'should lead
>to a correct estimate.

You did call it "exactly the case when a wrong argument leads to the
right answer", though. Or did I misread your meaning?

Whymme

PS: Alex, please don't take this post the wrong way. As you may have gathered, I love this site and I respect your knowledge of mathematics and your drive in making this all accessible for the general internet public. It's just that in this case I don't agree with your reasoning, but that doesn't diminish my respect in any way.
I did not mention what was wrong (IMHO) with your example of three colours because I have the idea that you like to figure it out for yourself and I didn't want to deny you that pleasure. If I'm wrong here, I'd like to apologise.
You might want to delete this PS; it is meant as a personal message anyway and not as something that is needed to be communicated to anyone who reads these discussions.

Hint for the three colour pancake mistake: look at the situation with two colours: do we have three pancakes B/b, G/b, G/g or do we have four pancakes B/b, G/b, B/g and G/g?

You again mix right reasoning with wrong. The above is unquestionably right. But it does not explain why adding two rows in your table makes sense, especially because the probabilities add up to 200%!

Also, the experiment would lead to a correct estimate regardless of whether your addition is right or wrong. The fact that it does does not prove the addition is meaningful as it presented.

>I may be mistaken, however,
>it's been years since I studied maths or statistics other
>than just as a hobby.

That's OK.

>If you correct the mistake, you'll find that my addition
>still works

Does not mean anything. It may work for a wrong reason.

>
>You did call it "exactly the case when a wrong argument
>leads to the
>right answer", though. Or did I misread your meaning?
>

Why, this is how it reads.

>PS: Alex, please don't take this post the wrong way.

I do not know what that might mean.

>It's just that
>in this case I don't agree with your reasoning,

That's OK.

>but that
>doesn't diminish my respect in any way.

Just this once: please do not make it personal.

>You might want to delete this PS;

I can't. I can only delete whole messages.

>it is meant as a personal
>message anyway and not as something that is needed to be
>communicated to anyone who reads these discussions.

Sorry about that.

I'm sorry I've started this. I did not expect this will venture
outside of pure probabilities.

What makes me feel uncomfortable with the cakes is the fact that
when added across colors the probality goes over 1 (2/3+2/3 for 2
colors). Maybe it includes twice some combined colors probability?
There must be something.

Also, my mistake on asigning prob 1 to conditional P(B/notBB).
I wrongly thought that if we see Brown color, then it is 1. I am
corrected it is 1/4 (there are 4 sides on 2 not BB cakes and only
one side is B).

So it turns out that the probability in question will always be
2 / ( #of colors+1 ), which corresponds to having 2 cakes of the
same color on both sides and cakes of this color on one side and all
other colors on the other side ( AA,AA, AB, AC, AD, AE, AF -> 2/7 for
six colors A,B,C,D,E,F)

Bob G.

>>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?

Well, because of the following:

Let's take the following statements as correct:
1. The probability of having chosen a certain pancake is:
B/B B/G G/G
2/3 1/3 0/3 if we see a brown side

2. The probability of having chosen a certain pancake is:
B/B B/G G/G
0/3 1/3 2/3 if we see a golden side

3. If we randomly grab a pancake and check one of its sides, there is a 50% chance of seeing a golden side and 50% chance of seeing a brown side.

If we assume the above, if we grab just any pancake and show one side of it, the chance that we've taken the B/B pancake is (2/3)*50% + (0/3)*50%. Together this is a chance of 1/3.
The chance that we've taken the G/B pancake is (1/3)*50% + (1/3)*50%. Together this is a chance of 1/3.
The chance that we've taken the B/B pancake is (0/3)*50% + (2/3)*50%. Together this is a chance of 1/3.

Herewith the computation I've shown in a previous post is proven. I probably shouldn't have called it "addition"; that was a bit too simplistic.

We can work from the other side as well. If we start with the following assumptions:
1. There is a 50% chance of seeing a golden side and 50% chance of seeing a brown side.
2. We choose a pancake randomly, with an equal chance of each pancake to be chosen.
There is thus a 2/3 chance that we choose a monochromatic pancake.

Furthermore, the chance that we choose a certain pancake is X(brown)*50% + X(gold)*50%,
where X(colour) is the chance that a certain pancake is taken, and that colour is shown.
We have stated that this chance is 1/3 for each pancake.

In the case of the B/B pancake, the total chance of having grabbed that one is 1/3. We can see that for this pancake, the chance X(gold) is zero. So X(brown)*50% + 0 = 1/3
Therefore, it follows that X(brown) = 2/3

We can make the same computations for the other pancakes. This will produce the scheme above the line in the "addition" above.

And this is why.

Whymme