## Pancake Problem

A forum for good logic/math puzzles.

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H2SO4
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### Pancake Problem

You have three pancakes. The sides of them are colored as followed because you can't cook.

Black/Black
Black/White
White/White

You put these in the oven for later to keep them warm. Later, you reach into the oven and randomly pick a pancake, putting it down on the plate without looking at the other side. You see that the upside is black. What is the probability that the other side is black?

Doozer
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One half.

Edit: No.

Edit2: Two thirds?

Xanthia
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ooo...can you use... a "!"to work it out? They're called factorials...right? or does that not work to take into account both sides?
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ulnevets
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why's it always gotta be the blacks and the whites? can't we all just get along?

Peshmerga
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1/2 .... I think... D:
i hurd u liek mudkips???

no-genius
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Once you pick up a pancake, it either has black the other side or white - you can't pick up both sides of the black/black pancake, so this doesnt matter. so the probability is 1/2.
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GreedyAlgorithm
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Have a pancake.
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moopanda
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I prefer my pancakes neither black nor white, but a goooolden brown...

Bigspring
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A noble attempt at recreating the Monty Hall problem though.

moopanda
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You have three pancakes. The sides of them are colored as followed because you can't cook.

Goat/Goat
Goat/Car
Car/Car

mmmm.... Goat pancakes...

SpitValve
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moopanda wrote:You have three pancakes.

You have two cows.

(no let's not go there...)

aaronspook
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It's 2/3. Put it this way: There are 6 pancakes. Each pancake is linked to another one, A-B, C-D, and E-F. So if you get Pancake A you also get Pancake B, and so on. A-C are white and D-F are black. If you grab a pancake at random, and it happens to be black, what are the chances the other pancake you get will also be black?

Well, you got one of D, E, or F, with equal probability. Of those, 2 will result in you getting another black pancake, and 1 will get you a white pancake, so the probability must be 2/3.

The important leap of logic in the original problem is that, knowing you're looking at a black side, you are not just as likely to be looking at the second pancake as the third, since the third has 2 black sides and the second only has 1.

Peshmerga
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Are we taking into account that the specific "sides" of the pancake affects the probability? Like, there's a "down" and an "up" side to each pancake? Or can the sides be transversed. Iono, it seems to me like the answer remains 1/2... then again that is probably the most logical, intuitive answer that most people assume.

I'm sorry I'm normally disinclined to theoretical probabilities
i hurd u liek mudkips???

bv2
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Sorry, I believe it is 1/2.

In general the conditional probability of an event A, given that event B has already occurred is given by:-

Pr (A | B) = Pr(A ∩ B) / Pr(B)

Whereupon Pr(A ∩ B) notates the probablity that both occur together.

Therefore,

Let Pr(Black underneath) = Pr(BuN) = 1/3
Let Pr(Black upside) = Pr(BuP) = 2/3

Pr(Black underneath|Black upside) = Pr( BuN∩BuP) / Pr(BuP)
= (1/3)/(2/3)
= 1/2

EDIT: Just saw solutions. Hah, boy was I wrong. Assumed the sides were set with the given information as the former is above and latter is below, my bad.

moopanda
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bv2 wrote:Let Pr(Black underneath) = Pr(BuN) = 1/3
Let Pr(Black upside) = Pr(BuP) = 2/3

How do you figure? The probability of either side being black is 1/2 (3 black sides, 3 white sides). Note that this is without any additional information.

(1/3)/(1/2) = 2/3 as required

Gemini25RB
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I hate the Monty Hall type problems. In fact, I really dislike probability. My gut says its 1/2, but previous experience tells me it's 2/3. I think.
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moopanda
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Gemini25RB wrote:experience tells me it's 2/3

You must be a horrible cook.

no-genius
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bv2 (corrected) wrote:Sorry, I believe it is 1/2.

In general the conditional probability of an event A, given that event B has already occurred is given by:-

Pr (A | B) = Pr(A ∩ B) / Pr(B)

Whereupon Pr(A ∩ B) notates the probablity that both occur together.

Therefore,

Let Pr(Black underneath) = Pr(BuN) = 3/6=1/2
Let Pr(Black upside) = Pr(BuP) = 2/3

Pr(Black underneath|Black upside) = Pr( BuN∩BuP) / Pr(BuP)
= (1/2)/(1/3)
= 2/3

Thanks though, I'd forgotten the Pr(A ∩ B) formula
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The Mighty Thesaurus wrote:Why? It does nothing to address dance music's core problem: the fact that it sucks.

Gemini25RB
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moopanda wrote:You must be a horrible cook.

Hey, I can make EasyMac, Spaghettios, and Ramen Noodles. And hot chocolate. What else could you want?
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offramp13
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1/2

bv2
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It turns out the answer is 2/3 if the sides are not predetermined by the given information. However, if it is assumed that the given information alludes to the former being the upside and the latter being the bottom side, then the answer is 1/2.

IE.

If black/white = black and white on EITHER sides of the pancake, meaning that it may be switched, and similarly described for the rest of the pancakes

BUT, if black/white = black on top, white on bottom, and similarly described for the rest of the pancakes

Zink
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Another (different!) explanation for this.
Assume you are blind, and you have a friend with you that can see. You pick out a random pancake, and let your friend look at it's upside. Your friend says: "I can see the colour of the upside of this pancake. I shall now denote the colour as C."

What is the probability that the pancake's downside is of colour C?
I think it is agreeable that the probability is 2/3. The probability that for the pancake you chose, the downside is the same colour as the upside, is the probability of choosing either the black/black or white/white pancake = 2/3.

Let us now assume that our friend didn't denote the colour as C, but rather, say, Black. Our friend tells us that he sees that the upside is Black, and therefore the downside has probability 2/3 of being Black.

I am quite positive that this proof is valid, though it feels a bit like cheating, as by saying that the upside is black, you take out the case white/white which we considered in calculating the probabilites. But black is just a denotion, and we could just as well switch the colours on all of the balls.

I prefer looking at this problem as three boxes, each box containing two balls, coluored similiarly to the pancakes.

SpitValve
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Zink wrote:...each box containing two balls, coluored similiarly to the pancakes.

That's what she said!

Penguin
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Zink wrote:... and we could just as well switch the colours on all of the balls.

Wow, your subject changed halfway through that sentence from pancakes to balls.
<3!

Zink
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Penguin wrote:
Zink wrote:... and we could just as well switch the colours on all of the balls.

Wow, your subject changed halfway through that sentence from pancakes to balls.

Actually, I changed the subject in a different sentence. Notice that the first sentence of the paragraph refers to pancakes, while the second refers to balls.

This is why one shouldn't add sentences to his post after finishing typing it. Damned chronological order...

demeteloaf
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Easiest way to look at it in my mind.

With 3 pancakes, there are 6 "sides" overall: 3 of the sides are white, 3 of the sides are black.

Of the 3 black sides, 2 out of the 3 have the other side being black as well.

Therefore, the probability is 2/3 that the other side is black.

Gemini25RB
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Wait...

If a pancake has a white side, doesn't that generally mean that it isn't fully cooked? (I've never seen a white pancake, but I have seen golden yellowish.) So, assuming that we have pancakes such that 3 sides are undercooked and 3 sides are burnt.

Now consider the undercooked pancake. It is, by definition, gloopy. Hence, the pancake cannot retain the same form for long. And even if it does, the sides are somewhat sticky, so it has some adhesive properties.

Now when we stick the three stated pancakes into the oven, the important question becomes what objects were they placed on? If they were put onto the metal bars, then one pancake will have glooped through to the bottom of the oven, and then no one would ever find it. If the pancakes were stacked, the stack will exhibit an adhesiveness that makes me question whether one may take only one pancake out (unless the B/B pancake was on top, resting on the B side of the B/W pancake).

(Later, when you grab the pancake and put it onto a plate, the dropping mechanics would definitely tell you whether the pancake was gloopy on one side or not. If it was, the pancake would stick to the plate, and exhibit a negligible "bounce" off the plate. If it were black, the pancake will bounce a little bit.)

So, in conclusion, I must say that since the B/W cannot be logically stacked or rested on anything in the oven so that the pancake can easily move AND it is black side up, the probability of the seized B/? pancake being B/B is actually 1.
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LE4dGOLEM
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... Imagine it's H/T on a coin then. This is not a discussion about pancakes, rather a discussion about probabilities in an imaginary perfectly logical world where, for some reason or another, pancakes can be overcooked on one side AND undercooked on another side.
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skegg
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### Re: Pancake Problem

okay... this is my first time on the boards, i was just looking for the meaning of #472 and this thread drove me nuts, it's such an easy problem... but everyone is messing it up

proof(who am i kidding, i don't know what form a proof should be in)

if you pull a pancake with one black side.... your options are limited to two pancakes... one has a black downside and the other has a white downside.... your knowledge of the side you can see eliminates the third option.... the probability is 1/2... and anyone that says differently is thinking way too hard

peace.... oh and if anyone knows what #472 is about i'd love to know.

PeteSF
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### Re: Pancake Problem

Hi skegg,
skegg wrote:if you pull a pancake with one black side.... your options are limited to two pancakes... one has a black downside and the other has a white downside...

Yes, your options are limited to two pancakes, but not with equal probability. Knowing that one side is black not only means eliminates the white/white pancake, it also means that you are twice as likely to have chosen the black/black pancake than the black/white pancake.

oh and if anyone knows what #472 is about i'd love to know.

It's apparently a parody of a post-postmodern novel, "House of Leaves". I'm still a little freaked out (by the comic - I haven't read the book, and now I don't dare to!)

ConMan
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### Re: Pancake Problem

pollywog wrote:
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I want to learn this smile, perfect it, and then go around smiling at lesbians and freaking them out.

bogglesteinsky
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### Re:

Peshmerga wrote: Like, there's a "down" and an "up" side to each pancake?

What are you talking about? Pancakes don't have any "down" sides. They're awesome.

But yeah, 2/3

skegg
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### Re: Pancake Problem

ugh... i knew i hated probability for a reason...

now isn't there such a thing as "aparent odds" versus "actual odds"?

like if i get a black pancake.... to me there are two equally likely outcomes.... wait, my prob and stat class is coming back to me now... is equally likely the right phrase? all outcomes are equally likely.... that sounds right

Moonbeam
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### Re:

For anyone still struggling to grasp the idea of the answer being 2/3, I think that aaronspook's reply explains it in laymen terms quite well:

aaronspook wrote:It's 2/3. Put it this way: There are 6 pancakes. Each pancake is linked to another one, A-B, C-D, and E-F. So if you get Pancake A you also get Pancake B, and so on. A-C are white and D-F are black. If you grab a pancake at random, and it happens to be black, what are the chances the other pancake you get will also be black?

Well, you got one of D, E, or F, with equal probability. Of those, 2 will result in you getting another black pancake, and 1 will get you a white pancake, so the probability must be 2/3.

The important leap of logic in the original problem is that, knowing you're looking at a black side, you are not just as likely to be looking at the second pancake as the third, since the third has 2 black sides and the second only has 1.

Lord Aurora
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### Re: Pancake Problem

Possibilities:

1) You are looking at side 1 of the black/black pancake.
2) You are looking at side 2 of the black/black pancake.
3) You are looking at the black side of the black/white pancake.

Which of those possibilities plays out to the other side being black? 1 and 2. How many is that? 2. How many total possibilities were there? 3. What is the probability* that the other side is black? 2/3.

*Technically, it’s not a probability, as it’s already happened. The correct question is, “How certain can you be that the other side is black?” but I don’t want to confuse people any more than they already are.
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Puck
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### Re: Pancake Problem

I knew the answer was 2/3 as soon as I saw the question, but I'm going to explain it in slightly different terms, so that it might resonate with some of the people that are still confused.

Forget the all-white pancake, we're looking at a black side, so there could have been 50 all-white pancakes in the (enormous) oven, we didn't pick one of them.

So, we picked, uniformly at random, either a B/B pancake or a B/W pancake.

For either pancake, we also put it with one side or the other facing up, uniformly at random.

If we picked the B/B pancake, what is the probability that we pulled it out black side up?

If we picked the B/W pancake, what is the probability that we pulled it out black side up?

So, to summarize, we could have been looking at any one of six "sides" when we pull the pancake from the oven. Three of those are white; we eliminate those, since we see black. Of the three black sides, two of them belong to the B/B pancake.
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Stev
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### Re: Pancake Problem

Since you took a pancake with a black side, the choice is restricted to:

Black/Black pancake (B/B)
Black/White pancake (B/W)

You're looking at one side so it can be either

side 1 of B/B
side 2 of B/B
side 1 of B/W

so in 2 cases of 3 the other side is black.
The probability is 2/3.

This assuming you don't paid attention to which side are you looking at.

If you somehow know that you're looking at 1st side, the probability is 1/2
If you know that you're looking at the 2nd one, then the probability is 1 (100%)
But I guess this is not the case.

cygnus
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### Re: Pancake Problem

What if the oven burnt any white side that was facing up in the oven?
This complicates things...

Stev
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### Re: Pancake Problem

What if the oven burnt any white side that was facing up in the oven?
This complicates things...

The B/B pancake remains B/B
The B/W pancake becomes B/B with 50% chance
The W/W pancake becomes B/W

So half of the time you have: B/B, B/W, B/W. So if you're looking a black face the probability of the other one being black is 1/2.
The other half of the time you have: B/B, B/B, B/W. So the probability is 4/5.

In general you can expect the other face being black (1/2 + 4/5)/2 = (5/10 + 8/10)/2 = 13/20 of the time.

Is this right?

Buttons
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### Re: Pancake Problem

cygnus wrote:What if the oven burnt any white side that was facing up in the oven?
This complicates things...

No, it makes it simpler. Then the question is, of all the times you see a black side face up (i.e. all the time), what percentage of the time is the bottom side black? The bottom side is simply one face chosen with uniform probability from all six possible faces, half of which are black, so the answer is 50%.

Stev: Your error came from assuming that which side you see is independent from which side got burned. This is clearly false: the side you see is always the side that gets burned.