What is the probability that the top pancake is burned on both sides?

[richardwander]

Hi.

I've done a search already to see if this question has been answered but haven't found anything. Hope it's okay to ask it here.

I have this brainteaser that's been bugging me a little bit. It's not homework, but is a question that I heard from someone who's already had an interview at a job I'm applying for. It was asked in the interview and goes like this:

You have three pancakes. One of them is burned on both sides, one of them is burned on one side, and one of them is perfectly cooked and burned on no side. You stack the pancakes on a plate and note that the visible side of the top pancake is burned. What is the probability that the top pancake in the stack is burned on both sides?

To me, this sounds like a Bayes' rule problem. I define the event X to be that the top pancake is burned on both sides, and the event Y to be that the top pancake is burned on at least one side. I then apply Bayes' rule to get

$$P(X|Y) = \frac{P(Y|X)P(X)}{P(Y)}$$

I have that P(Y|X) = 1 since if the top pancake is burned on both sides, the probability that it's burned on at least one side is 1.

I also have that P(X) = 1/3 since the marginal probability that the top pancake is burned on both sides is one in three.

I calculate the marginal probability P(Y) as follows:

$$P(Y) = P(Y|\textrm{no sides})P(\textrm{no sides}) + P(Y|\textrm{one side})P(\textrm{one side}) + P(Y|\textrm{both sides})P(\textrm{both sides})$$

where the conditionals refer to the burned sides on the pancakes. I find that

$$P(Y) = 0\times \frac{1}{3} + 1\times\frac{1}{3} + 1\times\frac{1}{3} = \frac{2}{3}$$

This means that the probability that the top pancake is burned on both sides given that we know it's burned on at least one side is given by

$$P(X|Y) = \frac{1\times\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$$

However, I don't think this is correct. Intuitively, it seems that the answer should be 2/3. Can anyone point out where I'm going wrong?

 

[micromass]

Hi RichardWander!

I don't think your going wrong somewhere. In my intuition, the chance should indeed be 1/2. If I give you a pancake that's burnt on one side, then it could either be the one burnt on two sides or the one burnt in exactly one side. Thus we have 1 chance in two that it's the one burnt in exactly two sides.

So, I do think your 1/2 chance is alright. Anybody correct me if I'm wrong!

 

[Hurkyl]

the event Y to be that the top pancake is burned on at least one side.

This is not the same event as "the top pancake is burnt on the visible side".

 

[parnandi]

I got 2/3.

Here is how I did using bayes rule. I know my notations are not the best, please bear with me on that-

B2 = burned on both sides
B1 = burned on one side
B0 = burned on no side

P(B2 being on top of the stack) = P(B2 = top) = 1/3 (all three have equal chances of being on top), hence
P(B2 = top) = 1/3
P(B1 = top) = 1/3
P(B0 = top) = 1/3

Let P(B2/burned) be the Probability of the top one being B2 given the top face is burned
We need to find P(B2/burned) = ?

Let P(burned/B2) = prob of top face being burned given the pancake is B2 (burned on both sides)
Now we know that P(burned/B2) = 1 (because both sides are burned)
P(burned/B1) = 1/2 (because one side is burned)
P(burned/B0) = 0 (because no sides are burned)

Using Bayes rule:

P(B2/burned) = P(burned/B2)*P(B2)/P(burned)

P(burned) = prob of the top face being burned
P(burned) = P(burned/B2)*P(B2) + P(burned/B1)*P(B1) + P(burned/B0)*P(B0)
P(burned) = 1*1/3 + 1/2*1/3 + 0*1/3

Hence, P(B2/burned) = (1*1/3)/(1*1/3 + 1/2*1/3 + 0*1/3) = 2/3

 

[ych22]

This is simply a variant of a classical problem, isn't it? If I remember correctly, the answer is 2/3.

Many people do give the intuitive answer, 1/2.

 

[HallsofIvy]

No, the correct answer is 1/2. Seeing that the pancake is burnt on the top, you know that it is one of the two pancakes that are burnt on at least one side and so you can ignore the pancake that is not burnt at all. The probability that the pancake is burnt on the other side is 1/2, the probabilty it is not burnt on the other side is 1/2.

 

[chronon]

The correct answer is 2/3. There are six pancake sides, each with an equal probability of being on top. Two of these belong to the pancake burned on both sides, one to that burned on one side.

 

[HallsofIvy]

The correct answer is 2/3. There are six pancake sides, each with an equal probability of being on top.

No, that is not true. The problem said "You stack the pancakes on a plate and note that the visible side of the top pancake is burned." The pancake which is burned on both sides has 0 probability of being on top.

Two of these belong to the pancake burned on both sides, one to that burned on one side.

 

[chronon]

No, that is not true. The problem said "You stack the pancakes on a plate and note that the visible side of the top pancake is burned." The pancake which is burned on both sides has 0 probability of being on top.

OK, perhaps my post was a bit too short. The equal probability was prior to noticing that the top side is burned. When that is taken into account it is still twice as likely that the top pancake will be the one burned on both sides as the one burned on one side.

Think of a large number of such stacks with the pancakes arranged at random. In 1/6 of the stacks the top side will be the burned side of the pancake burned on one side, in 1/3 of the stacks it will be a side of the pancake burned on one side and in 1/2 of the stacks it won't be burned. Thus in 2/3 of the stacks where the top side is burned, the pancake burned on both sides will be on top.

 

[Rogerio]

Of course the correct answer is 2/3.
There are 3 burned sides, and 2 of them belong to the completely burned pancake.

 

[micromass]

OK, denote by (B,B|N,B|N,N) that the bottom pancake is burnt on both sides, the middle is burtn on one side and the burnt side is flipped up and the upper pancake is not burnt. We have the following possibilities:

(B,B|N,B|N,N)
(B,B|B,N|N,N)
(B,B|N,N|N,B)
(B,B|N,N|B,N)
(N,N|N,B|B,B)
(N,N|B,N|B,B)
(N,N|B,B|N,B)
(N,N|B,B|B,N)
(B,N|N,N|B,B)
(B,N|B,B|N,N)
(N,B|N,N|B,B)
(N,B|B,B|N,N)

All these things happen with probability 1/12. Now, we scratch the possibilities in which the upper pancake is not burnt, this gives us

(B,B|N,N|N,B)
(N,N|N,B|B,B)
(N,N|B,N|B,B)
(N,N|B,B|N,B)
(B,N|N,N|B,B)
(N,B|N,N|B,B)

This leaves us with 6 cases. In 4 of them, both sides are burnt. So the chance that both sides are burnt is indeed 2/3.

I didn't really expect this...

 

[ych22]

Ahhh...this is indeed a timeless problem!

 

[Nyxie]

Assuming that all you know is that there are 6 unrelated things (the pancake sides), 3 of which are burnt, the prob. is 1/2. If each pancake had 1 burnt side 1 unburnt side then 1/2 would be right. But each pancake individually has a different probability than the others. (Related events get multiplied while unrelated ones are added.)
Each pancake comes with its own probability for having a visible burnt side.

If the top pancake's visible side is burnt then obviously it's not the unburnt pancake, so disregard it.
One of the two remaining pancakes has a 1/2 prob. of being burnt-side-up and the other has a 1 prob. Each have a 1/2 prob. of being on top. Prob. the burned side of half-burnt pancake is visible is (1/2 that it's on top) * (1/2 that its burnt side is visible) = 1/4.
So the answer is 3/4, not 2/3! (from 1 - (1/4) )

micromass - you can't count the unburnt pancake because we already know that it is not the one on top. Unburnt pancake has zero prob. that it could usurp the fully burnt pancake's place, so only the two with burnt sides provide the sample space.

Again [ (half-burned pancake on top) AND (half-burned pancake burnt side up) ] OR [burnt pancake on top] = (1/2)(1/2) + (1/2) = (1/4) + (1/2) = 3/4.

 

[praeclarum]

Assuming that all you know is that there are 6 unrelated things (the pancake sides), 3 of which are burnt, the prob. is 1/2. If each pancake had 1 burnt side 1 unburnt side then 1/2 would be right. But each pancake individually has a different probability than the others. (Related events get multiplied while unrelated ones are added.)
Each pancake comes with its own probability for having a visible burnt side.

If the top pancake's visible side is burnt then obviously it's not the unburnt pancake, so disregard it.
One of the two remaining pancakes has a 1/2 prob. of being burnt-side-up and the other has a 1 prob. Each have a 1/2 prob. of being on top. Prob. the burned side of half-burnt pancake is visible is (1/2 that it's on top) * (1/2 that its burnt side is visible) = 1/4.
So the answer is 3/4, not 2/3! (from 1 - (1/4) )

micromass - you can't count the unburnt pancake because we already know that it is not the one on top. Unburnt pancake has zero prob. that it could usurp the fully burnt pancake's place, so only the two with burnt sides provide the sample space.

Again [ (half-burned pancake on top) AND (half-burned pancake burnt side up) ] OR [burnt pancake on top] = (1/2)(1/2) + (1/2) = (1/4) + (1/2) = 3/4.

No the correct answer is 2 / 3.

It's not the case that the unburned pancake is on top, and it's not the case that the side of the 1-side-burned pancake is on top, so we have 3 other options, each (we assume) with equal probability. In two of the cases, it's always the case that the other side is also burned. So the answer is 2/3.

 

[Nyxie]

No the correct answer is 2 / 3.

It's not the case that the unburned pancake is on top, and it's not the case that the side of the 1-side-burned pancake is on top, so we have 3 other options, each (we assume) with equal probability. In two of the cases, it's always the case that the other side is also burned. So the answer is 2/3.

It totally depends on whether you count the two faces of the burnt pancake as distinct or not. If you count them as different you are right.
But I counted them as the same because they both refer to the same pancake. =P And that was the question after all.

The three options do not have equal probability when you take into account that 2 burnt sides are associated with one pancake, and only 1 burnt side with the other.

Simplifying using B1, B2 for the two burnt pancake's sides and H for the half burnt pancake with the burnt side up...Any of these could be on top:

B1
B2
H

From here it looks like the answer is 2/3 but this disregards which pancake we are talking about. You have to account for the probability that one of the pancakes is on top.

prob(B1) = prob(B2)
prob(B1)+prob(B2) = 2[prob(B1)]

2[prob(B1)] + prob(H) = 1.

2[prob(B1)] = prob(burnt pancake on top) = 1 - prob(H)/2

prob(H) = 1/2 because it's either H or B1 (=B2).

It's still 3/4

 

[Nyxie]

Also notice in the above that I start by counting both pancake sides as different in the sample space (in order to get the correct probability) then count them as the same (in order to make them be the same outcome)

For the probability each side of the burnt pancake contributes to its probability. But the question asks for either of two outcomes (either side)

 

[Hurkyl]

Again [ (half-burned pancake on top) AND (half-burned pancake burnt side up) ] OR [burnt pancake on top] = (1/2)(1/2) + (1/2) = (1/4) + (1/2) = 3/4.

This is a wonderful calculation of the following problem:

There are two pancakes, one burnt on both sides, the other burnt on one side. You stack the two pancakes randomly (uniformly chosen from all arrangements). What is the probability the visible side is burnt?​

(wonderful given the presumption it's sufficiently obvious that the "OR" is between two mutually exclusive events)


But the problem you solved has little relation to the problem that was asked.

 

[Nyxie]

If we use Baye's Theorem, say A is "we get the pancake burnt on both sides" and B is "we see a burnt side".

P(A|B) is what we're looking for. P(A|B) is the conditional probability of A given that B is true.

With P(B|A) = 1, P(A) = 1/2, P(B) = 3/4
P(A|B) = 2/3

P(B) = 3/4 would be because of the 4 sides of the two pancakes with at least one burnt side, 3 of those sides are burnt.

But this gives the probability where each side of each pancake is considered distinct from the others. Yet we are not looking for the probability that a given side of pancake A is upright. We are looking for either side being upright.

Thus P(B) = 2/3, counting pancake A as one occurence and both sides of the other pancake as 2 other occurences. 2/3 of those occurences will show a burnt side. This gives 3/4 again, so what's going on?

 

[micromass]

micromass - you can't count the unburnt pancake because we already know that it is not the one on top. Unburnt pancake has zero prob. that it could usurp the fully burnt pancake's place, so only the two with burnt sides provide the sample space.

That's not what I did. I wrote down the 12 possibilities to make it easier on myself. The sample space is actuallly

(B,B|N,N|N,B)
(N,N|N,B|B,B)
(N,N|B,N|B,B)
(N,N|B,B|N,B)
(B,N|N,N|B,B)
(N,B|N,N|B,B)

And you see from that that the probability is 2/3.

 

[micromass]

If we use Baye's Theorem, say A is "we get the pancake burnt on both sides" and B is "we see a burnt side".

P(A|B) is what we're looking for. P(A|B) is the conditional probability of A given that B is true.

With P(B|A) = 1, P(A) = 1/2, P(B) = 3/4
P(A|B) = 2/3

I'm at loss at how you obtained these probabilities. P(A)=1/3 and P(B)=1/2 to me, and then P(A|B)=2/3, like expected.

P(B) = 3/4 would be because of the 4 sides of the two pancakes with at least one burnt side, 3 of those sides are burnt.

No, that's not true. P(B) denotes the probability that we see a burnt side regardless of anything else. Since there are 6 sides with 3 burnt sides, the probability is 1/2. You can't just eliminate the pancake with no burnt sides like that! If you do that then you're conditioning B over some other event, and then you're not calculating P(B).

 

[Karlx]

Let's name the pancakes as

$B_{0}$= Pancake with no burnt sides
$B_{1}$= Pancake with one burnt side
$B_{2}$= Pancake with two burnt sides

The probability a priori, for a pancake of being at the top after stacking them is 1/3 for each one. I think there is no doubt about this.
Now, let's take into account that the top side is burnt.
This let us with only $B_{1}$ and $B_{2}$ as being the pancake on the top.
But they don't have the same probability at all.
If the pancake at the top is pancake $B_{2}$, it can appear in 2 possible ways, depending on which of its sides (both burnt) is up.
On the other hand, if the pancake on the top side is pancake $B_{1}$, then it can appear in only 1 way: its burnt side up.
So, the probability that the pancake in the top is $B_{2}$, knowing that the top side is burnt, is clearly 2/3.

 

[kaleidoscope]

there are only 3 burnt sides and 2 sides belong to the completely burnt pancake, so the probability is 2/3.