Omega


Total Posts: 434 
Joined: Jul 2006 


3 pancake. one is with both sides burned, one is with one side burned, the last one with no sides burned. Combined them in a plate. The top side is burned. what is the probability of the other side is burned?
I either have 1/2 or 3/4...but googling also bring up 2/3
My logic is top side is burnt which is with prob 2/3. Given this info (1/2)/(2/3)=3/4
I suspect I'm wrong but happy to hear where I've gone wrong in this logic 




pmrnt


Total Posts: 6 
Joined: Mar 2008 


the prob is 1/2 = (2/6) / (2/3) = P(bottom burned  top burned) =P(bottom burned & top burned) /P (top burned) You can draw the decision tree to visualize this: there are only 6 arrangements 



ahd


Total Posts: 4 
Joined: May 2017 


Disagree, but perhaps I misunderstand the assumptions. 3 pancakes, call them A, B and C. A is burned on both sides, B on one side only (call it side 1) and C is not burned. I assume that the stack is equally likely apriori to have A, B or C in any position and that either side is equally likely to be face up or down for a given pancake. So there are 48 equally likely configurations before we start conditioning(6 orderings of A, B, C times 8 ways of choosing up and down sides for a given ordering of 3 pancakes). But we're told that the top is burnt which tells us that the top is one of (A_side1, A_side2, B_side1), bringing us down to 3 choices for top side *2 orderings of remaining two pancakes * 4 ways of choosing up and down sides for a given ordering of those two (middle and bottom) pancakeds = 24 possible configurations. Of these 24, 8 also have a burnt side on the bottom. So the answer is 1/3 just by enumeration. Alternatively, Total prob is P(A is top pancake  top is burnt) P(B is bottom pancake  A is top) P(B is burnt side down) + P(B is top pancake  top is burnt) * P(A is bottom pancake) = 2/3 * 1/2 * 1/2 + 1/3*1/2 = 1/3.






I get 2/3
Unconditional P(burnt both sides) =P(ABC) +P(ACB) = 1/3
Unconditional P(burnt top side only) =0.5 (P(BAC) +P(BCA)) =1/6
Answer = 1/3 / ( 1/3 + 1/6)=2/3




ahd


Total Posts: 4 
Joined: May 2017 


Are we solving 1) for the probability that the bottom of the 3pancake stack is burnt, given that the top of the 3pancake stack is burnt? Or are we solving 2) for the probability that the underside of the top pancake in the stack is burnt, given that it's top is burnt? I was assuming 1) but I think other people are solving 2). What's the question being asked? 





I read it as "other side of the top pancake" but it is not unambiguous 



ahd


Total Posts: 4 
Joined: May 2017 


I agree that the answer to problem 2 (prob that the underside of the topmost pancake is burnt, given that the top is burnt) is 2/3 




AB12358


Total Posts: 61 
Joined: Apr 2014 


From a previous discussion on this topic elsewhere:
"You have three pancakes. The sides of them are colored as followed because you can't cook.
Goat/Goat Goat/Car Car/Car " 



chiral3

Founding Member

Total Posts: 5073 
Joined: Mar 2004 


Nice teaser Monty. 
Nonius is Satoshi Nakamoto. 物の哀れ 



chiral3

Founding Member

Total Posts: 5073 
Joined: Mar 2004 


After some thought this isn't really a monty hall. Monty Hall involves introducing new information by opening a door in response to an initial guess, based on 1/3rds prob, which in turn alters the probability from 1/3 to 2/3. In this problem all of the info is known and set a priori. 
Nonius is Satoshi Nakamoto. 物の哀れ 



I think it's 1/2. A = burnt both sides B = burnt on top

"Plausible regularities may be present but swamped by changes in attendant circumstances."
Ole Peters 




@billWalker  similar to my calculations but I get P(B) = 0.5 (by symmetry) 



punx120


Total Posts: 2 
Joined: Feb 2017 


I agree with @silverside, P(B) = 0.5 = 3 side burn / 6 total side.
I also think @pmrnt : P(bottom burned  top burned) =P(bottom burned & top burned) / P (top burned), as it simply rely on conditional probability formula (and not bayes theorem). but got P(top burned) wrong like @billWalker.
So the answer is 2/3.





pmrnt


Total Posts: 6 
Joined: Mar 2008 


Agreed P(B) = 1/2 and not 2/3.
The earlier assumption was that the pancakes cannot be flipped (2 with top burned and hence 6 arrangements) 



lmog


Total Posts: 134 
Joined: Mar 2010 


Just to be clear, if x=burned and o=not burned, aren't these the only stacks with top burnt?
x x x x x x o o x x x x o x o o o x o x o o x o
x o x o o o x o o x o o
P(bottomoftoppancake=burned) = 4/6 = 2/3 (EDIT: doh wrote 1/3)
no? 




mj


Total Posts: 1049 
Joined: Jun 2004 


I am interpreting the question take one of the three pancakes at random. Flip it randomly. Its top side is burnt. What is the probability the bottom is burnt?
The unconditional probability the top is burnt is 0.5. Since have the pancake sides are burnt.
The unconditional probability that both sides are burnt is 1/3.
The conditional probability is the ratio of these, by definition, which is 2/3.

More mathematical finance has been published.


