Pancakes and Bayes' Rule


by richardwander
Tags: bayes, pancakes, rule
micromass
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#19
Jun9-11, 07:41 AM
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Quote Quote by Nyxie View Post
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
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#20
Jun9-11, 07:49 AM
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Quote Quote by Nyxie View Post
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
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#21
Jun17-11, 03:34 PM
P: 75
Let's name the pancakes as

[itex]B_{0}[/itex]= Pancake with no burnt sides
[itex]B_{1}[/itex]= Pancake with one burnt side
[itex]B_{2}[/itex]= 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 [itex]B_{1}[/itex] and [itex]B_{2}[/itex] 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 [itex]B_{2}[/itex], 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 [itex]B_{1}[/itex], then it can appear in only 1 way: its burnt side up.
So, the probability that the pancake in the top is [itex]B_{2}[/itex], knowing that the top side is burnt, is clearly 2/3.
kaleidoscope
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#22
Jun17-11, 07:32 PM
P: 67
there are only 3 burnt sides and 2 sides belong to the completely burnt pancake, so the probability is 2/3.


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