Talk:Puzzles/Statistical puzzles/The Pentagon/Solution

The solution offered is not correct. There is a bigger chance that you will see 2 sides then 3. I SUSPECT the actual chance decreases the further you move away, approaching 50%, but I haven't calculated this (I haven't calculated anything, because the maths on this are pretty complex).

Now I hear you guys/girls shouting: WHERE IS YOUR PROOF? Well, very simple really: While I agree all the triangles have a 36° corner at their point, they do NOT share the same origin. Thus, a cross-section for a green triagle (3 side view) will be smaller then one from a yellow (2 side view) triangle at the same distance from the pentagon center. The author actually admits this: "When the distance from the Pentagon becomes large, the initial difference between the yellow and green sectors becomes negligible, and so the probability of being in a yellow sector (and seeing exactly two walls) is 50%"

The puzzle states that the distance should be larger then the 'only able to see 1 side' distance (it even says: "You are standing near The Pentagon."). To make this puzzle work it should state something like: You are looking at the Pentagon from a great distance. What is the probability that you are only seeing 2 sides.

Also, the last line offering 'proof' is clearly wrong ("If you see two sides of building, you friend at opposite side sees three of them. For each spot where you can see two sides, there is one spot where you can see three sides, and for each spot where you can see three sides, there is one spot where you can see two sides. This gives us exacly 50%."). Explanation: Alice takes position exactly left of the Pentagon (as seen in the diagram on the solution page), at a large distance (at least outside the 'red' part). She is in the yellow triangle. Bob takes position exactly right of the Pentagon (as seen in the diagram on the solution page), at a large distance (at least outside the 'red' part). He is in the yellow triangle. You can draw a straight line from Alice to Bob. Alice and Bob have found a spot where they both see 2 sides and are directly opposite each other.

ShadowLord 14:18, 27 Jun 2005 (UTC)


 * Oh, solution really is wrong. I just took it from some book, so i didn't really think if it correct or not. Then this thing is probably much more complex, and probably should be removed. --Divinity 19:31, 27 Jun 2005 (UTC)

The solution is somewhat correct. The problem is you're looking at a fraction of an infinite plane. If you start with a disk with as center the center of the pentagon, then indeed as the radius of the circle increases, the two fractions are the same (the seemingly extra bit of the yellow area is a zero-fraction of the plane, just like the pink areas). So the probability would be 50-50.

However, you could also look at an ellipse, instead of a disk. With the pentagon in on of the foci, and the other focus either in a green or red area. Now if you increase the size of the ellipse (along the large axis) and take the limit to infinity, then the probability gets skewed tremendously towards whatever area the second focus is in.

And that's just two possible approaches to find 'the' probability. Infinity is not a concept to mess around with indiscriminitely.

For fun compare this to taking the sum over all integer numbers. You can go 0 + 1 + -1 + 2 + -2 + 3 + -3 ... which has a cauchy limit of 0. But there is no reason why you couldn't do  0 + 1 + 2 + -1 + 3 + 4 + -2 ... which goes to infinity in the limit. Despite that in both cases you add every single number from the set once.

-- towr 17:44, 28 Jun 2005 (UTC)

Yes, the 50% probability is a limit of the probability as we choose random points in increasingly large circles centred at the Pentagon. If you have a better interpretation of "the probability", please share it.

--Borbrav 19:15, Jun 30, 2005 (UTC)
 * I'm not entirely convinced that is the only valid interpretation of 'the probability', although it certainly is the most natural, and probably the only one to give a sensible answer (but then not all good puzzles have answers, let alone sensible ones). If you want the probability for when you're at a given (be it unknown) distance, that I'd agree it's unambiguously 50%. But what the stochastic variable actually is in this case seems ambiguous to me. But maybe I'm just being daft :P -- towr


 * Actually, at any given fixed radius (larger than the red triangles) you have a probability of seeing two sides in excess of 50%, but this difference decreases as the circle's radius increases inversely to the radius. I too agree the puzzle is ambiguous. Borbrav July 6, 2005 04:43 (UTC)

Also, another thing in defense of the proof: look at the diagram from a very large distance above it. All you will see will be a point with five yellow and five green triangles, which will all appear identical.

--Borbrav 19:22, Jun 30, 2005 (UTC)