Best Answer bonanova, 28 May 2013 - 05:58 AM

Spoiler for Seems too simple but here is a first analysis

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I assume N 3 points means 3 points.

First point is fixed at 12:00 WOLOG.

Color the right half of the table (between 12:00 and 6:00 red.

Color the left half of the table (between 6:00 and 12:00) blue.

The table stands if the 2nd and 3rd points fall in different color regions.

The probability is thus 0.5.

Well, not quite: that analysis is too simple to be true.

It falls for points at 12:00, 11:00 and 1:00.

The restrictions have to be broadened:

It must be true no matter which point is taken as the first point.

In the 11-12-1 case it fails if either 11 or 1 is taken as the first point.

That makes the probability less than 0.5.

I'm guessing the added restriction is that the last two points also must lie one in the upper half and the other in the lower half (one between 9:00 and 3:00, the other between 3:00 and 9:00) thus imposing another factor of 1/2, for an overall probability of 1/4.

Hmmm but that set of constraints fails for 12, 1 and 8:00.

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First point is at 12:00 WOLOG.

Second point is red: it lies CW at an angle

Draw diameters from the first two points.

Their opposite ends define a blue arc length covering an angle

The third point must fall on this arc, for the table to be stable.

That probability is

The overall probability of stability is

First point is fixed at 12:00 WOLOG.

Color the right half of the table (between 12:00 and 6:00 red.

Color the left half of the table (between 6:00 and 12:00) blue.

The table stands if the 2nd and 3rd points fall in different color regions.

The probability is thus 0.5.

Well, not quite: that analysis is too simple to be true.

It falls for points at 12:00, 11:00 and 1:00.

The restrictions have to be broadened:

It must be true no matter which point is taken as the first point.

In the 11-12-1 case it fails if either 11 or 1 is taken as the first point.

That makes the probability less than 0.5.

I'm guessing the added restriction is that the last two points also must lie one in the upper half and the other in the lower half (one between 9:00 and 3:00, the other between 3:00 and 9:00) thus imposing another factor of 1/2, for an overall probability of 1/4.

Hmmm but that set of constraints fails for 12, 1 and 8:00.

**Edit:**(here is a real proof that p = 1/4.)---------------------------------------------------------------------

First point is at 12:00 WOLOG.

Second point is red: it lies CW at an angle

*: 0 <***a***< 180***a**^{o}WOLOGDraw diameters from the first two points.

Their opposite ends define a blue arc length covering an angle

**a**The third point must fall on this arc, for the table to be stable.

That probability is

*/360***a**^{o}.The overall probability of stability is

*/360***a**^{o}averaged over the interval 0<*<180.***a***=***p****0.25**.