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 [quote="bonanova"][quote="referee"]bonanova, probabilities can't be negative...[/quote][[spoiler] -1 + 2log2 = -1 + 1.386294361... = 0.386294361... Half that is 0.1931471806... [/spoiler]][quote="Jack_Ian"]I expect, what you need to do, would be to prove that for 3 sticks of length x, y and z respectively, the probability p(x/y) given z remains the same regardless of the process used.[/quote][[spoiler]For the sequential break case, after a first break at point [b]a[/b] <.5. the second break must be in the central region of width [b]a[/b] within the larger 1-[b]a[/b] piece. If you average [b]a[/b]/(1-[b]a[/b]) over all values of [b]a[/b] from 0 to .5 you get the expression above. If you specify two random break points on the original stick, the probability can be seen to be .25. I thought the three-break case might give a third result, similar to the manner in which "length of a random chord" in a circle has at least three answers. I found it interesting that it did not.[/spoiler]][/quote]
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referee
Posted: Sat Oct 27, 2012 5:56 am    Post subject: 1

Ah, that would make it positive, yes.
bonanova
Posted: Fri Oct 26, 2012 8:55 am    Post subject: 0

referee wrote:
no. 1-2log2 = 1 - .60106 = -.39894 and half of that is -.19947.
ln
referee
Posted: Thu Oct 25, 2012 11:12 pm    Post subject: -1

no. 1-2log2 = 1 - .60106 = -.39894 and half of that is -.19947.
bonanova
Posted: Thu Oct 25, 2012 7:33 pm    Post subject: -2

referee wrote:
bonanova, probabilities can't be negative...
[ -1 + 2log2 = -1 + 1.386294361... = 0.386294361... Half that is 0.1931471806... ]
Jack_Ian wrote:
I expect, what you need to do, would be to prove that for 3 sticks of length x, y and z respectively, the probability p(x/y) given z remains the same regardless of the process used.
[For the sequential break case, after a first break at point a <.5. the second break must be in the central region of width a within the larger 1-a piece. If you average a/(1-a) over all values of a from 0 to .5 you get the expression above. If you specify two random break points on the original stick, the probability can be seen to be .25. I thought the three-break case might give a third result, similar to the manner in which "length of a random chord" in a circle has at least three answers. I found it interesting that it did not.]
referee
Posted: Thu Oct 25, 2012 4:43 pm    Post subject: -3

bonanova, probabilities can't be negative...
bonanova
Posted: Thu Oct 25, 2012 4:11 pm    Post subject: -4

Jack_Ian wrote:
Only the relative lengths are important and the process always ends up with 3 sticks or various lengths.
I expect, what you need to do, would be to prove that for 3 sticks of length x, y and z respectively, the probability p(x/y) given z remains the same regardless of the process used.

Picking two points at random on the stick gives three pieces but a different probability.
bonanova
Posted: Thu Oct 25, 2012 4:04 pm    Post subject: -5

Chuck wrote:
By simulation, making three pieces by breaking the stick at random and then breaking the longer piece at random gives a little over 38.6% chance of making a triangle, twice as much as when you select one of the two at random.

Right. Also, you get a different answer by picking two points at random on the original stick [ .25 ]. So all scenarios aren't equivalent. The 4 piece discard one case didn't at first seem equivalent to the sequential three stick case. But it must be.i think lostdummy's argument shows that. And picking the longer stick rather than random pick of two sticks does double the result since the shorter stick always fails.
Chuck
Posted: Thu Oct 25, 2012 2:19 pm    Post subject: -6

By simulation, making three pieces by breaking the stick at random and then breaking the longer piece at random gives a little over 38.6% chance of making a triangle, twice as much as when you select one of the two at random.
Jack_Ian
Posted: Thu Oct 25, 2012 1:23 pm    Post subject: -7

Only the relative lengths are important and the process always ends up with 3 sticks or various lengths.
I expect, what you need to do, would be to prove that for 3 sticks of length x, y and z respectively, the probability p(x/y) given z remains the same regardless of the process used.
lostdummy
Posted: Thu Oct 25, 2012 9:14 am    Post subject: -8

bonanova wrote:

Can it be shown they are equivalent problems?

When you randomly remove one of four sticks, you can look at remaining length of 3 sticks as initial length of stick in original problem with 3 parts.

One stick that remained from pair (where other pair was tossed out) will be in this analogy stick that remains in original problem that is not again divided. Other two sticks are analogous to original problem where 'half' of stick was again divided with second split.

Above should work as analog because in both cases all choices were random.

More complicated to find analogy with 4 stick parts would be for original problem with 3 stick parts where you randomly divide stick, then split LONGER stick again.
bonanova
Posted: Thu Oct 25, 2012 6:46 am    Post subject: -9

There is a closed form solution for the two-break problem. [(-1+2log2)/2]
It turns out that the three-break problem simulates to the same result, which I found interesting.
Can it be shown they are equivalent problems?
Chuck
Posted: Thu Oct 25, 2012 1:41 am    Post subject: -10

By computer simulation I'm getting a little over 19.3% after 36 million tries.
bonanova
Posted: Wed Oct 24, 2012 6:36 pm    Post subject: -11

There's a puzzle that asks, when a stick is randomly broken into three pieces, the probability the pieces could form a triangle. The answer depends on the random process. One process is to choose a point at random for the first break, then randomly choose one of the pieces and break it at a random point. Let's change the puzzle slightly.

What is the probability of forming a triangle if in the second step both pieces are broken at random points and a random choice of the four resulting pieces is discarded?