1-dimensional geometric probability

Agnishom · 2015-01-05 21:26:53

What is the probability that a rope of length l cut into 3 random pieces can form a triangle?

Bob · 2015-01-06 02:07:56

hi Agnishom,

Here's my effort on this. I did a simulation that seems to confirm it.

Call the first length a, the second b and the third c.

c= 1 - a - b

By the triangle inequality you want

a+b > c => a + b > 1 - a - b => a+b > 0.5. This is equivalent to c < 0.5

and

c+b > a => 1 - a - b + b > a => a < 0.5

and similarly

b < 0.5

Assuming the distributions are uniform

P(a<0.5) = 0.5 and similarly for the other two.

For all three constraints, P(a triangle) = 0.5 x 0.5 x 0.5 = 0.125

Bob

anonimnystefy · 2015-01-06 02:38:56

But, you are counting cases in which a+b+c<1.

Agnishom · 2015-01-06 03:50:41

The answer should be 0.25

Bob · 2015-01-06 05:03:24

OK. I'll have a re-think.

Bob

Agnishom · 2015-01-06 05:14:24

I found this!

Bob · 2015-01-06 23:19:32

hi

But, you are counting cases in which a+b+c<1.

Yes, I accept that is wrong. But that means the actual answer would be lower. ???

I have re-done my simulation as follows:

Choose a at random in the range (0,1)

Choose b at random in the range (a,1)

Calculate c = 1 - a - b

I have then flagged all cases where a,b,c < 0.5 and counted them.

Running this 1 000 000 times I get P = 0.019275

Typical flagged results ( 0.235791, 0.47383, 0.290379) and (0.124516, 0.42401, 0.451474)

and typical un-flagged results (0.797796, 0.08612, 0.116084) and ( 0.107225, 0.74065, 0.152125)

This simulation isn't getting results anywhere close to 0.25. So, can anyone see what's going wrong ?

Bob

anonimnystefy · 2015-01-07 01:52:38

Shouldn't you be picking b from (0,1-a)?

Bob · 2015-01-07 06:19:03

Yes, I just said it wrong. If you look at my sample data you'll see it does pick as you say.

Bob

anonimnystefy · 2015-01-07 06:54:08

I think we can look at this geometrically. If we plot in R³ which points satisfy the conditions (i.e. x+y+z=1) it's an equilateral triangle with a side of

and the ones which also satisfy x,y,z<0.5 make up an equilateral triangle of side . So, the answer should be 1/4. The simulation I did confirms it:

l = Select[Table[{#1, #2, 1 - #1 - #2} &[RandomReal[], RandomReal[]], {100000}], #[[3]] > 0 &];
l1 = Select[l, #[[1]] < 0.5 && #[[2]] < 0.5 && #[[3]] > 0 && #[[3]] < 0.5 &];
Length[l1]/Length[l] // N

Bob · 2015-01-07 12:02:12

hi Stefy and Agnishom,

Oh, boy. This question really has me beat. First of all, Stefy, I like your solution best of all; neat way to do it.

So now for my simulation. I found two errors (after lots of puzzling) and what follows is my best effort. It doesn't give 0.25. It seems to want to give 0.025, which is tantalisingly close but I cannot account for the factor of ten. .

I used Excel. Sorry. The command RAND() returns a number in the range 0,1.

So I used it to pick two points, P and Q along the line from 0 to 1.

If P>Q, I compute P-Q and call this a. Otherwise Q-P

If P>Q, I store OQ in b. Otherwise OP.

I compute c = 1 - a - b

Then I test for all three less than 0.5 and flag this with a 1, else 0.

Then I add up all the 1s, and compute the experimental probability.

I also have done the following checks.

(1) Check that a + b + c = 1 for every case.

(2) Calculate the expected (mean average) value a, b and c.

(3) Look at some cases to check that 1 really does mean the triangle is drawable and 0 means it isn't. This has checked out for every case I've checked.

Here's a screen shot of the last two lines of 10000 repeats and the formulas that I used.

Can anyone see where I'm going wrong ?

Bob

anonimnystefy · 2015-01-07 12:06:41

Have you tried doing it a few more times?

Bob · 2015-01-07 12:07:50

Yes, loads. It always gives numbers like 0.02.....

Bob

anonimnystefy · 2015-01-07 12:10:23

I have to look at it a bit more.

Glad we're finally online at the same time. It's a bit late over there, isn't it?

Bob · 2015-01-07 12:14:17

Yes it is. I ought to go to bed. Got a busy day tomorrow. Bye for now.

Bob

anonimnystefy · 2015-01-07 12:18:00

Good night.

When you get the chance, please tell us what you used to get the mean of the 1's and 0's.

Bob · 2015-01-07 21:21:02

Good question.

The command in cell F10002 is =SUM(F2:F1001)/(ROW(A10001)-ROW(A2))

I had used 10000 in place of (ROW(A10001)-ROW(A2)), but decided to let Excel count the rows in case I was doing something stupid with the number of trials. Mrs B had a similar thought so she looked at the rows and is convinced that I really had 10000.

FIRST EDIT: Got it! Just looked at what I have typed in. I'll just adjust that to =SUM(F2:F10001)/(ROW(A10001)-ROW(A2)) Thanks.

SECOND EDIT: Now I'm getting 0.25............. Hurray!!!!!! Many many thanks.

Bob

Last edited by Bob (2015-01-07 21:24:46)

anonimnystefy · 2015-01-07 21:32:14

Nice work!

I guess we can safely say the answer is 1/4.

Bob · 2015-01-07 21:35:17

I'm happy to say that. I may still see if I can make my original method work, now I've got a clear procedure.

Bob

anonimnystefy · 2015-01-07 21:38:06

If you mean the one in post #7, that would be hard, because if you pick like that, the distribution of b over (0,1) is not uniform. You would need to be picking it from (0,1-a) with a non-uniform distribution, I think.

Bob · 2015-01-07 21:43:23

Yes. I realise that now. It may take a while but it will be good for my brain to try.

Got to stop now. See you tomorrow.

Bob

anonimnystefy · 2015-01-07 21:44:16

See you and good luck.

Math Is Fun Forum

#1 2015-01-05 21:26:53

1-dimensional geometric probability

#2 2015-01-06 02:07:56

Re: 1-dimensional geometric probability

#3 2015-01-06 02:38:56

Re: 1-dimensional geometric probability

#4 2015-01-06 03:50:41

Re: 1-dimensional geometric probability

#5 2015-01-06 05:03:24

Re: 1-dimensional geometric probability

#6 2015-01-06 05:14:24

Re: 1-dimensional geometric probability

#7 2015-01-06 23:19:32

Re: 1-dimensional geometric probability

#8 2015-01-07 01:52:38

Re: 1-dimensional geometric probability

#9 2015-01-07 06:19:03

Re: 1-dimensional geometric probability

#10 2015-01-07 06:54:08

Re: 1-dimensional geometric probability

#11 2015-01-07 12:02:12

Re: 1-dimensional geometric probability

#12 2015-01-07 12:06:41

Re: 1-dimensional geometric probability

#13 2015-01-07 12:07:50

Re: 1-dimensional geometric probability

#14 2015-01-07 12:10:23

Re: 1-dimensional geometric probability

#15 2015-01-07 12:14:17

Re: 1-dimensional geometric probability

#16 2015-01-07 12:18:00

Re: 1-dimensional geometric probability

#17 2015-01-07 21:21:02

Re: 1-dimensional geometric probability

#18 2015-01-07 21:32:14

Re: 1-dimensional geometric probability

#19 2015-01-07 21:35:17

Re: 1-dimensional geometric probability

#20 2015-01-07 21:38:06

Re: 1-dimensional geometric probability

#21 2015-01-07 21:43:23

Re: 1-dimensional geometric probability

#22 2015-01-07 21:44:16

Re: 1-dimensional geometric probability

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