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PostPosted: Sat Jul 31, 2010 15:55 
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post totally edited:
trakgalvis wrote:
Steve wrote:
trakgalvis wrote:
Take a finite line element (or thin rod, say), and cut it in two random places, assuming independence, and find the probability that you can make a triangle out of the pieces.

Can this be reduced to "what is the probability that the sum of two numbers is equal or greater than a single number (all from a uniform distribution)"?


I will tell you, when you provide an answer! :) Remember the two cuts are in the same line element!

I think I was slightly wrong before. I think the reduction is "the probability that the sum of the two smallest numbers (of three) are greater or equal to the largest one".
So 0.5!

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PostPosted: Sat Jul 31, 2010 16:21 
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Steve wrote:
Isn't that the probability of one number being greater than one other? (as opposed to one number being greater the sum of two)


It's the same thing. Assuming you start by cutting the rod into two pieces (unless you make both cuts simultaneously, but the principle remains the same)
you now have two pieces. You then cut one of them into two. If you cut the longer of the two pieces then you can make a triangle, but if you cut the shorter of the two then you can't.

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PostPosted: Sat Jul 31, 2010 16:40 
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Pete317 wrote:
Steve wrote:
Isn't that the probability of one number being greater than one other? (as opposed to one number being greater the sum of two)


It's the same thing. Assuming you start by cutting the rod into two pieces (unless you make both cuts simultaneously, but the principle remains the same)
you now have two pieces. You then cut one of them into two. If you cut the longer of the two pieces then you can make a triangle, but if you cut the shorter of the two then you can't.


Be careful! Both cuts of the original rod were independent.

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PostPosted: Sat Jul 31, 2010 16:48 
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Pete317 wrote:
Steve wrote:
Isn't that the probability of one number being greater than one other? (as opposed to one number being greater the sum of two)


It's the same thing. Assuming you start by cutting the rod into two pieces (unless you make both cuts simultaneously, but the principle remains the same)
you now have two pieces. You then cut one of them into two. If you cut the longer of the two pieces then you can make a triangle, but if you cut the shorter of the two then you can't.

You probably didn't see that I edited that post (before you submitted yours) as I realised what I said within it wasn't relevant. I have given my new answer within it (and we seem to be in agreement with the final figure :) ).

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PostPosted: Sat Jul 31, 2010 16:52 
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oh, oops!. The probability must be less than 0.5, because, even if you cut the longer of the two pieces you can still end up with one piece longer than the sum of the other two. I'll work it out.

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PostPosted: Sat Jul 31, 2010 17:11 
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It must be 0.25 :)

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PostPosted: Sat Jul 31, 2010 17:15 
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Ah, crap, yes, I did did a spreadsheet to prove my thoughts but I proved myself wrong. Lo and behold I again end up as the same answer as Pete (0.25, but I don't quite know why). Ho hum!

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PostPosted: Sat Jul 31, 2010 17:35 
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Prob = 1/4, is what I recall from a long time back, so I am glad that is what you are getting.

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PostPosted: Sat Jul 31, 2010 17:40 
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If we can include riddles, here's one which surprisingly catches a few people out for a short time..

I am behind you but you are also behind me. How can that be?

No need to reply but try it on a few friends you know and see the look on their face as they look pensive for a while or even come up with some remarkable answers :)

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PostPosted: Sat Jul 31, 2010 17:49 
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Big Tone wrote:
I am behind you but you are also behind me. How can that be?


Ah yes, reminds me of that old story about a driver who stopped to ask a farmer how far it is to the village which he'd (unknowingly) just passed through :D

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PostPosted: Sat Jul 31, 2010 19:08 
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Steve wrote:
Ah, crap, yes, I did did a spreadsheet to prove my thoughts but I proved myself wrong. Lo and behold I again end up as the same answer as Pete (0.25, but I don't quite know why). Ho hum!


Yep. I modelled it and get 0.25. But can't understand it intuitively

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PostPosted: Sat Jul 31, 2010 19:12 
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I would be nice to find a method that makes it look obvious, though it might not exist.

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PostPosted: Sat Jul 31, 2010 19:21 
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An alien measures my height in alien units, and I have no idea what these units are. In normal scientific decimal notation, what is the probability that my height starts with a 1, in these alien units?

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PostPosted: Sat Jul 31, 2010 19:27 
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I was just about to ask how many fingers the aliens had, then I saw you said "decimal" :wink:

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PostPosted: Sat Jul 31, 2010 19:36 
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I make it 1/9

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PostPosted: Sat Jul 31, 2010 19:43 
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Pete317 wrote:
I make it 1/9


Wrong!

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PostPosted: Sat Jul 31, 2010 20:28 
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Pete317 wrote:
I make it 1/9


So do I
In 0-9 there is 1: 1/10 = .1
In 10-99 there are 10: 10+1/100 = .11
In 100 -999 there are 100: 100+10+1/1000 = .111
In 1000-9999 there are 1000: 1000+100+10+1/10000 = .1111

Convergent on .1 recurring or 1/9

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Last edited by dcbwhaley on Sat Jul 31, 2010 21:49, edited 1 time in total.

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PostPosted: Sat Jul 31, 2010 20:40 
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dcbwhaley wrote:
Steve wrote:
Ah, crap, yes, I did did a spreadsheet to prove my thoughts but I proved myself wrong. Lo and behold I again end up as the same answer as Pete (0.25, but I don't quite know why). Ho hum!


Yep. I modelled it and get 0.25. But can't understand it intuitively

That's what I like about (us) engineers; we can still find a way to get the right answer even when normal methods are beyond us :D

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PostPosted: Sat Jul 31, 2010 20:41 
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1/10

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PostPosted: Sat Jul 31, 2010 21:07 
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graball wrote:
1/10


Wrong!

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