Triangle From a Ruler
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Fred had a ruler that was exactly 12 inches long. His second cousin, Pop, was practicing with her Samurai sword and made two straight slashes at arbitrary spots on the ruler, cutting it into three pieces. What is the probability that the pieces of Fred's ruler can form a triangle?
HintGraphing it is easiest.
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Answer
To find the answer, we will use two random variables, a and b, on a common graph with rectangular axes. Both variables are uniformly distributed between 0 and 12, so we have a probability square which each side equal to 12. We will find regions on the square that lead to three lengths that can form a triangle.
Let us suppose that the two cuts are made at some lengths a and b from one end. If b > a, then the three lengths will be a, ba, and 12b. In order to create a triangle, any two of these lengths needs to be greater than the third. So,
a + ba > 12b or b > 6
ba + 12b > a or a < 6
a + 12b > ba or b < a+6
When we plot these on our graph, we get a triangle with points at (6,6), (0,6), and (6,12). The triangle formed has an area of 1/8th of the square.
If we then consider the case of a > b, then we get a similar triangular region reflected over the x=y diagonal of our graph. That triangle also has an area of 1/8th of the square.
Therefore the total area of our two regions is given by 1/8 + 1/8 = 1/4 and so the probability of Pop having created pieces capable of making a triangle is 25%.
The calculation of the probability that Fred lost a finger or two is left up to the reader.
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Comments
Miss_Xuni
Nov 17, 2005
 Meh, Geometry.
I did finally figure it out though, which made me happy. 
smarty_blondy
Nov 18, 2005
 Great teaser, however eh... i can't help but ask, if you look at your hands... how many fingers do you see? 
BOOKLOVER1
Nov 18, 2005
 boring 
cdrock
Nov 18, 2005
 Good one Phrebh, took me a while 
swimmergirl_mrh
Nov 19, 2005
 no offence but that is kinda stupid. I mean Who Cares about Freds ruler???? 
suyolcusu
Dec 01, 2005
 Cutting a stick into three parts to form a triangle
The solution presented is not exactly right, because it assumes the generation of two uniformly distributed random numbers between 0 and 1. The ruler of unit length is then cut at these positions. The probability of forming a triangle is then 0.25 as given in the answer.
A more realistic way is to cut the ruler first at location x (x is uniformly distributed between 0 and 1) then cut the remaining piece at location y (y is uniformly distributed between 0 and "1x").
The resulting three pieces have lengths "x", "y", and "1xy". The probability of having a triangle is then
P= integral from 0 to 0.5 {integral from(0.5x) to (0.5) [dy/(1x)]}dx
0.5
P={xln(1x}
0
P=0.193147
The same answer can be obtained by simulation. A simple BASIC program using one million trials is shown below:
10 REM "Cutting a ruler into three pieces to form a triangle"
11 N=1000000
12 S=0
13 RANDOMIZE
14 FOR I=1 TO N
15 R1=RND(1)
16 A=R1
17 B=1R1
18 L1=A
19 L2=B
20 R2=RND(1)
21 C=L2*R2
22 L3=L2C
23 L2=C
24 IF (L1+L2)>L3 THEN 26
25 GOTO 29
26 IF (L1+L3)>L2 THEN 28
27 GOTO 29
28 IF (L2+L3)>L1 THEN S=S+1
29 NEXT I
30 P=S/N
31 PRINT "Probability ="; P
32 END 
phrebh
Dec 01, 2005
 I believe you misread the answer. The second cut is taken from the remainder of the ruler. It's just that its length is shown in relation to the first cut. 
choptlivva
Dec 11, 2005
 Oh, my head hurts!!!
Good teaser I'm sure, Phreb.... but I wiggled through college Math by the skin of my teeth!! I have no idea how, except that the TA that taught the class liked me.... 
choptlivva
Dec 11, 2005
 BTW.... I thought all you needed was three straight lines to form a triangle!! Geez... I feel stupid!!! 
gaiapeach1
Dec 16, 2005
 I don't get it. Do you mean three straight edges, with the inside being a triangle? Or do you have three pieces that fit together in a puzzlepiece manner to create a solid triangle? I'm a bit confues. 
ChessNut
Dec 18, 2005
 wow...I think I need to go back to playing Chess...you are doing too much Math for me. great job! 
sebastian
Dec 19, 2005
 huh??????????????? 
SilverMoonBee
Jan 21, 2006
 To be honest, i never liked probability questions. Yours i did enjoy though. 
udoboy
May 26, 2006
 If b>6, you can in no way form a triangle.
If b = 6, you form a "straightline" triangle.
It seems to me then that, if the first cut is perfectly even, then 100% of the time you have the straightline triangle.
If not, then the second cut, if made on the shorter length, will never produce a triangle. This should be 50% of the time.
If neither of those take place, then the third cut, randomly made on the longer piece, needs to not be further than 6" from either endpoint.
The probability of this depends on the length of the piece. 
Kiroho
Sep 05, 2006
 if any segment > 6 then triangle is not possible
if any segment is exactly = 6 then a triangle is not possible, only a straight line is.
if all segments are < 6 inches, a triangle is always possible.
So the answer is slightly less than 50% 
phrebh
Sep 05, 2006
 No, it isn't. You are ignoring the fact that you have to look at two instances (one for each cut), not just one, so you multiply the two 50% probabilities to arrive at 25% (1/2 * 1/2 = 1/4). 
Kiroho
Sep 08, 2006
 I just ran it through Exel and you're right. It is about 25%. 
opqpop
Sep 29, 2010
 Hurray for geometric probability! It's very powerful 
abharath27
May 13, 2013
 You are making distinction between the first cut and second cut...
So a cut of pieces of size 4,4,4 will be counted twice..
This isn't correct right?
The correct answer then should be 12.5% 
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