The Logical Puzzle

I posed the problem to my mother and she came up with a slightly easier analysis. Again, we're going to double an integral (and it will work out to be a similar integral, but the process is slightly easier to understand). This time we'll assume that x is the length of the shorter stick after the first, random, cut. So we're integrating from x equals 0 to 0.5 and then doubling the result. Now, if the first cut is in the first half of the stick, we figure out where the second cut can go such that we can still make a triangle. The second cut can't be before 0.5, because then the "last" piece would be longer than 0.5. However, the "middle" piece can't be longer than 0.5, so the cut can't be after x + 0.5. So the range for the second cut is from 0.5 to x + 0.5, and the probability of making the cut in that range (given that you could cut the sticks anywhere) is x!

Thus, the probability that you can make a triangle is twice the integral of x from 0 to 0.5. The latter is 0.125, thus the probability is 0.25.

I agree with your initial analysis. However, I don't think 50% is the right answer. My initial analysis was wrong (I created a simple simulation in Excel to generate a lot of random tests quickly so I could validate my intuition).

Here's my analysis. The first cut is completely random. As a result, our analysis is equivalent to twice the integral from 0.5 to 1 of the probability that we will end up with one stick that is 0.5 or greater given that the longer stick is the length from 0.5 to 1.

Once we have a longer stick of length x, the first probability is that we will cut the shorter stick. That probability is 1-x. The second probability is that given the longer stick, we will never the less cut a stick longer than .5 out of it. If you visualize the cut, there are two ranges where you could make that cut - one at each end of the stick. The length of each range is x-0.5. The probability that you will make the cut in each range is (x-0.5)/x, so the probability that you will end up with a stick longer than 0.5 if you select the longer stick for the cut is 2*(x-0.5)/x. Of course, that gets multiplied by the probability that you will pick the longer stick, which is x (since the probability you would pick the shorter stick is 1-x).

So, the answer is 2 time the integral from 0.5 to 1 of:

1-x + x*2*(x-0.5)/x

which is

1-x + 2*(x-0.5)

which is

1-x + 2x-1

which is

x

The integral from 0.5 to 1 of x is x^2/2 evaluated at 1 minus the same evaluated at 0.5, which is 0.5 - 0.125, or 0.375, but then we still have to double that (remember that we have to account for switching the sticks), which gives us 0.75, or 3/4s! Or, rather, the odds of ending up with three pieces that make a triangle are only one in four!

I'm sure there's an easier way to do this, though.

Sorry, my line breaks disappeared -
So:
x < y + (1-x-y);
x < 1-x;
2x < 1;
x < 1/2;

Sorry, my line breaks disappeared -
So:
x < y + (1-x-y);
x < 1-x;
2x < 1;
x < 1/2;

You cut a stick (of length = 1 unit) into three pieces randomly. You can only make a triangle out of them if the longest piece (x) is LESS than the combined length of the other two pieces (y and (1-x-y))

So: x < y + (1-x-y)
x < 1-x
2x < 1
x < 1/2;

Therefore, the longest piece has to be less than half the length of the whole stick. You have (just under) a 50% chance of cutting a piece too long. (If it is exactly 1/2 the length, I wouldn't count that as a triangle.)