The Mad Carpenter
October 8, 2006 am31 8:13 am
The Mad Carpenter builds stools. He attached legs to a seat. But he is totally, certifiably insane. Instead of planning his carpentry, he attaches legs at random locations around the edge of the seat. After attaching a leg, he attempts to stand the stool up. If it stands, he starts a new stool. If it falls, he attaches another leg.
What is the average number of legs on the Mad Carpenter’s stools?
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JD2718 
This question looks lovely. My initial guess is 4, but that isn’t based on anything other than the knowledge that the answer must be larger than 3 and the fact that nice problems often have nice answers. I don’t really have a grasp on how to attack it, yet — I think I’m going to end up doing a bunch of calculus and then try to figure out how to figure it out without that.
Also, congratulations on your limmerick prize :)
I should have asked first: What is the probability that a stool will stand with just 3 legs?
I think there is a 25% chance that 3 legs is enough.
In order for a stool to stand with 3 legs (assuming your mad carpenter isn’t so wacky as to be attaching legs to the top or sticking out the sides of the seat), then the legs need to be non-collinear. Are you saying that the carpenter only attaches them to the perimeter of the seat? That constrains the problem in a much more plausible and solvable way. If so, then I think the model you’re looking for is that the three legs need to be distributed in such a way that there’s no arc between any two adjacent legs measuring greater than 180 degrees. (The legs have to be spread out beyond one half of the seat in order to support it.)
So, going with these assumptions, let’s drop the first leg anywhere on the circle. Call that zero degrees. The other two legs go at x and y degrees (assuming the carpenter can’t make legs exist at the exact same location). Now, what is the chance that there exists an arc greater than 180 between any two of these three? Is that the question?
Nice restatement, mrc.
I’m agree with shuusaku. I’ve also calculated the probability that a stool with 4 or 5 legs stands up. Unfortunately, the only way I have to do this is with ugly calculus, although I have a definite memory of having once seen a non-calculus solution. Any hints, Jonathan?
Maybe we could think about it like this: The placement of the first leg is unimportant. Wherever it happens to be placed, we then draw a line from that leg through the exact center of the stool to the other side. This diameter divides the stool in half. Now the question is, what are the chances that the two remaining legs will be randomly placed on the same half of the circle? I think that’s 1/2. So half the time, the stool will fall over with 3 legs, and half the time it will stand. (As long as you agree with this interpretation of the physical situation.)
Now, back to the original question… what is the expected value of the number of random legs needed to make the stool stand?
I solved this first years ago. I still have the files (pieces of paper), but do not remember the solution. I am trying to solve this along with you (if I/we are genuinely stuck I will open the files.)
Now, I think that calculus can be avoided, but ultimate we will need a series and some nifty factoring (poor man’s calculus).
The simpler question, what is the probability that a 3-legged stool will stand, is 1/4. I created a Leg 2 vs. Leg 3 graph (how’s this for non-calculus?) and shaded the areas corresponding to a stool standing.
(I get two right triangles, the first with vertices (0,180), (180,180), and (180,360), the second with vertices (180,0), (180,180) and (360,180). The area is one fourth of the total)
I used calculus.
Pacing two legs at angle alfa the probability for the stool to stand by placing a third leg is alfa/(2*pi)
Then integrate(sum) over all alfa from 0 to pi which gives pi/4.
Probability one matches integrating one over 0 to pi which is pi, this gives (pi/4)/pi = 1/4.
For 4,5,… the integral get more complicated as JBL suggested.
mrc, having one leg on each side isn’t good enough: imagine that one leg is on each side, but they’re both very close to your first leg.
Jonathan, that geometric probability works very nicely for the case with 3 legs, but I’d like to see you shade the regions of a 4-dimensional cube which work when you have 5 legs :-)
By the way, I think that the probability it falls down with n legs is n / 2^(n – 1)
oh, darn, I knew there had to more to it!
Working from the assumption that my guess is correct, I can carry on to the end of the question as follows:
If n/ 2^(n – 1) fall down with n legs and (n – 1)/ 2^(n – 2) fall down with (n – 1) legs (where n has to be at least 2 for these to make sense), the difference = (n – 2)/2^(n – 1) is the number which stand up for the first time with n legs. Then the expected value we need is the sum
n(n – 2) / 2^(n – 1)
where n starts at 2 and runs over all positive integers. Unfortunately, (maybe predictably) my initial guess was a little low, although the principle was correct: the answer is actually 5.
This is of course still based on my un-proven guess.
Did it occur to anyone other than me that it *will* stand on simply ONE leg? It will, ya know.
Don’t over-think things. Simplify. Put the leg on the outer rim and stand the stool like a lolipop.
It’ll work.
And for those die-hards, one leg, long enough, will hold the stool up at a 45 degree angle, *DEPENDING* on how that leg is configured. If the leg has a base long enough and of equal weight to the seat itself, it will cantilever itself.
sorry I ran away for so long. JBL is correct.