Lumberjack Analysis

gingertoad

New member
Joined
Dec 7, 2006
Messages
1
Here's a question from a couple years ago I've never managed to figure out.

There are two parts to it, with the first being pretty simple and the second quite a bit deeper.


We start out with a lumberjack chopping down trees. It takes four successful chops to take down a tree. (Small trees...) But there's a chance each swing can miss. How many swings on average will it take to down each tree?


{Call the miss chance "m". (1-m) hits per swing, 4 hits needed --> 4/(1-m) swings needed.}


Now things get complicated...

Everyone wants to be a lumberjack! To capitalize on this abundance of labor, the logger boss is considering doubling up lumberjacks to get those trees down faster. When doubled up, the two lumberjacks swing at exactly the same time, with exactly the same chance of missing. Though it still takes four hits to down a tree, the catch is that one lumberjack can miss when the other hits, meaning five successful chops might land before a tree falls.

The boss wants to know 1) how many swings on average will it take to down each tree (one "swing" here involves both of them swinging) and 2) how often will you get a 5th "wasted" hit landing?


{My intuition tells me the answer to 1) is not 2/(1-m); it should be larger, thanks to those wasted hits. Beyond that, I'm stuck.}
 
Consider the entire distribution of each double-whack...

I suppose the whack probabilities are independent.

\(\displaystyle Pr(Two\;Misses)\;=\;m^{2}\)

\(\displaystyle Pr(One\;Miss)\;=\;2*m*(1-m)\)

\(\displaystyle Pr(Two\;Hits)\;=\;(1-m)^{2}\)

It shouldn't take too long to construct a small tree. (pun intended)
 
Top