May 2014
The answer for the general n bullet problem is zero if n is odd (n=2m+1) and \Pi_{i=1}^m ((2*i-1)/(2*i)) when n is even (n=2m).
So for n=20, the answer is (1*3*5*7*9*11*13*15*17*19) / (2*4*6*8*10*12*14*16*18*20) = 46189/262144 = 0.1761970520.
However, it is not easy to prove this, and several solvers sent us incorrect proofs.
Note, for example, that the same order of bullet speeds can yield different collision results:
For example, let's look at two cases of four bullet speeds:
A. 0.5, 0.65, 0.9, 0.6
B. 0.5, 0.8, 0.9, 0.6
In both cases, the permutation order is the same. The first bullet is the slowest, the second is the second fastest, the third is the fastest, and the last is the second-slowest.
But if we examine the collisions pattern we see that
A. 0.5, 0.65, 0.9, 0.6 will yield zero bullets (first the 0.9 will run into the 0.65 and then the 0.6 will collide with the first 0.5 bullet);
and
B. 0.5, 0.8, 0.9, 0.6 will give two live bullets forever (first the 0.8 will annihilate the 0.5, then the remaining 0.9 and 0.6 will run forever).
As a result, we are keeping the challenge open for a while to let you send us your proofs.
If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com