Shooting bullets

Question

This is from http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/challenges/May2014.html

Every second, a gun shoots a bullet in the same direction at a random constant speed between 0 and 1.

The speeds of the bullets are independent uniform random variables. Each bullet keeps the exact same speed and when two bullets collide, they are both annihilated.

After shooting $n$ bullets, prove that the probability that eventually all the bullets will be annihilated is zero if $n$ is odd and $\prod_{i=1}^{n/2} \frac{2i-1}{2i}$ when $n$ is even.

I tried to write recursion without success and Markov chain's but I don't see how them helps here. The case of $n\equiv 1 \pmod 2$ seems to be trivial.

I was going to say that IBM post the solution at the end of the month, but http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/solutions/May2014.html says not, they have not added any successful names since early June — Henry, Nov 27 '14 at 01:05
For just $4$ bullets it seems at first glance, so unchecked, that $\frac{13}{24}$ of cases will fail and $\frac{8}{24}$ will succeed, without considering actual speeds. For the remaining $\frac{3}{24}$ of cases it is harder, so this may not be a good or scalable approach. — Henry, Nov 27 '14 at 08:44
@BrianM.Scott: I found a link on the main problem page, but you are correct there is no solution given. — Ross Millikan, Nov 27 '14 at 15:29
@Henry: I get $7$ permutations that succeed unconditionally and $4$ that can go either way: $2431,3421,4213$, and $4312$ (where larger numbers are faster, and motion is left to right). Each of these succeeds half the time, so we get the correct result, $\frac9{24}=\frac38$. — Brian M. Scott, Nov 27 '14 at 16:41
@Brian M.Scott: so I missed one. But why do they each succeed half the time? Perhaps you could pair 2431 with 4312 and 3421 with 4213 to add up to $2$ and so make the overall expectation $\frac{7+2}{24}$ — Henry, Nov 27 '14 at 18:05
@Henry: I’ve not actually checked the details, but the speeds are uniformly distributed, and in each of these cases the outcome depends on which of two collisions occurs first suggests a $50$-$50$ split. I could be wrong, though. — Brian M. Scott, Nov 27 '14 at 18:32
@Brian M.Scott: I do not think the ordered speeds are uniform (more likely to be Beta distributed) — Henry, Nov 27 '14 at 18:45
@DanielFischer Thank you for the reversion, that user's edits are ... incredibly weird. — pjs36, Dec 16 '15 at 19:10
@pjs36 I looked at a few others and those seemed a bit overdone, but decent. Should I take a longer look? — Daniel Fischer, Dec 16 '15 at 19:12
@DanielFischer The two oddest themes are that \large fonts are used fairly frequently when none were present, and the command \: (that I'd never seen before) is used for all kinds of spacing; at least once in lieu of the align environment. Generally not terrible edits, just achieved in weird ways! — pjs36, Dec 16 '15 at 19:17

score -1 · Answer 1 · answered Aug 03 '15 at 17:45

I'm wondering if we actually need any more information about those velocities.

Let's take the case in which $n=2.$ The formula says the probability of total annihilation should be $\frac38=\frac9{24}.$

The velocities are $v_1<v_2<v_3<v_4,$ and I'll refer to these velocities by their subscripts.

There are 24 possible orders in which the bullets are fired.

In 7 of those 24 orders, annihilation is certain: 2143, 3142, 3241, 3412, 4132, 4231, 4321.

In 13 of the 24, annihilation will not happen (which includes all 1xxx and xxx4 orders).

But that leaves the following four orders: 2431, 3421, 4213, 4312.

In each of those four, whether total annihilation occurs or not depends on which order the collisions take place. And that depends on what the velocities are.

Is there any way to partition down these four events so they split evenly? Maybe, but I haven't worked out the details.

For instance, if 31 happens before 43 then 2431 becomes "no" and 4312 becomes "yes", leaving the other two undecided. That sort of thing.

Shooting bullets

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