Probability that at least one of the bullets will go on forever.

Question

There's a gun located on an infinite line, let's say at 0 on number line. It starts shooting bullets along that line, +X axis, at the rate of one bullet per second.

Each bullet has a velocity in the range [0, 1] m/s randomly chosen from a uniform distribution.
If two bullets collide (are at the same spot at the same time) they explode and disappear.

What is the probability that at least one of the infinite bullets will infinitely fly without colliding with another bullet?

Hint: what happens in there is a long subsequence that is monotonic increasing in velocity? — Stella Biderman, Jan 28 '16 at 04:52
Intuitively I think the probability is 0, there will always be infinitely faster bullets. — CAGT, Jan 28 '16 at 04:53
Further hint: There is a zero-one law that is applicable here — Stella Biderman, Jan 28 '16 at 04:55
@StellaBiderman This question was posted in a group-chat on whatsapp, i tried to solve it using geometric approach, my answer is 0.50. Please help ^_^ — ABcDexter, Jan 28 '16 at 04:58
@Logophobic Actually the person who sent it was referring to this berkeley puzzles — ABcDexter, Jan 28 '16 at 05:01
For the case of a finite number of bullets see this question — Ross Millikan, Jan 28 '16 at 05:32
@RossMillikan Yes, thanks :-) I saw that question is the plausible cases of the "question already exists". — ABcDexter, Jan 28 '16 at 05:39

score 1 · Accepted Answer · answered Jan 28 '16 at 09:48

1

Let $X_n\stackrel{\mathrm{iid}}\sim U(0,1)$. For all positive integers $n$ and $j$ we have $$\mathbb P\left(\bigcap_{k=n+1}^{n+j}\{X_k\leqslant X_n\} \right) = 2^{-j}, $$ and $$\bigcap_{k=n+1}^{n+j}\{X_k\leqslant X_n\}\supset \bigcap_{k=n+1}^{n+1+j}\{X_k\leqslant X_n\}, $$ and therefore \begin{align} \mathbb P\left(\bigcap_{k=n+1}^{\infty}\{X_k\leqslant X_n\} \right) &= \mathbb P\left(\bigcap_{k=n+1}^{\infty}\{X_k\leqslant X_n\} \right)\\ &= \lim_{j\to\infty}P\left(\bigcap_{k=n+1}^{n+j}\{X_k\leqslant X_n\} \right)\\ &= \lim_{j\to\infty}2^{-j}\\ &=0. \end{align} Hence $$ \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n+1}^\infty\{X_k\leqslant X_n\} \right) \leqslant \sum_{n=1}^\infty \mathbb P\left(\bigcap_{k=n+1}^\infty\{X_k\leqslant X_n\} \right)=0, $$ so that $$\mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n+1}^\infty\{X_k> X_n\} \right) = 1 - \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n+1}^\infty\{X_k\leqslant X_n\} \right)=1. $$ For each bullet $n$, there is a bullet $k>n$ traveling faster than bullet $n$ with probability $1$. It follows that the probability of at least one bullet flying forever without collision is zero.

answered Jan 28 '16 at 09:48

Math1000

38,041

1

Your answer is most probably correct. I'm waiting for someone to show otherwise (am y other answers) . – ABcDexter Jan 28 '16 at 12:47
1

It's possible that a bullet will travel forever even if there's a later, faster bullet (because some third bullet interposes itself between them). – Micah Jan 28 '16 at 14:35
This answer to the question is counter-intuitive :-) Should I write my approach ? And then mark it as community wiki so that everyone can critique. – ABcDexter Jan 29 '16 at 04:15
1

@ABcDexter By all means. – Math1000 Jan 29 '16 at 05:15
@Math1000 Posting my answer... – ABcDexter Feb 01 '16 at 13:50

Probability that at least one of the bullets will go on forever.

1 Answers1