A gun shots bullets with different speed each second.
The bullets form a straight line.
The bullets keep their speed (not getting slower or faster).
But if 2 of the bullets collide, 2 of them are destroyed (annihilated)
The gun stops after 12 bullets fired.
What is the probability that all the bullets will be destroyed ?
There is a beautiful math formula for every even bullets fired, find it!
This is a puzzle from a good website, I will show the link after the puzzle solved.
I am adding another answer since my first try was way off the mark.
Key thing to notice is that, for 12 bullets to be eliminated, there has to be 6 collisions in total involving any bullets in any order.
That means we don't have to workout all possible relative speeds and their respective probability of total elimination. We just need to calculate the probability of 6 succesive collisions to happen.
If the first bullet is the fastest, it will escape and the probability of this happening is 1/12. But if first bullet is not the fastest, there is guaranteed to be at least one collision (it doesn't matter that the first bullet may escape this collision, we will account for it in later collisions).
So the probability of at least one collision happening when 12 bullets are involved is:
$$p_{12} = 1 - \frac{1}{12}$$
For the second collision, we only have 10 bullets left. Following the same logic, if the first among the surviving bullets is not the fastest, there is chance for at least one more collision happening. So the independent probability of second collision is $p_{10} = 1 - \frac{1}{10}$.
In general, so for a scenario with n bullets, the independent probability of at least one collision happening is:
$$p_{n} = 1 - \frac{1}{n}$$
Looking back at our key insight, the probability of total annihilation is same as that of 6 successive collision events to happen, which is:
\begin{align} p &= p_{12}*p_{10}*p_8*p_6*p_4*p_2 \\\\ &= (1 - \frac{1}{12})*(1 - \frac{1}{10})*(1 - \frac{1}{8})*(1 - \frac{1}{6})*(1 - \frac{1}{4})*(1 - \frac{1}{2}) \\\\ &= 0.2256 \end{align}
General formula for n bullets would be: \begin{align} p &= (1 - \frac{1}{n})*(1 - \frac{1}{n-2})...(1 - \frac{1}{4})*(1 - \frac{1}{2}) \end{align}
Let us try to solve this by solving smaller problems and gradually adding more complexity to it.
First we estimate the probability of two consecutive bullets colliding. We can plot the relative time-distance graph for first bullet as a 45 degree line. Since both time and distance are infinite, we can adjust the scale to make it 1:1.
Now the second bullet won't hit the first bullet as long as its speed is lower than that of first, that is the bottom half of our plot. The probability can be arrived by dividing the area under the line with the total area of the chart
$p = 0.5$
This is the probability of drawing 11 lines of any angle (speed) as long as they are less than 45 degrees. Since speed of each shot is purely random and independent of prior shot, the actual probability is a conditional probability of the non-event happening 11 times (the number of possible encounters with the following bullets) in succession.
$p = (0.5)^{11}$
This is simple. It is merely the inverse probability of previous event. Since this event is directly related to the answer we want, let us index this as p1.
$p1 = 1 - (0.5)^{11}$
For 2nd bullet (the cardinal position is relative, number 2 referring to the next non-destroyed bullet), it has 9 possible encounters (a pair lost to prior collision). So $p2 = (0.5)^9$. Similarly probabilities for remaining bullets can be calculated as:
$p2 = 1 - (0.5)^9$
$p3 = 1 - (0.5)^7$
$p4 = 1 - (0.5)^5$
$p5 = 1 - (0.5)^3$
$p6 = 1 - (0.5)^1$
Now I consider this as the probability of above 6 events happening successively.
$p = p1 \times p2 \times p3 \times p4 \times p5 \times p6$
$p = (1−(0.5^{11}))\times(1−(0.5^9))\times(1−(0.5^7))\times(1−(0.5^5))\times(1−(0.5^3))\times(1−(0.5^1))$
$p = 0.4195$
So there is a 42% chance that all bullets will be lost.
There may be some holes in my logic, particularly with respect to chaining events $p1$-$p6$. I considered the 6 possible collisions as independent events, I am not sure whether my assumption is valid.
