I took a shot at this problem by coding up a simulation in R. I followed the first part of @Dmitry Kamenetsky's solution but took @justhalf's comment about how colliding works into account.

I'll start with an example of how we can determine when two bullets will collide. Below is an example of two bullets, one fired at 4 seconds with a speed of ~0.29 and the other fired at 6 seconds with a speed of ~0.93. They collide at ~6.9 seconds.


(Horizontal red line is the x-axis. Vertical red line is where the lines intersect)
[![Two example bullets][1]][1]


  [1]: https://i.sstatic.net/xrvQl.png


We can determine when they will collide by estimating what time (if any) satisfies the following: 

>! 0.29 * (time - 4) = 0.93 * (time - 6)

Using this process, we can determine when and if each bullet collide and determine the order in which they collide. Here is my simulation process:

>! 1. Simulate the bullets fired by drawing from a Bernoulli distribution with 10 trials and probability of 0.6. Assign a speed to each bullet (uniform distribution from 0 to 1). If there are an odd number of bullets, then stop (because then there will always be at least one bullet remaining).

>! 2. Estimate, for all combinations of bullets, when they would collide (or not collide), ignoring all other bullets. (See table below for an example)

>! 3. Determine which two bullets are the first to collide. Then, ignoring bullets we know already collided, determine which two bullets collide next. 

>! 4. Repeat step 3 until all the bullets have collided or no more bullets collide.

Here is an example of the collision table, which records when each bullet would collide: (The bottom diagonal is blank because it would just mirror the top diagonal)

|     | Bullet 1  | Bullet 2 | Bullet 3  |    Bullet 4 |
| ---- | ------ | ------ | ----- | ------ |
|Bullet 1 |   | 3.615827  |  | 10.787499 |
|Bullet 2 |    |   |     |     |
|Bullet 3  |   |   |   | 8.563391  |
|Bullet 4 |   |    |  |        |

The first two to collide are bullets 1 and 2. Ignoring bullets 1 and 2 because they have already collided, the next to collide are bullets 3 and 4. As it turns out, bullets 1 and 3 would never collide (even without bullet 2), but bullets 1 and 4 could collide (without bullets 2 and 3).

Using this process, I performed 100,000 simulations. The proportion of times when there were no bullets or all the bullets disappeared was:

>!  0.21162