Chronos does normally know his stuff, but I still don’t get this.
Large star, lower entropy than 2 smaller
OK
To make 2 smaller would require increase in entropy of another system…
Why is that? 1 star becoming 2 would increase the entropy of the system. Why should this need to be offset in any way?
…greater than the decrease of this one
It’s not a decrease 
You have to add energy to lift the smaller stars out of each others’ gravity wells. Without that addition, there’s no separation.
No, the argument is that the entropy of the large star would be larger than the entropy of the two smaller stars, thus the spontaneous fission would decrease entropy, and hence, be thermodynamically impossible. As I said above, this is exactly the case for a black hole (which is simply the object of a given volume for which the entropy is maximal, hence the limiting case for this sort of thing). Since energy must be conserved, and the energy of a black hole scales with its radius, the sum of the radii of the resultant black holes must equal the radius of the original one; but entropy scales with the area, so the two black holes would have a much smaller area, and hence, a lower total entropy, than the original one.
EDIT: Ah, I think Chronos maybe typed ‘lower’ where he meant to say ‘larger’. Otherwise, I’m also confused…
I think you can make strong arguments that splitting a star into two stars of comparable masses greatly reduces the maximum entropy of the system, which itself is suggestive that it simply is not a process that occurs naturally.
Stellar fission can feasibly can occur when the ‘star’ in question is a cloud of protostellar gas and indeed it is thought to be a common process due to the very large number of binary star systems observed.
Yeah, I made a thinkographical error there. When I wrote that a single star has lower entropy than two stars, I should have said higher. Thanks for keeping me honest.
You haven’t addressed the most important part. Perhaps there might be a temporary state where two subsets of the original star are separated from each other. What keeps them from blowing up?
Stars are in a robust equilibrium involving temperature, gravity, and nuclear reactions. It’s covered in more detail in various textbooks and other discussions about stars, so I’m not going to go into detail here. The main point is that the temperature of a star is a function of the total mass.
Now in this situation, you are taking the mass from a large star and subdividing it into two or more smaller stars. The stuff is still going to have the temperature of the large star. It’s not going to cool off in to any significant degree in the time this partition takes. But the appropriate temperature for those small stars is significantly less than for the large star. So they’re no longer in the equilibrium mentioned above. They’ll expand until that’s reached. For a small star, that will most likely result in the star simply blowing apart.
In order to make this work, you have to assume there’s a star-size Maxwell’s Demon segregating the particles so that none of the resulting stars ends up at too high a temperature. Which brings us to the entropy point Chronos brought up.
As the star expands, it cools. As the interaction stretches the star’s mass in one direction more than the other two, it will expand in those other two directions. We’re not simply starting with two half-mass stars. Even if we were, I don’t agree with your assessment that the stars will blow apart. They will expand, certainly, and oscillate for a while, but I don’t believe most stars, if any, would have near the energy to “blow apart”, leaving no star behind. Do you really think that would happen?
As I said earlier, I’ve got lots of parameters to play with. You’ve got a hand-waving argument that doesn’t consider any of those parameters. What mass of star are you assuming? You aren’t. What mass of black hole are you assuming? You aren’t. And so forth. But somehow you think you’ve shown that no combination of all those parameters I listed earlier can pull the star apart in one dimension without it also being pulled apart in the other two. You’re nowhere near showing that.
Try adding more velocity; that’s the most important free parameter, surely. If the energy of impact greatly exceeds the gravitational binding energy of both stars, you’ll end up with a huge smear of gas that might eventually collapse into an arbitrary number of stars.
That’s why they came up with stellar nurseries 
But really, is that not caused by a supernova explosion, the death of one star leads to the creation of many?
I ran some numbers for the Sun.
I get the RMS velocity of Hydrogen in the Sun’s Core as about 607 km/s. I base that on the calculation found here, among others. That has 1.92 km/s for Hydrogen molecules at 300K. The RMS velocity varies like sqrt(T/M), so for Hydrogen atoms at 15 million K, I get 1.92 * sqrt(15,000,000/300 * 2/1). That’s the velocity at the very center, and it drops by the edge of the core (the temperature there is about half as hot). These velocities won’t change when we simply remove 1/2 the Sun’s mass (removed uniformly from throughout the Sun.
Second, I calculate the escape velocity of the Sun’s core. It was hard to find a reliable mass for just the Sun’s core, but I found a dataset here. I used the first link, which gives density versus radius, allowing me to get its mass at any radius. The Sun’s core is commonly defined as the inner 1/4 of its radius. I get 0.48 solar mass inside 0.25 solar radius. I also get 0.68 solar mass inside 1/3 solar radius.
The Sun’s escape velocity is 617.7 km/s at its surface. Escape velocity varies as sqrt(M/R), so for just the core of half of the Sun, the escape velocity will be smaller by a factor of sqrt((0.48/2) / (1/4)), giving 605 km/s. This is the escape velocity only considering the mass inside of 1/4 the Sun’s radius. So if all we had was 1/2 of the Sun’s core (i.e. 0.24 of the Sun’s mass), and none of the non-core, we might lose a substantial portion of that mass.
At 1/3 solar radii, the escape velocity is near its peak, different from the Sun’s by a factor of sqrt((0.68/2) /(1/3), giving a slightly higher 624 km/s. The core nuclei would have to pass 1/3 solar radius on their way out. Thus, even the center RMS velocity isn’t quite high enough to escape, although it’s close.
Of course, since the highest velocity atoms are at the center, they will lose speed (due to gravity) as they move outward to 1/3 solar radii, so they’ll be farther below the escape velocity at that radius than those numbers show. The outer part of the core already has a smaller temperature, hence smaller RMS velocity. So most of the Sun’s core’s mass won’t have a high enough velocity to escape.
Further inhibiting the escape of core nuclei is that the non-core half of the Sun’s mass is in the way, and the core gas will lose velocity as it plows into it. And of course, the non-core half of the Sun will have much lower velocities, and never be in danger of escaping.
tl/dr:
Cutting the Sun’s mass in half, and leaving its temperature and mass profiles unchanged, won’t cause the Sun to blow apart. Almost all of its mass will remain gravitationally bound, able to reform a star.
(Warning: This was pulled squarely out of my ass)
I think you almost need 3 separate stars in order to make a 4th star.
The first two stars that I think of is a white dwarf leeching material from a Red Giant spilling over the Roche limit. A more massive star (2-5x the Red Giant’s size, but in its main sequence phase) that is passing by this binary system would pull the transferring material out of the gravity “saddle” between the binary pair, and if the third star was passing by fast enough…would that be enough material liberated from the binary pair to reform into a small dwarf star?