Okay, here is the my full analysis for my last puzzle:

The key to figure the whole thing out is to work backwards.

Firstly, let's imagine what will happen if all 3 criminals have missed one shot (obviously Alan would have missed his 1st shot intentionally). Each of them must aim his next shot at the string of somebody else, or face certain death.

There are 6 possible orders:

(notation: A=Alan, B=Bob, C=Chuck, S(P)=Survival probability for P)

1. ABC: A will kill B and take a 50% chance to survive C's shot.

(If A kills C, he will only have a 20% chance to survive B's shot.)

=> S(A)=0.5, S(B)=0, S(C)=0.5

2. ACB: A will kill B and take a 50% chance to survive C's shot.

(If A kills C, he will only have a 20% chance to survive B's shot.)

=> S(A)=0.5, S(B)=0, S(C)=0.5

3. BAC: B will try to kill A because even if he kills C, he won't survive A's shot.

If B misses, he will die, and A will survive by killing C.

If B kills A, he will have a 50% chance to survive C's shot.

=> S(A)=0.2, B=0.8x0.5=0.4, C=0.8x0.5=0.4

4. BCA: B will try to kill A because even if he kills C, he won't survive A's shot.

If B misses, he will die, and A will have a 50% chance to survive C's shot.

If B kills A, he will have a 50% chance to survive C's shot.

=> S(A)=0.2x0.5=0.1, S(B)=0.8x0.5=0.4, S(C)=0.2x0.5+0.8x0.5=0.5

5. CAB: C will try to kill A because even if he kills B, he won't survive A's shot.

If C misses, he will die, and A will survive by killing B.

If C kills A, he will have a 20% chance to survive B's shot.

=> S(A)=0.5, S(B)=0.5x0.8=0.4, S(C)=0.5x0.2=0.1

6. CBA: C will try to kill A because even if he kills B, he won't survive A's shot.

If C misses, he will die, and A will have a 20% chance to survive B's shot.

If C kills A, he will have a 20% chance to survive B's shot.

=> S(A)=0.5x0.2=0.1, S(B)=0.5x0.8+0.5x0.8=0.8, S(C)=0.5x0.2=0.1

Now we can start analysing their decisions in the "1st round":

1. ABC: If A & B somehow both miss, C knows if he kills one of the others he has a 20% (or 0%) chance to survive.

But if C misses intentionally he will have a 50% chance (as listed above). Therefore C will miss intentionally.

If A somehow misses, B knows if he doesn't kill A right there with his 1st shot, A will have another shot to kill him.

Then B will be as good as dead meat. Therefore B will try his best to kill A in that situation.

Thus A knows that if he doesn't kill B right there with his 1st shot, B will try to kill him immediately.

Then the chance for A to survive won't exceed 20%.

Hence A will kill B with his 1st shot and take a 50% chance to survive C's 1st shot (and then kill C with his 2nd shot).

=> S(A)=0.5, S(B)=0, S(C)=0.5

2. ACB: If A & C somehow both miss, B knows he must kill A right there with his 1st shot, or die (as listed above).

If A somehow misses, C knows if he kills one of the others he has a 20% (or less) chance to survive.

But if C misses intentionally he will have the next shot against the survivor between A & B.

Therefore C will miss intentionally, and take a 50% chance to survive from his guaranteed 2nd shot.

Thus A knows that if he doesn't kill B right there with his 1st shot, B will get a shot to try to kill him.

Then the chance for A to survive won't exceed 20%.

Hence A will kill B with his 1st shot and take a 50% chance to survive C's 1st shot (and then kill C with his 2nd shot).

=> S(A)=0.5, S(B)=0, S(C)=0.5

3. BAC: If B & A somehow both miss, C knows if he kills one of the others he has a 20% (or 0%) chance to survive.

But if C misses intentionally he will have a 40% chance (as listed above). Therefore C will miss intentionally.

If B misses, A knows if he doesn't kill B right there with his 1st shot, B will get another shot to try to kill him.

Then the chance for A to survive won't exceed 20%.

Therefore A will kill B with his 1st shot and take a 50% chance to survive C's 1st shot (and then kill C with his 2nd shot).

Thus B knows if he doesn't kill A right there with his 1st shot, A will kill him immediately.

Hence B will try to kill A with his 1st shot.

If B misses, A will kill B immediately and take a 50% chance to survive C's 1st shot before killing C with his 2nd shot.

If B kills A, C will try to kill B with both his 1st and 2nd shots (provided he survives B's 2nd shot).

However if C misses both times he will die immediately, leaving B as the survivor.

=> S(A)=0.2x0.5=0.1, S(B)=0.8x0.5x0.8+0.8x0.5x0.2x0.5=0.36, S(C)=0.2x0.5+0.8x0.5+0.8x0.5x0.2x0.5=0.54

4. BCA: If B & C both miss, A knows if he doesn't kill B right there B will shoot him with his 2nd shot (as listed above).

Then the chance for A to survive won't exceed 20%. Therefore he will kill B and take a 50% chance to survive C's 2nd shot.

If B misses, C knows if he kills one of the others he has a 20% (or 0%) chance to survive.

But if C misses intentionally he will have the next shot against the survivor between A & B.

Therefore C will miss intentionally, and take a 50% chance to survive from his guaranteed 2nd shot.

Thus B knows if he doesn't kill A right there with his 1st shot, A will get a shot to kill him.

Hence B will try to kill A with his 1st shot.

If B misses, C will intentionally miss, then A will kill B and take a 50% chance to survive C's 2nd shot.

If B kills A, C will try to kill B with both his 1st and 2nd shots (provided he survives B's 2nd shot).

However if C misses both times he will die immediately, leaving B as the survivor.

=> S(A)=0.2x0.5=0.1, S(B)=0.8x0.5x0.8+0.8x0.5x0.2x0.5=0.36, S(C)=0.2x0.5+0.8x0.5+0.8x0.5x0.2x0.5=0.54

5. CAB: If C & A somehow both miss, B knows if he misses he will only have a 40% chance to survive (as listed above).

If B kills C he will be killed by A immediately. But if B kills A he will have a 50% chance to survive C's 2nd shot.

Therefore B will try to kill A in that situation.

If C misses, A knows if he doesn't kill B right there with his 1st shot, B will to try to kill him immediately.

Then the chance for A to survive won't exceed 20%.

Therefore A will kill B with his 1st shot and take a 50% chance to survive C's 2nd shot.

Thus C knows if he kills one of the others he has a 20% (or less) chance to survive.

But if C misses intentionally he will have the next shot against A.

Hence C will miss intentionally, and take a 50% chance to survive from his guaranteed 2nd shot against A.

=> S(A)=0.5, S(B)=0, S(C)=0.5

6. CBA: If C & B both miss, A knows if he intentionally misses he will only have a 10% chance to survive (as listed above).

If A kills C, he will only have a 20% chance to survive B's 2nd shot.

Therefore A will kill B and take a 50% chance to survive C's 2nd shot.

If C misses, B knows if he doesn't kill A right there with his 1st shot, A will kill him immediately.

Therefore B will try his best to kill A in that situation.

Thus C knows if he kills one of the others he has a 20% (or less) chance to survive.

But if C misses intentionally he will have the next shot against the survivor between A & B.

Hence C will miss intentionally, and take a 50% chance to survive from his guaranteed 2nd shot.

If B misses, A will kill B immediately and take a 50% chance to survive C's 2nd shot.

If B kills A, he will have a 50% chance to survive C's 2nd shot.

=> S(A)=0.2x0.5=0.1, S(B)=0.8x0.5=0.4, S(C)=0.2x0.5+0.8x0.5=0.5

So collectively,

Combined survival probability for Alan = (0.5+0.5+0.1+0.1+0.5+0.1)/6=0.3

Combined survival probability for Bob = (0+0+0.36+0.36+0+0.4)/6=1.12/6=14/75=0.186667

Combined survival probability for Chuck = (0.5+0.5+0.54+0.54+0.5+0.5)/6=3.08/6=77/150=0.513333

And the ratio of their survival probability is 45:28:77.

My 200th post!