The truel problem, why aren't the better players intentionally missing? I am working on this problem: https://puzzle.dse.nl/teasers/index_us.html#the_truel
Following the solution here: https://puzzle.dse.nl/teasers/the_truel_us.html
But I don't understand the part where: 

P(A,BCA) < 1/2 (since B will definitely shoot at A, because P(B,AB) = 0!)

If B shoot A then $P(B, BCA) = 0.8 \times 4/9 + 0.2 P(B, CAB)$
If B shoot C then $P(B, BCA) = 0.2 P(B, CAB)$
If B misses intentionally then $P(B, BCA) = P(B, CAB)$
While it is obvious that B should prefer shooting A over C. I can't see why shouldn't B miss intentionally. Also, if B shoots A, why is $P(A, BCA) < 1/2$?
I have the same question for: 

P(A,CBA) < 1/2 (since B will definitely shoot at A, because P(B,ABC) =
  0!)

Thank you in advanced.
 A: *

*The answer to first question, why B shouldn't miss intentionally when it's his turn to shoot, and both C and A are still alive, is given by the analysis of the position CAB, which is the one that will arise if B does miss (whether deliberately or not).  This is done in row CAB of the table, where it's established that $\ P\left(B,CAB\right)=0\ $.  Thus, you can simplify your set of probabilities for this case to:
If B shoots at A, then: $$ P(B, BCA) = 0.8 \times 4/9 + 0.2 P(B, CAB)=\frac{16}{45}\ .$$
If B shoots at C, then:
$$ P(B, BCA) = 0.2 P(B, CAB)=0\ .$$
If B misses intentionally, then:
$$P(B, BCA) = P(B, CAB)=0\ .$$

*The answer to the second question, why $\ P(B, BCA)<\frac{1}{2}\ $ when B shoots at A, follows from the fact that if he does so, he will kill A with probability $\ \frac{4}{5}\ $. Thus, A's probability of survival cannot be more than $\ \frac{1}{5}\ $ in this case.

*The answer to the third question, why $\ P(A, CBA)<\frac{1}{2}\ $, is similar to those to the first two.  Row CBA of the table estsblishes that C will deliberately choose to miss from this position, thus leaving the position BAC, from which B will fire at A, killing him with probability $\ \frac{4}{5}\ $. 
Is the above argument circular?  While I'm fairly sure it's the argument which is being relied upon to justify this purported solution, I agree with the OP's nice observation in the comment below that it appears to be circular.  In fact, if the second condition—"they will continue firing at each other in this order until only a single person is alive"—given here in the description of the problem is taken strictly literally, the purported solution, in which one of the players is allowed not to fire at anyone else, would appear to me to violate it.
On the other hand, if the second condition is interpreted more loosely, as allowing any of the players to deliberately not fire at any of the other players, then I believe the problem no longer remains well-posed, because there's no clear indication of what happens if all three choose to keep  missing deliberately.  Do they all keep firing until two of them die of starvation, or are they allowed to call a halt to the whole thing as being pointless to continue?  In any case, the best one can say of the proposed solution is that it's a Nash equilibrium, but it certainly doesn't appear to be the only one, even for the two-player cases, because the strategy pairs or triples in which all players choose to miss deliberately seems to me to be another.
