Report For Bushindo - (others try at your own risk) in New Logic/Math Puzzles Posted March 18, 2013 Let me try and bring consensus here by wearing everyone down with a lengthy, unnecessarily tedious, and tiresome detailed solution. Outside of solving inequality, which I think is more revealing than equality, I don't see any significant difference in the results. I did use a little shortcut without giving an adequate explanation/justification for it. Therefore, I feel compelled to clarify the point where the reasoning ends and algebra begins. I'll stick to my nomenclature: probability of Bobby hitting target = b; probability of Cole hitting target = c. (To avoid typing subscripts.) If Cole shoots in the air, then Bobby shoots at Alex. 1) If Bobby misses, Alex kills Bobby and Cole gets just one shot at Alex. 2) If Bobby kills Alex then it is a duel between Cole and Bobby, with Cole shooting first. Cole's chance in that duel (D) is: D = c + (1-c)(1-b)D; From which we find: D = c/(1 - (1-b)(1-c)) = c/(b+c-bc). Thus Cole's survival probability if he shoots in the air (A) is:A = (1-b)c + bD = (1-b)c + bc/(b+c-bc). Now, here is the justification for the shortcut I used. If it suits my purposes, I can substitute A as following:A = (1-c)A + cA. (It is perfectly legitimate algebraic substitution.) If Cole shoots at Alex, then: 1) Cole hits the air instead at the probability of (1-c). Thereafter, it is the same scenario as when Cole hit the air on purpose. With overall Cole's survival probability in this variation: (1-c)A. 2) Cole hits Alex at the probability of c. Then Bobby gets the first shot at Cole. We are not interested in variation where Bobby hits Cole, since we are calculating Cole's survival chances. So if on his first shot Bobby misses Cole at the probability of (1-b), then it is the same duel with Cole's first shot. Thus the probability of Cole's survival in this situation is c(1-b)D = c2(1-b)/(b+c-bc). Overall, Cole's probability of survival when shooting at Alex (S) is:S = (1-c)A + c(1-b)D = (1-c)((1-b)c + bc/(b+c-bc)) + c2(1-b)/(b+c-bc). Now we must compare the two strategies. And find that when A > S, we shall advise Cole to shoot in the air. A > S; (1-c)A + cA > (1-c)A + c(1-b)D. Here we notice that we can eliminate the term (1-c)A from both sides of equation. That was the shortcut I used. cA > c(1-b)D. A > (1-b)D, (since c is positive, not equal zero.) Now I drop all reasoning and just use algebra: (1-b)c + bc/(b+c-bc) > (1-b)c/(b+c-bc) (1-b) + b/(b+c-bc) > (1-b)/(b+c-bc), again, since c is positive not equal zero. 1-b > (1-2b)/(b+c-bc). Note, that for b and c between zero and 1, b+c-bc > 0, therefore: (1-b)(b+c-bc) > 1-2b c-2cb+cb2> 1-3b+b2 c(1-2b+b2) > 1-3b+b2c > (1-3b+b2)/(1-b)2 (given that (1-b)2 > 0). (Same as Bushindo, except for the b2 instead of b3 in the numerator.) c > ((1-b)2 - b))/(1-b)2c > 1 - b/(1-b)2 Let's study 1 - b/(1-b)2, or (1-3b+b2)/(1-b)2. We know that 0 < b < 1. As b increases the function decreases. Solving for zero, and omitting the greater than 1 root, we get:b = (3-sqrt(5))/2. That is the value of b above which the expression yields negative numbers. And since we know that c is positive, we can draw the conclusion that for the values of b above that boundary, Cole is always at advantage when making his first shot in the air. Approximating that as a percentage yields 38.1966%. When Bobby is a is a better shot than 38.2%, Cole must shoot in the air without giving it a second thought. Another boundary condition is where the expression yields values greater than b. Since we know that c < b, past the boundary of b = (1-3b+b2)/(1-b)2, the values of c satisfying the inequality contradict the condition that Bobby is a better shot than Cole. Solving cubic equation, we find that the approximate value b=0.317672 or 31.7672% is that boundary. If Bobby shoots worse than 31.7672%, Cole must try and shoot Alex. If Bobby shoots in between those boundary conditions (31.7672% < b < 38.1966%), Cole must evaluate the expression 1 - b/(1-b)2 and compare his shooting prowess c to the result. If c is better, Cole must shoot in the air, if c is less – aim at Alex. If c happens to be exactly equal, for example b=35% and c=17.1598%, then Cole must decide who he hates more: Alex or Bobby. If Cole shoots in the air first, Alex's survival is (1-b)(1-c), if Cole aims at Alex, Alex's survival chance becomes (1-c)(1-b)(1-c). If Cole hates Alex and Bobby exactly equally, then he must shoot in the air – a noble gesture.At no time there is any uncertainty, save for the wacky cases c=0, or b=1. Hopefully, this solves all tiebreak situations and removes the uncertainty. Reading from the OP: Given that it [Cole's strategy] is uncertain, what can we determine regarding Bobby's shooting accuracy? From your reasoning regarding Cole's strategy, what is the answer? I agree about semantics. I think the underlying crux of the discussion is the interpretation of 'uncertain' From above, it is easy to derive the expression for the chance of Cole winning if he shoots into the air and if he shoots at Alex. Let's denote those functions Wshoot_air(b,c) and Wshoot_at_Alex(b,c), respectively. Prime and bonanova seems to interpret 'uncertain' as being unable to tell whether Wshoot_air(b,c) < Wshoot_at_Alex(b,c) or Wshoot_air(b,c) > Wshoot_at_Alex(b,c) without applying those functions. Indeed, between .31 < b < .39, Cole must be able to apply those function in order to see which strategy gives a higher winning probability. My interpretation is that, since Wshoot_at_Alex(b,c) and Wshoot_air(b,c) are easily derivable and b and c are known, Cole is 'uncertain' if the two strategies give precisely the same survival probability. That is, Wshoot_air(b,c) = Wshoot_at_Alex(b,c). I have no objection to the interpretation of Prime and bonanova, except that it should be qualified to prevent confusion. In post #4, Prime did indicate that between .31 < b < .39, Cole should compute Wshoot_air(b,c) and Wshoot_at_Alex(b,c) to determine the better strategy. Presenting the result without the qualification as in post #8 may mislead the reader into thinking that the two strategies are the same within .31 < b < .39.