bonanova Posted March 15, 2013 Report Share Posted March 15, 2013 A twist on the "truel" puzzle - a duel among three participants, where shots are taken in turn. Alex always hits his man. Bobby is not perfect, but he shoots better than Cole does.The referee thus gives Cole the first shot, followed by Bobby and Alex in that order.When a man is hit, he drops out [or drops dead].The shooting continues in sequence until only one man is unhit. Cole does not want to take Bobby out of the competition, then Alex will hit Cole. Game over.But it may not be wise for Cole to hit Alex either. Cole is disadvantaged in an ordinary duel. So Cole's first-shot strategy is uncertain.Given that it is uncertain, what can we determine regarding Bobby's shooting accuracy? Assume each knows the others' accuracy level. p_{x} = probability that x will hit his man. p_{Alex} = 1 > p_{Bobby} > p_{Cole}. For some values of p_{Bobby}, Cole's first-shot strategy is certain.For example, if p_{Bobby} = 0.0001, Cole tries to hit Alex and have a long almost-50% duel with Bobby.. For what values of p_{Bobby} is Cole uncertain of what to do on his first shot? Quote Link to comment Share on other sites More sharing options...

0 Prime Posted March 15, 2013 Report Share Posted March 15, 2013 (edited) I am uncertain, what uncertain means in this context. Cole's first shot strategy is clear. He must shoot himself in the foot thus exiting the duel with a non-fatal injury. (Hopefully, he does not miss.) Coles probability to hit the target is c; Bobby's probability -- is b. Let's calculate the probability of Cole's survival in one on one duel with Bobby, where Cole shoots first. X = c + (1-c)(1-b)X X = c/(b+c-bc) If Cole's first shot is in the air, Bobby shoots at Alex. If Bobby misses, Alex kills Bobby and Cole has one shot at Alex. If Bobby hits Alex, then a duel between Cole and Bobby ensues with Cole going first. Cole's probability of survival in this scenario is:(1-b)c + bc/(b+c-bc). If Cole shoots at Alex and misses, thereafter the probability is the same as as with shooting in the air. If Cole hits Alex, then he has a duel with Bobby shooting first. If survival chances in such duel are worse compared to shooting in the air, it is to Cole's advantage to shoot in the air. If Bobby misses Cole on his first shot, it becomes a duel with Cole shooting first. Therefore Cole's surviving probability(air shot) > probability(Bobby-Cole duel) is as following:(1-b)c + bc/(b+c-bc) > (1-b)c/(b+c-bc) Solving we get: c > 1 - b/(1-b)^{2} When b >= 39% the expression yields a negative number, meaning Cole must shoot in the air. For b <= 31% the expression is greater than 31%, which contradicts the condition that Cole is a worse shot than Bobby. That means Cole must try and shoot Alex. For b between 31% and 39%, Cole must carry out calculations and comparison before shooting. Edited March 15, 2013 by Prime Quote Link to comment Share on other sites More sharing options...

0 bushindo Posted March 15, 2013 Report Share Posted March 15, 2013 Probably not what you were looking for, but here's what I recommend Cole do Cole should shoot into the air (i.e. not shoot Alex or Bobby) on his turn. Bobby and Alex will shoot one another in sequence. Cole then has the nice advantage of having the first shot on whoever survives between Alex and Bobby. So, clarification please. Can any participant choose to shoot into the air instead of the other duelists? Quote Link to comment Share on other sites More sharing options...

0 bushindo Posted March 15, 2013 Report Share Posted March 15, 2013 A twist on the "truel" puzzle - a duel among three participants, where shots are taken in turn.Alex always hits his man. Bobby is not perfect, but he shoots better than Cole does.The referee thus gives Cole the first shot, followed by Bobby and Alex in that order.When a man is hit, he drops out [or drops dead].The shooting continues in sequence until only one man is unhit.Cole does not want to take Bobby out of the competition, then Alex will hit Cole. Game over.But it may not be wise for Cole to hit Alex either. Cole is disadvantaged in an ordinary duel.So Cole's first-shot strategy is uncertain.Given that it is uncertain, what can we determine regarding Bobby's shooting accuracy? Assume each knows the others' accuracy level.p_{x} = probability that x will hit his man.p_{Alex} = 1 > p_{Bobby} > p_{Cole}.For some values of p_{Bobby}, Cole's first-shot strategy is certain.For example, if p_{Bobby} = 0.0001, Cole tries to hit Alex and have a long almost-50% duel with Bobby..For what values of p_{Bobby} is Cole uncertain of what to do on his first shot? Assuming each participant is forced to shoot only at remaining duelists, then here's what I get Intuition says that C should shoot A first. Let's go through the game tree and in what situation does C become uncertain. Let's denote the participants by A, B, and C. Let's write down the probability of C winning at the beginning of the game as P( C, B, A), where the identity of the remaining participants are listed as arguments of the function P( ), and the underlined C indicates that C has the first shot. If there are only two participants left, then we can write out the corresponding win percentages. Note that P( . ) always indicate the probability of C winning. P( C, B ) = ( 1-p_{b})p_{c}/[ 1 - (1 - p_{c})(1-p_{b}) ] P( C, B ) = p_{c}/[ 1 - (1 - p_{c})(1-p_{b}) ] P( C, A ) = p_{c} P( C, A ) = 0. Now, we start off the the logic that A will always shoot the remaining person(s) with the highest hit probability. Therefore, B will always shoot the remaining participant with highest hit probability as well. I drew out two game tree for two cases- one where C chooses to shoot B first and one where C shoots A first. The game trees are here Combining the game trees above with the two-participant results in the beginning, we can write down the winning chances for C from the two strategies. Let W_{A} be the winning chance for C if he shoots A first, and let W_{B} be the winning chance if he shoot B first, the probabilities are W_{B} = (1-p_{c})p_{c} p_{b}/[ 1 - (1 - p_{c})(1-p_{b}) ] + (1-p_{c}) (1-p_{b}) p_{c} W_{A} = (1-p_{c})p_{c} p_{b}/[ 1 - (1 - p_{c})(1-p_{b}) ] + (1-p_{c}) (1-p_{b}) p_{c} + p_{c} p_{c} (1-p_{b})/[ 1 - (1 - p_{c})(1-p_{b}) ] or W_{A} = W_{B} + p_{c} p_{c} (1-p_{b})/[ 1 - (1 - p_{c})(1-p_{b}) ] So, we see that shooting A first increases the winning probability by the amount p_{c} p_{c} (1-p_{b})/[ 1 - (1 - p_{c})(1-p_{b}) ]. Now, the only way to make C uncertain is to let p_{c} = 0, at which point his winning chance is the same regardless of who he shoots. But if his hitting probability is 0, he's screwed either way. Quote Link to comment Share on other sites More sharing options...

0 bushindo Posted March 16, 2013 Report Share Posted March 16, 2013 A twist on the "truel" puzzle - a duel among three participants, where shots are taken in turn. Alex always hits his man. Bobby is not perfect, but he shoots better than Cole does.The referee thus gives Cole the first shot, followed by Bobby and Alex in that order.When a man is hit, he drops out [or drops dead].The shooting continues in sequence until only one man is unhit. Cole does not want to take Bobby out of the competition, then Alex will hit Cole. Game over.But it may not be wise for Cole to hit Alex either. Cole is disadvantaged in an ordinary duel. So Cole's first-shot strategy is uncertain.Given that it is uncertain, what can we determine regarding Bobby's shooting accuracy? Assume each knows the others' accuracy level. p_{x} = probability that x will hit his man. p_{Alex} = 1 > p_{Bobby} > p_{Cole}. For some values of p_{Bobby}, Cole's first-shot strategy is certain.For example, if p_{Bobby} = 0.0001, Cole tries to hit Alex and have a long almost-50% duel with Bobby.. For what values of p_{Bobby} is Cole uncertain of what to do on his first shot? Assuming each participant is forced to shoot only at remaining duelists, then here's what I get Intuition says that C should shoot A first. Let's go through the game tree and in what situation does C become uncertain. Let's denote the participants by A, B, and C. Let's write down the probability of C winning at the beginning of the game as P( C, B, A), where the identity of the remaining participants are listed as arguments of the function P( ), and the underlined C indicates that C has the first shot. If there are only two participants left, then we can write out the corresponding win percentages P( C, B ) = ( 1-p_{b})p_{c}/[ 1 - (1 - p_{c})(1-p_{b}) ] P( C, B ) = p_{c}/[ 1 - (1 - p_{c})(1-p_{b}) ] P( C, A ) = pc P( C, A ) = 0. Now, we start off the the logic that A will always shoot the remaining person(s) with the highest hit probability. Therefore, B will always shoot the remaining participant with highest hit probability as well. I drew out two game tree for two cases- one where C chooses to shoot B first and one where C shoots A first. The game trees are here IMG_20130315_135409.jpg Combining the game trees above with the two-participant results in the beginning, we can write down the winning chances for C from the two strategies. Let W_{A} be the winning chance for C if he shoots A first, and let W_{B} be the winning chance if he shoot B first, the probabilities are WB = (1-p_{c})p_{c} p_{b}/[ 1 - (1 - p_{c})(1-p_{b}) ] + (1-p_{c}) (1-p_{b}) p_{c} WA = (1-p_{c})p_{c} p_{b}/[ 1 - (1 - p_{c})(1-p_{b}) ] + (1-p_{c}) (1-p_{b}) p_{c} + p_{c} p_{c} (1-p_{b})/[ 1 - (1 - p_{c})(1-p_{b}) ] or WA = WB + p_{c} p_{c} (1-p_{b})/[ 1 - (1 - p_{c})(1-p_{b}) ] So, we see that shooting A first increases the winning probability by the amount p_{c} p_{c} (1-p_{b})/[ 1 - (1 - p_{c})(1-p_{b}) ]. Now, the only way to make C uncertain is to let pc = 0, at which point his winning chance is the same regardless of who he shoots. But if his hitting probability is 0, he's screwed either way. Okay, so earlier we found that not allowing an airshot on Cole's turn reduces to a trivial solution. So, let's assume that on the first shot, Cole has the option of shooting A, B, or none of those two. We already found that the probability of Cole winning if he shoots at A first is W_{A} = (1-p_{c})p_{c} p_{b}/[ 1 - (1 - p_{c})(1-p_{b}) ] + (1-p_{c}) (1-p_{b}) p_{c} + p_{c} p_{c} (1-p_{b})/[ 1 - (1 - p_{c})(1-p_{b}) ] By the same derivation process, we can see the probability of Cole winning if he shoots at the air first is W_{N} = p_{b}p_{c}/[ 1 - (1-p_{b})(1-p_{c}) ] + (1-p_{b}) p_{c} So, if we set W_{N} = W_{A} and solve for p_{b} and p_{c}, we see that both strategies are equally beneficial for Cole (hence the uncertainty, I suppose) when p_{c} = ( p_{b}^{3} - 3p_{b} + 1)/( p_{b}-1)^{2} Quote Link to comment Share on other sites More sharing options...

0 Prime Posted March 17, 2013 Report Share Posted March 17, 2013 Okay, so earlier we found that not allowing an airshot on Cole's turn reduces to a trivial solution. So, let's assume that on the first shot, Cole has the option of shooting A, B, or none of those two. We already found that the probability of Cole winning if he shoots at A first is W_{A} = (1-p_{c})p_{c} p_{b}/[ 1 - (1 - p_{c})(1-p_{b}) ] + (1-p_{c}) (1-p_{b}) p_{c} + p_{c} p_{c} (1-p_{b})/[ 1 - (1 - p_{c})(1-p_{b}) ] By the same derivation process, we can see the probability of Cole winning if he shoots at the air first is W_{N} = p_{b}p_{c}/[ 1 - (1-p_{b})(1-p_{c}) ] + (1-p_{b}) p_{c} So, if we set W_{N} = W_{A} and solve for p_{b} and p_{c}, we see that both strategies are equally beneficial for Cole (hence the uncertainty, I suppose) when p_{c} = ( p_{b}^{3} - 3p_{b} + 1)/( p_{b}-1)^{2} If it was p_{b}^{2} (square instead of cube,) our results would match. p_{c} = ( p_{b}^{2} - 3p_{b} + 1)/( p_{b}-1)^{2} Quote Link to comment Share on other sites More sharing options...

0 bushindo Posted March 17, 2013 Report Share Posted March 17, 2013 Okay, so earlier we found that not allowing an airshot on Cole's turn reduces to a trivial solution. So, let's assume that on the first shot, Cole has the option of shooting A, B, or none of those two. We already found that the probability of Cole winning if he shoots at A first is W_{A} = (1-p_{c})p_{c} p_{b}/[ 1 - (1 - p_{c})(1-p_{b}) ] + (1-p_{c}) (1-p_{b}) p_{c} + p_{c} p_{c} (1-p_{b})/[ 1 - (1 - p_{c})(1-p_{b}) ] By the same derivation process, we can see the probability of Cole winning if he shoots at the air first is W_{N} = p_{b}p_{c}/[ 1 - (1-p_{b})(1-p_{c}) ] + (1-p_{b}) p_{c} So, if we set W_{N} = W_{A} and solve for p_{b} and p_{c}, we see that both strategies are equally beneficial for Cole (hence the uncertainty, I suppose) when p_{c} = ( p_{b}^{3} - 3p_{b} + 1)/( p_{b}-1)^{2} If it was p_{b}^{2} (square instead of cube,) our results would match. p_{c} = ( p_{b}^{2} - 3p_{b} + 1)/( p_{b}-1)^{2} Indeed it should be a quadratic expression instead of cubic. My mistake. Good catch, Prime. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted March 18, 2013 Author Report Share Posted March 18, 2013 I am uncertain, what uncertain means in this context. Cole's first shot strategy is clear. He must shoot himself in the foot thus exiting the duel with a non-fatal injury. (Hopefully, he does not miss.) Coles probability to hit the target is c; Bobby's probability -- is b. Let's calculate the probability of Cole's survival in one on one duel with Bobby, where Cole shoots first. X = c + (1-c)(1-b)X X = c/(b+c-bc) If Cole's first shot is in the air, Bobby shoots at Alex. If Bobby misses, Alex kills Bobby and Cole has one shot at Alex. If Bobby hits Alex, then a duel between Cole and Bobby ensues with Cole going first. Cole's probability of survival in this scenario is:(1-b)c + bc/(b+c-bc). If Cole shoots at Alex and misses, thereafter the probability is the same as as with shooting in the air. If Cole hits Alex, then he has a duel with Bobby shooting first. If survival chances in such duel are worse compared to shooting in the air, it is to Cole's advantage to shoot in the air. If Bobby misses Cole on his first shot, it becomes a duel with Cole shooting first. Therefore Cole's surviving probability(air shot) > probability(Bobby-Cole duel) is as following:(1-b)c + bc/(b+c-bc) > (1-b)c/(b+c-bc) Solving we get: c > 1 - b/(1-b)^{2} When b >= 39% the expression yields a negative number, meaning Cole must shoot in the air. For b <= 31% the expression is greater than 31%, which contradicts the condition that Cole is a worse shot than Bobby. That means Cole must try and shoot Alex. For b between 31% and 39%, Cole must carry out calculations and comparison before shooting. Yes, that is the answer. Nice job. Cole must not shoot first at Bobby, Alex and Bobby must always shoot the higher ranking opponent. Cole shoots first at Alex. If Cole's survival probability for hitting Alex is the same as for missing Alex, then Cole's strategy is uncertain: should he try to hit Alex? Or to miss him? The critical probabilities are P(B C) and P(B C). Duels that involve Alex are trivial. If q_{X} = 1-p_{X} then P(B C) = (1 - q_{B}) / (1 - q_{B}q_{C}) P(B C) = (1 - q_{C}) / (1 - q_{B}q_{C}) If Cole hits Alex, Cole survives with probability 1 - P(B C) If Cole misses Alex, Cole survives with probability p_{B}P(B C) + q_{B}p_{C}. Equating these and simplifying gives q_{C} = (1 - q_{B})/q^{2}_{B} Now q_{B} < q_{C}, and both must lie between 0 and 1. It boils down to how (in)accurate Cole is. The extremes are totally inept and as good as Bobby. Totally inept: q_{C} = 1 this gives q_{B} = .618. As good as Bobby: q_{C} = q_{B }this gives q_{B} = .683 . So, If Cole's strategy is uncertain, then we know that Bobby's shooting accuracy lies between .317 and .382. Quote Link to comment Share on other sites More sharing options...

0 bushindo Posted March 18, 2013 Report Share Posted March 18, 2013 I am uncertain, what uncertain means in this context. Cole's first shot strategy is clear. He must shoot himself in the foot thus exiting the duel with a non-fatal injury. (Hopefully, he does not miss.) Coles probability to hit the target is c; Bobby's probability -- is b. Let's calculate the probability of Cole's survival in one on one duel with Bobby, where Cole shoots first. X = c + (1-c)(1-b)X X = c/(b+c-bc) If Cole's first shot is in the air, Bobby shoots at Alex. If Bobby misses, Alex kills Bobby and Cole has one shot at Alex. If Bobby hits Alex, then a duel between Cole and Bobby ensues with Cole going first. Cole's probability of survival in this scenario is:(1-b)c + bc/(b+c-bc). If Cole shoots at Alex and misses, thereafter the probability is the same as as with shooting in the air. If Cole hits Alex, then he has a duel with Bobby shooting first. If survival chances in such duel are worse compared to shooting in the air, it is to Cole's advantage to shoot in the air. If Bobby misses Cole on his first shot, it becomes a duel with Cole shooting first. Therefore Cole's surviving probability(air shot) > probability(Bobby-Cole duel) is as following:(1-b)c + bc/(b+c-bc) > (1-b)c/(b+c-bc) Solving we get: c > 1 - b/(1-b)^{2} When b >= 39% the expression yields a negative number, meaning Cole must shoot in the air. For b <= 31% the expression is greater than 31%, which contradicts the condition that Cole is a worse shot than Bobby. That means Cole must try and shoot Alex. For b between 31% and 39%, Cole must carry out calculations and comparison before shooting. Yes, that is the answer. Nice job. Cole must not shoot first at Bobby, Alex and Bobby must always shoot the higher ranking opponent. Cole shoots first at Alex. If Cole's survival probability for hitting Alex is the same as for missing Alex, then Cole's strategy is uncertain: should he try to hit Alex? Or to miss him? The critical probabilities are P(B C) and P(B C). Duels that involve Alex are trivial. If q_{X} = 1-p_{X} then P(B C) = (1 - q_{B}) / (1 - q_{B}q_{C}) P(B C) = (1 - q_{C}) / (1 - q_{B}q_{C}) If Cole hits Alex, Cole survives with probability 1 - P(B C) If Cole misses Alex, Cole survives with probability p_{B}P(B C) + q_{B}p_{C}. Equating these and simplifying gives q_{C} = (1 - q_{B})/q^{2}_{B} Now q_{B} < q_{C}, and both must lie between 0 and 1. It boils down to how (in)accurate Cole is. The extremes are totally inept and as good as Bobby. Totally inept: q_{C} = 1 this gives q_{B} = .618. As good as Bobby: q_{C} = q_{B }this gives q_{B} = .683 . So, If Cole's strategy is uncertain, then we know that Bobby's shooting accuracy lies between .317 and .382. I believe the above solution conflates "trying to hit Alex" with "actually hitting Alex". Here is my reasoning In the beginning before even taking a shot, Cole has two choices for strategy, either "shoot first at Alex" or "shoot into the air (deliberately missing)". (Note that if Cole is constrained to only shoot at Alex and see the shot outcome, then he doesn't really have a strategy). The solution of the puzzle is to compute and solve the following equation [1] P( Cole win | Cole shoot first at Alex) = P( Cole win | Cole shoot first into air ) However, the solution above I think conflates "attempting to shoot at Alex" with "actually hitting Alex". From above, I annotate the given terms with what they actually mean >> If Cole hits Alex, Cole survives with probability 1 - P(B C) => P( Cole win | Cole shoot first at Alex AND Cole hits Alex ) >> If Cole misses Alex, Cole survives with probability p_{B}P(B C) + q_{B}p_{C} => P( Cole win | Cole shoot first into air ) So, the solution above actually solves the following equality [2] P( Cole win | Cole shoot first at Alex AND Cole hits Alex ) = P( Cole win | Cole shoot first into air ) But this is not the quantity that we need to solve for, and the equality in [2] is comparing apples to oranges. The term on the left should be P( Cole win | Cole shoot first at Alex ), which can be expressed as P( Cole win | Cole shoot first at Alex ) = p_{c} * P( Cole win | Cole shoot first at Alex AND Cole hits Alex ) * (1- p_{c}) P( Cole win | Cole shoot first at Alex AND Cole does not hit Alex ) P( Cole win | Cole shoot first at Alex ) = (1-p_{c})p_{c} p_{b}/[ 1 - (1 - p_{c})(1-p_{b}) ] + (1-p_{c}) (1-p_{b}) p_{c} + p_{c} p_{c} (1-p_{b})/[ 1 - (1 - p_{c})(1-p_{b}) ] So, if we replace the proper term P( Cole win | Cole shoot first at Alex ) into equation [1], we should get a different answer, which Prime gave in I saw that the above analysis also uses a different meaning for P(.) compared to its usage in so I want to clarify that so we don't have any unnecessary confusion when going back and forth between the analyses. In Post 3, P(.) is always assumed to be the probability of Cole winning, regardless of who has the first shot. In the above quoted analysis, P(.) seems to give the probability of winning for whoever has the first shot. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted March 18, 2013 Author Report Share Posted March 18, 2013 I don't see exactly were we disagree, mainly because I'm logic-notation challenged. In English my thinking is: On his first shot, Cole will not jeopardize Bobby. After Cole's first shot, either Alex is hit or Alex in not hit. Cole decides what he will attempt to do by comparing his survival chances in those two cases. If Cole's survival is more likely if Alex is not hit, Cole will of course fire into the air. If Cole's survival is more likely if Alex is hit, Cole will try to hit Alex. If those chances are the same, Cole does not know what to do. OP asks the values of p_{B} for case 3. Do we agree this far? Quote Link to comment Share on other sites More sharing options...

0 Prime Posted March 18, 2013 Report Share 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 = c^{2}(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)) + c^{2}(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+cb^{2}^{}> 1-3b+b^{2} c(1-2b+b^{2}) > 1-3b+b^{2}c > (1-3b+b^{2})/(1-b)^{2} (given that (1-b)^{2} > 0). (Same as Bushindo, except for the b^{2} instead of b^{3} in the numerator.) c > ((1-b)^{2} - b))/(1-b)^{2}c > 1 - b/(1-b)^{2} Let's study 1 - b/(1-b)^{2}, or (1-3b+b^{2})/(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+b^{2})/(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 â€“ 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 â€“ 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. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted March 18, 2013 Author Report Share 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 = c^{2}(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)) + c^{2}(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+cb^{2}> 1-3b+b^{2} c(1-2b+b^{2}) > 1-3b+b^{2}c > (1-3b+b^{2})/(1-b)^{2} (given that (1-b)^{2} > 0). (Same as Bushindo, except for the b^{2} instead of b^{3} in the numerator.) c > ((1-b)^{2} - b))/(1-b)^{2}c > 1 - b/(1-b)^{2} Let's study 1 - b/(1-b)^{2}, or (1-3b+b^{2})/(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+b^{2})/(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 â€“ 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 â€“ 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? Quote Link to comment Share on other sites More sharing options...

0 Prime Posted March 18, 2013 Report Share Posted March 18, 2013 (edited) 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 = c^{2}(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)) + c^{2}(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+cb^{2}> 1-3b+b^{2} c(1-2b+b^{2}) > 1-3b+b^{2}c > (1-3b+b^{2})/(1-b)^{2} (given that (1-b)^{2} > 0). (Same as Bushindo, except for the b^{2} instead of b^{3} in the numerator.) c > ((1-b)^{2} - b))/(1-b)^{2}c > 1 - b/(1-b)^{2} Let's study 1 - b/(1-b)^{2}, or (1-3b+b^{2})/(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+b^{2})/(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 â€“ 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 â€“ 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? We are arguing semantics. To me uncertain means cannot be determined, like division by zero. I just do not see the choice between two equally good (bad) values as an uncertainty. And I have given several good ways to decide the tiebreak. I don't think it's all that important in our three way duel. I figure, Bushindo's objection was to omitting inclusion of all of the variations into the equation. I think we are in the right to do that. And after algebraic simplifications the full equations would come to the same thing. And we all use different nomenclature. (Mine is the easiest to type, although not as formal.) I found it interesting that in Bonanova's analysis, using probability of missing (q), simplifies the solution. In particular the solution of the cubic equation. I say, this duel is solved. Alternatively, shooting himself in the foot may be the best option for Cole. Edited March 18, 2013 by Prime Quote Link to comment Share on other sites More sharing options...

0 bushindo Posted March 18, 2013 Report Share 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 = c^{2}(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)) + c^{2}(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+cb^{2}> 1-3b+b^{2} c(1-2b+b^{2}) > 1-3b+b^{2}c > (1-3b+b^{2})/(1-b)^{2} (given that (1-b)^{2} > 0). (Same as Bushindo, except for the b^{2} instead of b^{3} in the numerator.) c > ((1-b)^{2} - b))/(1-b)^{2}c > 1 - b/(1-b)^{2} Let's study 1 - b/(1-b)^{2}, or (1-3b+b^{2})/(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+b^{2})/(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 â€“ 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 â€“ 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 W_{shoot_air}(b,c) and W_{shoot_at_Alex}(b,c), respectively. Prime and bonanova seems to interpret 'uncertain' as being unable to tell whether W_{shoot_air}(b,c) < W_{shoot_at_Alex}(b,c) or W_{shoot_air}(b,c) > W_{shoot_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 W_{shoot_at_Alex}(b,c) and W_{shoot_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, W_{shoot_air}(b,c) = W_{shoot_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 W_{shoot_air}(b,c) and W_{shoot_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. Quote Link to comment Share on other sites More sharing options...

0 Prime Posted March 18, 2013 Report Share 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 = c^{2}(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)) + c^{2}(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+cb^{2}> 1-3b+b^{2} c(1-2b+b^{2}) > 1-3b+b^{2}c > (1-3b+b^{2})/(1-b)^{2} (given that (1-b)^{2} > 0). (Same as Bushindo, except for the b^{2} instead of b^{3} in the numerator.) c > ((1-b)^{2} - b))/(1-b)^{2}c > 1 - b/(1-b)^{2} Let's study 1 - b/(1-b)^{2}, or (1-3b+b^{2})/(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+b^{2})/(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 â€“ 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 â€“ 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 W_{shoot_air}(b,c) and W_{shoot_at_Alex}(b,c), respectively. Prime and bonanova seems to interpret 'uncertain' as being unable to tell whether W_{shoot_air}(b,c) < W_{shoot_at_Alex}(b,c) or W_{shoot_air}(b,c) > W_{shoot_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 W_{shoot_at_Alex}(b,c) and W_{shoot_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, W_{shoot_air}(b,c) = W_{shoot_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 W_{shoot_air}(b,c) and W_{shoot_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. Now Bushindo is with Bonanova insisting that Win(shooting air)=Win(shooting Alex) means uncertainty. What's wrong with my tiebreaks? A true 3-way duel indeed. But all got the same answer to the problem. I can see where my guidlines to Cole in post#4 could be misinterpreted. My intention was to help Cole avoiding computations, where possible. For if he sits down with a calculator in the middle of the duel, the other two guys may get angry and just shoot him out of turn. Quote Link to comment Share on other sites More sharing options...

0 bushindo Posted March 18, 2013 Report Share Posted March 18, 2013 Now Bushindo is with Bonanova insisting that Win(shooting air)=Win(shooting Alex) means uncertainty. What's wrong with my tiebreaks? A true 3-way duel indeed. But all got the same answer to the problem. I can see where my guidlines to Cole in post#4 could be misinterpreted. My intention was to help Cole avoiding computations, where possible. For if he sits down with a calculator in the middle of the duel, the other two guys may get angry and just shoot him out of turn. I think I misinterpreted your position. I don't think bononova and I share the same interpretation of 'uncertain', though. Let me see if we have the following positions correct bonanova: 'Uncertain' means being unable to tell whether W_{shoot_air}(b,c) < W_{shoot_at_Alex}(b,c) or W_{shoot_air}(b,c) > W_{shoot_at_Alex}(b,c) without applying those functions. So, Cole is uncertain between .31 < b < .39. bushindo: 'Uncertain' means W_{shoot_air}(b,c) = W_{shoot_at_Alex}(b,c). Prime: 'Uncertain' means being unable to make a decision based on all available information. There is no situation when Cole is uncertain about what to do next. Cole should always choose the strategy that gives higher value between W_{shoot_air}(b,c) and W_{shoot_at_Alex}(b,c). I agree that under the standard dictionary definition of 'uncertain', Prime's interpretation would be most correct. I thought that bonanova had some different specialized meaning in mind, and apparently he does. It just isn't what I thought it was =). Quote Link to comment Share on other sites More sharing options...

0 Prime Posted March 18, 2013 Report Share Posted March 18, 2013 Now Bushindo is with Bonanova insisting that Win(shooting air)=Win(shooting Alex) means uncertainty. What's wrong with my tiebreaks? A true 3-way duel indeed. But all got the same answer to the problem. I can see where my guidlines to Cole in post#4 could be misinterpreted. My intention was to help Cole avoiding computations, where possible. For if he sits down with a calculator in the middle of the duel, the other two guys may get angry and just shoot him out of turn. I think I misinterpreted your position. I don't think bononova and I share the same interpretation of 'uncertain', though. Let me see if we have the following positions correct bonanova: 'Uncertain' means being unable to tell whether W_{shoot_air}(b,c) < W_{shoot_at_Alex}(b,c) or W_{shoot_air}(b,c) > W_{shoot_at_Alex}(b,c) without applying those functions. So, Cole is uncertain between .31 < b < .39. bushindo: 'Uncertain' means W_{shoot_air}(b,c) = W_{shoot_at_Alex}(b,c). Prime: 'Uncertain' means being unable to make a decision based on all available information. There is no situation when Cole is uncertain about what to do next. Cole should always choose the strategy that gives higher value between W_{shoot_air}(b,c) and W_{shoot_at_Alex}(b,c). I agree that under the standard dictionary definition of 'uncertain', Prime's interpretation would be most correct. I thought that bonanova had some different specialized meaning in mind, and apparently he does. It just isn't what I thought it was =). The way I see it, Bonanova has made his problem statement/position 100% clear: Uncertainty is when Cole's winning chance by shooting in the air = his winning chance by shooting at Alex. The question Bonanova wants us to solve is: For what values of b (accuracy of Bobby) such situation is possible? And solved it we have: And we all (including Bonanova) have found same equation yielding the same answer: When b is between 31.8% and 38.2% that "uncertainty" may occur. For the values of b outside those boundaries, a situation where W_{shoot_air}(b,c) = W_{shoot_at_Alex}(b,c) is impossible. Bushindo's view of uncertainty is the same as Bonanova's: W_{shoot_air}(b,c) = W_{shoot_at_Alex}(b,c). However, there seems to be some uncertainty as to what exactly Bonanova wants us to find. Prime picks on the usage of the word uncertain. Does not believe there is any uncertainty here at all. (When chances are equal, Cole must shoot Alex, because he hates him more than he hates Bobby.) Also, Prime promotes (unsuccessfully) an alternative strategy whereby Cole shoots himself in the foot with his very first shot. Quote Link to comment Share on other sites More sharing options...

0 bushindo Posted March 19, 2013 Report Share Posted March 19, 2013 Now Bushindo is with Bonanova insisting that Win(shooting air)=Win(shooting Alex) means uncertainty. What's wrong with my tiebreaks? A true 3-way duel indeed. But all got the same answer to the problem. I can see where my guidlines to Cole in post#4 could be misinterpreted. My intention was to help Cole avoiding computations, where possible. For if he sits down with a calculator in the middle of the duel, the other two guys may get angry and just shoot him out of turn. I think I misinterpreted your position. I don't think bononova and I share the same interpretation of 'uncertain', though. Let me see if we have the following positions correct bonanova: 'Uncertain' means being unable to tell whether W_{shoot_air}(b,c) < W_{shoot_at_Alex}(b,c) or W_{shoot_air}(b,c) > W_{shoot_at_Alex}(b,c) without applying those functions. So, Cole is uncertain between .31 < b < .39. bushindo: 'Uncertain' means W_{shoot_air}(b,c) = W_{shoot_at_Alex}(b,c). Prime: 'Uncertain' means being unable to make a decision based on all available information. There is no situation when Cole is uncertain about what to do next. Cole should always choose the strategy that gives higher value between W_{shoot_air}(b,c) and W_{shoot_at_Alex}(b,c). I agree that under the standard dictionary definition of 'uncertain', Prime's interpretation would be most correct. I thought that bonanova had some different specialized meaning in mind, and apparently he does. It just isn't what I thought it was =). The way I see it, Bonanova has made his problem statement/position 100% clear: Uncertainty is when Cole's winning chance by shooting in the air = his winning chance by shooting at Alex. The question Bonanova wants us to solve is: For what values of b (accuracy of Bobby) such situation is possible? And solved it we have: And we all (including Bonanova) have found same equation yielding the same answer: When b is between 31.8% and 38.2% that "uncertainty" may occur. For the values of b outside those boundaries, a situation where W_{shoot_air}(b,c) = W_{shoot_at_Alex}(b,c) is impossible. Bushindo's view of uncertainty is the same as Bonanova's: W_{shoot_air}(b,c) = W_{shoot_at_Alex}(b,c). However, there seems to be some uncertainty as to what exactly Bonanova wants us to find. Prime picks on the usage of the word uncertain. Does not believe there is any uncertainty here at all. (When chances are equal, Cole must shoot Alex, because he hates him more than he hates Bobby.) Also, Prime promotes (unsuccessfully) an alternative strategy whereby Cole shoots himself in the foot with his very first shot. You're right. My view of uncertainty is the same as bonanoba's. It's still too early for me to be senile, *sigh*. Good analysis, though. Your analysis definitely deserves the distinction of Best Answer. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted March 19, 2013 Author Report Share Posted March 19, 2013 Most people attribute the "truel" puzzle to Martin Gardner. Its charm, like many of his puzzles, it that doing the thinking and the math rewards you with a surprise. The surprise here is that the best shooter can end up the underdog, and the worst shooter can end up the favorite to win. If Alex shoots first we'll bet on him all day long. Having to shoot last is clearly the reason why he might be least likely to win. And that is not a greatly surprising result. But it is not because Cole may shoot first that he can be the most likely to win: it's because he is able to "volunteer" to shoot last -- along with his being the least desirable target. That is the genius of Gardner. I can't add anything to what you guys have said. This was enjoyable. Quote Link to comment Share on other sites More sharing options...

## Question

## bonanova

A twist on the "truel" puzzle - a duel among three participants, where shots are taken in turn.

Alex always hits his man.

Bobby is not perfect, but he shoots better than Cole does.

The referee thus gives Cole the first shot, followed by Bobby and Alex in that order.

When a man is hit, he drops out [or drops dead].

The shooting continues in sequence until only one man is unhit.

Cole does not want to take Bobby out of the competition, then Alex will hit Cole. Game over.

But it may not be wise for Cole to hit Alex either. Cole is disadvantaged in an ordinary duel.

So Cole's first-shot strategy is

uncertain.Given that it is uncertain, what can we determine regarding Bobby's shooting accuracy?

Assume each knows the others' accuracy level.

## Link to comment

## Share on other sites

## 18 answers to this question

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.