18 replies to this topic

[Prime]

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.

Spoiler for detailed solution

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+b²
c > (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)²
c > 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.

 

[bonanova]

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.

Spoiler for detailed solution

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+b²
c > (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)²
c > 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?

[Prime]

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.

Spoiler for detailed solution

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+b²
c > (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)²
c > 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?

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 by Prime, 18 March 2013 - 04:41 PM.

[bushindo]

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.

Spoiler for detailed solution

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+b²
c > (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)²
c > 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'

Spoiler for

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.

[Prime]

 

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.

Spoiler for detailed solution

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+b²
c > (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)²
c > 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'

Spoiler for

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.

[bushindo]

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

Spoiler for

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 =).

[Prime]

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

Spoiler for

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:

Spoiler for the answer

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.

[bushindo]

 

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

Spoiler for

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:

Spoiler for the answer

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.

[bonanova]

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.