18 replies to this topic

[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.
 

Spoiler for translation

 
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?
 

[bushindo]

Probably not what you were looking for, but here's what I recommend Cole do

 

Spoiler for

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?

[bushindo]

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.
 

Spoiler for translation

 
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

 

Spoiler for

 

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.

 

 

 

 

[Prime]

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.)

Spoiler for the honorable way

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 by Prime, 15 March 2013 - 10:24 PM.

[bushindo]

 

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.
 

Spoiler for translation

 
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

 

Spoiler for

 

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.

Spoiler for

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_bp_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_b³ - 3p_b + 1)/( p_b-1)²

[Prime]

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.

Spoiler for

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_bp_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_b³ - 3p_b + 1)/( p_b-1)²

Spoiler for correction

If it was p_b² (square instead of cube,) our results would match.
p_c = ( p_b² - 3p_b + 1)/( p_b-1)²

[bushindo]

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.

Spoiler for

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_bp_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_b³ - 3p_b + 1)/( p_b-1)²

Spoiler for correction

If it was p_b² (square instead of cube,) our results would match.
p_c = ( p_b² - 3p_b + 1)/( p_b-1)²

 

 

 

Indeed it should be a quadratic expression instead of cubic. My mistake. Good catch, Prime.

[bonanova]

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.)

Spoiler for the honorable way

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.

 

Spoiler for Analysis

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_Bq_C)

P(B C) = (1 - q_C) / (1 - q_Bq_C)

 

If Cole hits Alex, Cole survives with probability 1 - P(B C)

If Cole misses Alex, Cole survives with probability p_BP(B C) + q_Bp_C.

 

Equating these and simplifying gives q_C = (1 - q_B)/q²_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.

[bushindo]

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.)

Spoiler for the honorable way

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.

 

Spoiler for Analysis

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_Bq_C)

P(B C) = (1 - q_C) / (1 - q_Bq_C)

 

If Cole hits Alex, Cole survives with probability 1 - P(B C)

If Cole misses Alex, Cole survives with probability p_BP(B C) + q_Bp_C.

 

Equating these and simplifying gives q_C = (1 - q_B)/q²_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

 

Spoiler for 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_BP(B C) + q_Bp_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 post #6.

 

 

 

Spoiler for 'minor point about notation

I saw that the above analysis also uses a different meaning for P(.) compared to its usage in post #3, 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.

[bonanova]

Spoiler for In English

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.

1. If Cole's survival is more likely if Alex is not hit, Cole will of course fire into the air.
2. If Cole's survival is more likely if Alex is hit, Cole will try to hit Alex.
3. 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?