Talk:Truel

Expansion

There! Should do for now. Paidgenius 20:24, 21 November 2006 (UTC)

Thanks for your work on the article. The solution to the theoretical example needs a lot of work. Two things to start: as long as C is still alive, C is the "most dangerous opponent". B would never shoot at A if C is still alive: if B shoots A, then C shoots B; if B misses A, then C, who can shoot (and kill) either A or B, will certainly shoot B. Either way, if B shoots at A while C is still alive, B will die. So the solution as it stands is not correct (the conclusion might be, but the argument is not). -- Doctormatt 01:15, 24 November 2006 (UTC)

Okay, I fixed the argument by giving the probabilities of success of A's different strategies. -- Doctormatt 02:00, 24 November 2006 (UTC)

Well, everyone makes mistakes. That's cool though. Paidgenius 20:12, 8 December 2006 (UTC)

I am unclear why my last edit should be removed. I added a sentence to say that B would shoot at C rather than A. Doctormatt removed it, saying it was unsupported, and the question is A's strategy, rather than B's. I also think that the figures given for chances of survival are incorrect.

C's best choice is to shoot B if both B and C survive, because B has the best chance of killing him. B's best choice is to shoot C, as C will kill him if he shoots at him. If A kills B, then C will kill A. If A kills C, B has a 2/3 chance of killing A. So A shoots to miss.

Then B shoots at C. In 6/9 examples, B will kill C. It is then A's turn again: in 2/9 examples, A will kill B and survive. In 3/9 examples, B will miss C. C will then kill B. A will shoot C, and in 1/9 examples A will kill C and survive. In 2/9 examples, A will miss C, and C will then kill A. In 4/9 examples, B will shoot at A having killed C, and have a 2/3 chance of killing A.Abigailgem 16:59, 29 May 2007 (UTC)

I find your argument very confusing. You claim that if A kills C, then B has a 2/3 chance of killing A. This is only true on the first shot. In fact, if A kills C, then B and A will shoot back and forth at each other until one of them is dead. This results in B having a 6/7 chance of killing A. So, I'm not sure your calculations are valid: you need to take into consideration truels of unlimited length. In any case, this is all rather OR-ish, so why don't we just go to the library? Here are some papers to try:

• Kilgour, D. M., The simultaneous truel, Internat. J. Game Theory 1 (1971/72), 229--242
• Kilgour, D. M., The sequential truel, Internat. J. Game Theory 4 (1975), no. 3, 151--174
• Kilgour, D. M., Equilibrium points of infinite sequential truels, Internat. J. Game Theory 6 (1977), no. 3, 167--180

Cheers, Doctormatt 18:05, 29 May 2007 (UTC)

Having had a look at some of the previous revisions, I still would prefer a verbal as well as mathematical explanation of the solution. I am placing that here rather than making such an extreme revision. Does anyone have a comment?

Given a choice between shooting at A or B, C should shoot B, who has the best chance of killing C. Therefore, given a choice of shooting A or C, B should choose to shoot at C.

If A kills B, A will then be killed by C. If A kills C, he faces B and it is B's shot. So A shoots to miss.

In 1/3 of cases, B will miss, then C will kill B and A then has one shot at C, a one third chance of success.

In 2/3 of cases, B will kill C, so that A faces B and it is A's shot.

I do not propose to alter the calculations. Abigailgem 14:44, 28 July 2007 (UTC)

Your explanation fails to explain why A is better off missing. Your first claim "If A kills B, A will then be killed by C. If A kills C, he faces B and it is B's shot. So A shoots to miss." does not compare A's chances of success when shooting to miss with A's chance of success when aiming at C. Note that A is not a perfect shot, so shooting a C is not the same as killing C. One has to consider the probabilty of A's hitting C if one is considering whether or not A should shoot at C.

In general, one must calculate the probability of success with each possible strategy in order to choose the best one. In this example, the fact that C is a perfect shot makes the calculations simpler, and this makes people think that they can construct an intuitive solution in the general case. But, one can't. Please take a look at the referenced paper for lots of examples that show the non-intuitive nature of this problem.

I would like to change the example to one in which nobody is a perfect shot in order to avoid giving readers the impression that an intuitive solution is generally possible. Doctormatt 21:33, 28 July 2007 (UTC)

Indeed. But, if A shoots with the intention of hitting B, in one third of cases (according to the way the problem is drafted) A will kill B. This is an outcome against A's interests, as C, the perfect shot, will then kill A. This is part of the rules of the puzzle, as it is set. There is no need to state it in the solution. The point is that A does not want to kill B or C with his first shot, as that will harm his interests.

I consider changing the example would not be an improvement of the article. However, adding an additional example which was non-intuitive could be an improvement, giving other aspects of the problem. Abigailgem 15:31, 5 August 2007 (UTC)

I apologize, but I'm still not following your logic. Perhaps you could apply your argument to other cases, and tell me the results? This might help me see how you are using the probabilities in your argument. Consider, if you care to, these variations:

1. A is 40% accurate, B is 50% accurate, C is perfect
2. A is 30% accurate, B is 35% accurate, C is perfect
3. A is 20% accurate, B is 30% accurate, C is perfect
4. A is 10% accurate, B is 35% accurate, C is perfect

What is your impression of A's optimal strategy in these variations of the problem?

I agree that it is clear that A is not helped by shooting at B. That can be argued without calculation. So, in these problems, the question is should A intentionally miss, or should A shoot at C? I can say, based on calculations of the probability of A's survival with each strategy. I honestly don't see any other way to determine the optimal strategy. Cheers, Doctormatt 04:18, 6 August 2007 (UTC)

Necro- A clearly should not shoot at B in any case, as he can only be hurt. If A misses, intentionally or not, B will fire at C to kill, because if B does not kill C, C kills B. If A kills C, then it becomes a duel between B and A, with B having the first shot. If B kills C, it becomes a duel between A and B, with A having the first shot. If B misses C, then C kills B, and A gets one shot at C for the overall win.

Let a be the odds that A hits on a given shot that A makes, and likewise b and B. The odds of A winning the duel between A and B is the sum of the one of the infinite series a+a((1-a)*(1-b))^n (If A shoots first), or (1-b)*(a+a((1-a)*(1-b))^n (if B shoots first). [(1-a)(1-b)^n is the odds that the duel will continue for n full exchanges without resolution, a is the chance that A will win the duel on a given shot, provided that shot occurs, and 1-b is the change that A will not get a shot on round n+1, provided B fires first and round n completed without resolution. A is then (1-b) times as likely to win if B shoots first. Define u to be equal to the infinite series (a+a((1-a)*(1-b))^n; A wins with probability u if A fires first, or (1-b)u if B fires first.

Trivial: If C kills B, A wins the truel with probability a.

If A fails to kill C, A duels B with first shot with probability b, and duels C with first shot with probability 1-b. Odds of winning, are a(1-b)+b(u) [Odds that B misses and A wins the duel with C, plus odds that B hits and A wins the duel with B]

If A kills C, A duels B granting first shot, and wins with probability (1-b)u.

A is better off not killing C if a(1-b)+b(u)>(1-b)u.

Doing the algebra, A is better off not killing C when a-ab+2bu-u>0 a and b are between 0 and 1, with a<b, therefore a-ab>0. u is also between 0 and 1, so 2bu=u>0 for all b>0.5

Without even evaluating u, clearly A is better off missing whenever B is at least 50% accurate. Describing u as a function of a and b and then describing the curve a-ab+2bu-u=0 is left as an exercise for the student.

Treedel (talk) 17:48, 15 September 2011 (UTC)

Example: OR?

An editor removed the example of a theoretical truel, claiming that it is "blatant OR". My feeling is that this is not OR, as any undergraduate probability student can verify this example. It is not "research": it is merely a not completely trivial (but not difficult) computation. In mathematics articles there are numerous such examples of the results of computations. In terms of verifiability, anyone with a modicum of probability knowledge can verify this, without any other references, so I don't think any citations are needed for this. I'd be happy to hear what other think. I'd also be happy to typset the calculations, and perhaps put them on a separate "proof" page. If people really feel it's OR, I'll find a published example to copy. Doctormatt 22:57, 4 November 2007 (UTC)

• Doing any derivation on a Wikipedia page without providing a source or a published example of that precise formula is original research. You cannot assume that everyone has "a modicum of probability knowledge" and because of the requirements in WP:OR, you cannot simply create that material yourself. If you do find a published copy, do not recreate the formula with a reference on the end of it, but instead quote only the conclucions of that finding and cite that. Cumulus Clouds 01:09, 6 November 2007 (UTC)

Gardner's truel

In a book by Martin Gardner, there is a solved problem about a truel. I suspect this is the "original" truel problem, others being variations of this one. From memory: Cowboy Smith is a 100% sure shot, Brown 80% and Jones 50%. They shoot sequentially, the order set by a lottery from the start. They have infinitely many bullets, they can deliberately miss, and they keep on shooting till only one is alive. Last man standing wins the girl they are fighting about. As long as S and B are both alive, they will shoot at each others only, and J should deliberately miss - because if he shoots one of them, it will be the other one's turn, and he will die at the next shot with 80% or 100% certainty. If he waits till S or B has finished off the other one, it will be his turn next, and J has a 50% chance of winning at the next shot. Completing the probability tree (including a geometric series),

P(S wins) = 27/90 (approx. 30 %)

P(B wins) = 16/90 (approx. 18 %)

P(S wins) = 47/90 (approx. 52 %)

This contradicts the statement presently in the article that it is the best strategy to shoot at the best shooter - not so for Jones!--Noe (talk) 15:40, 24 November 2008 (UTC)

Merger proposal

I propose merging Three way duel (puzzle) into this article (Truel), as they seem to deal with essentially the same topic. I chose to propose a merge into this article because this article seems to have a longer history and more activity. However, one could make an argument for a merge in the other direction.--GregRM (talk) 04:37, 23 December 2008 (UTC)

Support - I see no reason not to do so.--Noe (talk) 13:12, 23 December 2008 (UTC)

• Support - don't see any reason to have separate articles here. Robofish (talk) 20:45, 5 May 2010 (UTC)
  • One person agreed with the proposer a day after the proposal was made and another waited until 4½ months ago to agree. However, despite this apparent strong support nothing was done. I am removing the tags from both articles in the hope that some one will be moved to do something. JimCubb (talk) 01:22, 23 September 2010 (UTC)

Merge from Three-way duel

OK, the following ought to be merged nicely into the article - I'll try to get back to this if not someone beats me to it ;-) --Nø (talk) 14:25, 23 September 2010 (UTC)

The three way duel is a logic puzzle proposed by Martin Gardner. [1]

Three men are in a pistol duel. Each man will shoot in turn. The three men are identified as A, B, and C. A is a poor shot, and hits only 50% of the time. B is an expert marksman, and always hits. C is a moderate shot, and hits 80% of the time. (In some variants of the problem, A's probability is 25% and C's is 50%; in practice this makes little difference as long as A's is lower than C's.)

The exact order in which the three men will take turns shooting is variable (although in some variants it is stated as being A, B, C).

As the problem is based on classical dueling it is assumed that a hit target is "killed", but this is not necessary and some variants of the problem alter the theme to a game where targets are unharmed, but required to withdraw, when hit.

The question is, what is the best strategy for A to follow to win the duel?

Solutions

The most common solution [2] is that A should deliberately fire into the ground until one of the other two men is dead, then shoot at him.

The reasoning for this is as follows: B and C are greater threats to each other than A is to either of them, and thus rationally should target each other first. If B fires first, he will certainly kill C; if C fires first, he may kill B, and if he does not, B will certainly kill him. In the first "round" of the duel one of these two interactions will occur between B and C and it is not in A's interest to disrupt it. If A kills one of the men, the other man will target him, probably killing him; if A fires but does not kill either man, he makes no difference. After either of these interactions completes, it will become A's turn with A never having been targeted and having a chance to kill the one remaining man and win the duel, and this is the best possible position for him.

Paradox

This solution is considered by some observers to expose a paradox in reasoning. They argue that, if A is permitted to fire into the ground, then B and C could do so too. Based on that, if survival is a duelist's only priority, the best strategy is for all three men to fire into the ground until the ammo runs out and then walk away. Intuitively and emotionally we tend to believe that if A dies by the end of the duel, it makes no difference how many people he killed before he died; but using this heuristic to judge duel strategies may lead to this strategy being considered optimal. If "winning the duel" has additional requirements other than survival (eg, killing the opponents), A's strategy above might no longer be considered the best because it fails to meet those other requirements. Although since B is going to be the first one to get shot anyway, he might as well shoot someone (C if he is still there), therefore C should shoot him, so this strategy does work only for A. If it is allowed, B could also shoot himself non-fatally, so as to make sure he was not the first shot, but whether or not this would actually make sense would depend on how much of a difference this makes in B's ability to aim.

On the other hand if one duelist (the first) hates one other (the second) so badly that he would being willing to die to kill him. (Otherwise why be in a duel in the first place?) Then he should shot at him right away. If he hits him then the third will shoot at the first guy and maybe miss. If he misses then the third will probably shoot at the second as the only armed man left. Or at least he has a 50-50 chance of not being the target. This assumes that they have one bullet as is common in traditional duels. If the accuracy of the guns gives them about a 50-50 chance of hitting their target (which also seems traditional) then he would have about a 75% chance of his hated person being killed and around a 50% chance of himself surviving. 168.137.100.23 (talk) 17:27, 8 November 2010 (UTC)

If we consider the case where all three truelists have perfect accuracy, then the only logical course of action is for everyone to intentionally miss all the time; whoever kills the first opponent will immediately be killed. (Insert obligatory Wargames quote) Treedel (talk) 17:54, 15 September 2011 (UTC)

Category:Logic puzzles

Uhhh...

You can't be serious about that math equation. I know that Wikipedia is full (F-U-L-L) of useful information, but sometimes, one could live without such nonsense. An explanation of the topic, which, depending on the cultural position and popularity of the topic itself, etc., etc., etc., can vary in depth; the history of the topic; and other information about it should be all that is needed. Obviously, most people don't want a math equation about a word that is similar to the word "duel". — Preceding unsigned comment added by Kaabii123 (talk • contribs) 21:10, 3 June 2012 (UTC)

What edit are you proposing? Thanks. SPECIFICO talk 19:06, 15 February 2015 (UTC)

The game theory is referenced. The lede needs expanding. Widefox; talk 19:57, 15 February 2015 (UTC)