This is problem 20 in Mosteller Fifty Challenging Problems in Probability and it also appears (with an almost identical solution) on page 82 in Larsen & Marx An Introduction to Probability and Its Applications.

Here's Mosteller's solution:

A is naturally not feeling cheery about this enterprise. Having the first shot he sees that, if he hits C, B will then surely hit him, and so he is not going to shoot at C. If he shoots at B and misses him, then B clearly {I disagree; this is not at all clear!} shoots the more dangerous C first, and A gets one shot at B with probability 0.3 of succeeding. If he misses this time, the less said the better. On the other hand, suppose A hits B. Then C and A shoot alternately until one hits. A's chance of winning is (.5)(.3) + (.5)^2(.7)(.3) + (.5)^3(.7)^2(.3) + ... . Each term corresponds to a sequence of misses by both C and A ending with a final hit by A. Summing the geometric series we get ... 3/13 < 3/10. Thus hitting B and finishing off with C has less probability of winning for A than just missing the first shot. So A fires his first shot into the ground and then tries to hit B with his next shot. C is out of luck.

As much as I respect Mosteller, I have some serious problems with this solution. If we allow the option of firing into the ground, then if all fire into the ground with every shot, each will survive with probability 1. Now, the argument could be made that a certain strategy for X that both allows them to survive with probability 1 and gives less than a probability of survival of less than 1 for at least one of their foes would be preferred by X. However, if X pulls the trigger and actually hits someone what would the remaining person, say Y, do? If P(X hits)=1, clearly Y must try to hit X, since X firing at Y with intent to hit dominates any other strategy for X. If P(X hits)<1 and X fires at Y with intent to hit, then P(Y survives)<1 (since X could have hit Y). Thus, Y must insure that X can not follow this strategy by shooting back at X (thus insuring that P(X survives)<1). Therefore, I would conclude that the ideal strategy for all three players, assuming that they are rational and value survival above killing their enemies, would be to keep firing into the ground. If they don't value survival above killing their enemies (which is the only a priori assumption that I feel can be safely made in the absence of more information), then the problem can't be solved unless the function each player is trying to maximize is explicitly given. -- -- clong@cnj.digex.com (Chris Long)

OK - I'll have a go at this.

How about the payoff function being 1 if you win the "duel" (i.e. if at some point you are still standing and both the others have been shot) and 0 otherwise? This should ensure that an infinite sequence of deliberate misses is not to anyone's advantage. Furthermore, I don't think simple survival makes a realistic payoff function, since people with such a payoff function would not get involved in the fight in the first place!

Now, denote each position in the game by the list of people left standing, in the order in which they get their turns (so the initial position is (A,B,C), and the position after A misses the first shot (B,C,A)). We need to know the value of each possible position for each person.

By definition
valA(A) = 1 valB(A) = 0 valC(A) = 0 valA(B) = 0 valB(B) = 1 valC(B) = 0 valA(C) = 0 valB(C) = 0 valC(C) = 1

Consider the two player position (X,Y). An infinite sequence of misses has value zero to both players, and each player can ensure a positive payoff by trying to shoot the other player. So both players deliberately missing is a sub-optimal result for both players. The question is then whether both players should try to shoot the other first, or whether one should let the other take the first shot. Since having the first shot is always an advantage, given that some real shots are going to be fired, both players

should try to shoot the other first. It is then easy to establish that
valA(A,B) = 3/10 valB(A,B) = 7/10 valC(A,B) = 0 valA(B,A) = 0 valB(B,A) = 1 valC(B,A) = 0 valA(B,C) = 0 valB(B,C) = 1 valC(B,C) = 0 valA(C,B) = 0 valB(C,B) = 5/10 valC(C,B) = 5/10 valA(C,A) = 3/13 valB(C,A) = 0 valC(C,A) = 10/13 valA(A,C) = 6/13 valB(A,C) = 0 valC(A,C) = 7/13

Now for the three player positions (A,B,C), (B,C,A) and (C,A,B). Again, the fact that an infinite sequence of misses is sub-optimal for all three players means that at least one player is going to decide to fire. However, it is less clear than in the 2 player case that any particular player is going to fire. In the 2 player case, each player knew that if it was sub-optimal for him to fire, then it was optimal for the other player to fire at him and that he would be at a disadvantage in the ensuing duel because of not having got the first shot. This is not necessarily true in the 3 player case.

Consider the payoff to A in the position (A,B,C). If he shoots at B, his

expected payoff is
0.3*valA(C,A) + 0.7*valA(B,C,A) = 9/130 + 0.7*valA(B,C,A)
If he shoots at C, his expected payoff is
0.3*valA(B,A) + 0.7*valA(B,C,A) = 0.7*valA(B,C,A)
And if he deliberately misses, his expected payoff is
valA(B,C,A)

Since he tries to maximise his payoff, we can immediately eliminate shooting at C as a strategy - it is strictly dominated by shooting at B. So A's

expected payoff is
valA(A,B,C) = MAX(valA(B,C,A), 9/130 + 0.7*valA(B,C,A))
A similar argument shows that C's expected payoffs in the (C,A,B) position are
For shooting at A: 0.5*valC(A,B,C) For shooting at B: 35/130 + 0.5*valC(A,B,C) For missing: valC(A,B,C)
So C either shoots at B or deliberately misses, and
valC(C,A,B) = MAX(valC(A,B,C), 35/130 + 0.5*valC(A,B,C))

Each player can obtain a positive expected payoff by shooting at one of the other players, and it is known that an infinite sequence of misses will result in a zero payoff for all players. So it is known that some player's strategy must involve shooting at another player rather than deliberately missing.

Now look at this from the point of view of player B. He knows that if it is sub-optimal for him to shoot at another player, then it is optimal for at least one of the other players to shoot. He also knows that if the other players choose to shoot, they will shoot at him. If he deliberately misses, therefore, the best that he can hope for is that they miss him and he is presented with the same situation again. This is clearly less good for him than getting his shot in first. So in position (B,C,A), he must shoot at another player rather than deliberately miss.

B's expected payoffs are
For shooting at A: valB(C,B) = 5/10 For shooting at C: valB(A,B) = 7/10

So in position (B,C,A), B shoots at C for an expected payoff of 7/10. This

gives us
valA(B,C,A) = 3/10 valB(B,C,A) = 7/10 valC(B,C,A) = 0

So valA(A,B,C) = MAX(3/10, 9/130 + 21/100) = 3/10, and A's best strategy is

position (A,B,C) is to deliberately miss, giving us
valA(A,B,C) = 3/10 valB(A,B,C) = 7/10 valC(A,B,C) = 0

And finally, valC(C,A,B) = MAX(0, 35/130 + 0) = 7/26, and C's best strategy

in position (C,A,B) is to shoot at B, giving us
valA(C,A,B) = 57/260 valB(C,A,B) = 133/260 valC(C,A,B) = 7/26

I suspect that, with this payoff function, all positions with 3 players can be resolved. For each player, we can establish that if their correct strategy is to fire at another player, then it is to fire at whichever of the other players is more dangerous. The most dangerous of the three players then finds that he has nothing to lose by firing at the second most dangerous.

Questions:

(a) In the general case, what are the optimal strategies for the other two players, possibly as functions of the hit probabilities and the cyclic order of the three players?

(b) What happens in the 4 or more player case?

-- David Seal <dseal@armltd.co.uk>

In article <1993Mar25.022459.10269@cs.cornell.edu>, karr@cs.cornell.edu (David Karr) writes:

The Good, the Bad, and the Ugly are standing at three equidistant

"P" "Q" "R" -- allow me these alternate names.

points around a very large circle, about to fight a three-way duel to see who gets the treasure. They all know that the Good hits with probability p=.9, the Bad hits with probability q=.7, and the Ugly hits with probability r=.5.

Yes, I know this sounds like decision/truel from the rec.puzzles archive. But here's the difference:

At some instant, all three will fire simultaneously, each at a target of his choice. Then any who survive that round fire simultaneously again, until at most one remains. Note that there are then four possible outcomes: the Good wins, the Bad wins, the Ugly wins, or all are killed.

A multi-round multi-person game can get complicated if implicit alliances are formed or the players deduce each other's strategies. For simplicity let's disallow communication and assume the players forget who shot at whom after each round.

Now the questions:

These are not easy questions, even with the simplifying assumptions I've made.

1. What is each shooter's strategy?

Each player has two possible strategies so there are eight cases to consider; unfortunately none of the players has a strictly dominant strategy:

P aims at Q aims at R aims at P survival Q survival R survival Noone lives --------- --------- --------- ---------- ---------- ---------- -----------

Q P P 0.0649 0.0355 0.7991 0.1005 Q P Q 0.1371 0.0146 0.6966 0.1517

2. Who is most likely to survive?

Good, Bad, or Ugly, depending on the strategies.

3. Who is least likely to survive?

Bad or Ugly, depending on the strategies.

4. Can you change p, q, and r under the constraint p > q > r so that

the answers to questions 2 and 3 are reversed? Which of the six possible permutations of the three shooters is a possible ordering of probability of survival under the constraint p > q > r?

Yes. Of the six possible survival-probability orderings,

five can be obtained readily

p q r P_surv Q_surv R_surv Order --- --- --- ------ ------ ------ ------- 0.255 0.25 0.01 0.408 0.413 0.172 Q P R 0.26 0.25 0.01 0.412 0.406 0.173 P Q R 0.75 0.25 0.01 0.675 0.076 0.242 P R Q 0.505 0.50 0.01 0.325 0.324 0.344 R P Q 0.505 0.50 0.02 0.314 0.320 0.353 R Q P

Unlike the p=.9, q=.7, r=.5 case we are given, the five cases in this table do have simple pure solutions: in each case p shoots at q, while q and r each shoot at p. (I've found no case with a "simple pure" solution other than this "obvious" p aims at q, q aims at p, r aims at p choice.)

5. Are there any value of p, q, and r for which it is ever in the

interest of one of the shooters to fire into the ground?

No. It can't hurt to shoot at one's stronger opponent. This is the easiest of the questions ... but it's still not easy enough for me to construct an elegant proof in English.

-- David Karr (karr@cs.cornell.edu)

Speaking of decision/truel, I recall a very interesting analysis (I might have seen it here in rec.puzzles) suggesting that the N-person "truel" (N-uel?) has a Cooperative Solution (ceasefire) if and only if N = 3. But I don't see this in the FAQL; anyone care to repost it?

-- James Allen

In article <1993Apr1.123404.18039@vax5.cit.cornell.edu> mkt@vax5.cit.cornell.edu writes:

In article <1993Mar25.022459.10269@cs.cornell.edu>, karr@cs.cornell.edu (David Karr) writes:

...?

5. Are there any value of p, q, and r for which it is ever in the

interest of one of the shooters to fire into the ground?

Yes, p=1, q=1, r=1. The only way for one to survive is to have the other two shoot at eachother. Shooting at anyone has no effect on ones personal survival.

I assume by "has no effect on" you mean "does not improve."

If all follow the same logic, they will keep shooting into the

ground and thus all live.

I was very pleased by this answer but I had to think about it. First of all, it assumes that continuing the fight forever has a positive value for each shooter. My preferred assumption is that it doesn't. But even if each shooter is simply trying to maximize his probability of never being shot, I wonder about the "has no effect" statement.

Suppose that in round 1 the Good fires into the ground and the Bad shoots at the Good. Then the Ugly lives if he shoots the Bad and dies if he does anything else. (The Bad will surely shoot at the Ugly if he can in round 2, since this dominates any other strategy.) So it definitely makes a difference to the Ugly in this case to shoot at the Bad.

But all this is under the assumption that no shooter can tell what the others are about to do until after all have shot. This isn't entirely unreasonable--we can certainly set up a game that plays

this way--but suppose we assume instead

All three start out with guns initially holstered. Each one is a blindingly fast shot: he can grab his holstered gun,

aim, and fire in 0.6 second.

A shooter can redirect his unholstered gun at a different target and

fire in just 0.4 second.

The reaction time of each shooter is just 0.2 second. That is, any

decision he makes to act can be based only on the actions of the other two up to 0.2 second before he initiates his own action.

The bullets travel between shooters in less than 0.1 second and

stop any further action when they hit.

Then I think the conclusion holds for p=q=r=1: The best strategy is to wait for someone else to grab for their gun, then shoot that person, therefore nobody will shoot at anyone. At least I haven't yet thought of a case in which you improve your survival by shooting at anyone. Of course this is only good if you don't mind waiting around the circle forever.

-- David Karr (karr@cs.cornell.edu)

In article <1993Apr5.210749.2657@cs.cornell.edu>, karr@cs.cornell.edu (David Karr) writes:

In article <1993Apr1.123404.18039@vax5.cit.cornell.edu> mkt@vax5.cit.cornell.edu writes:

In article <1993Mar25.022459.10269@cs.cornell.edu>, karr@cs.cornell.edu (David Karr) writes:

...?

5. Are there any value of p, q, and r for which it is ever in the

interest of one of the shooters to fire into the ground?

Yes, p=1, q=1, r=1. The only way for one to survive is to have the other two shoot at eachother. Shooting at anyone has no effect on ones personal survival.

I assume by "has no effect on" you mean "does not improve."

If all follow the same logic, they will keep shooting into the

ground and thus all live.

I was very pleased by this answer but I had to think about it. First of all, it assumes that continuing the fight forever has a positive value for each shooter. My preferred assumption is that it doesn't. But even if each shooter is simply trying to maximize his probability of never being shot, I wonder about the "has no effect" statement.

Suppose that in round 1 the Good fires into the ground and the Bad shoots at the Good. Then the Ugly lives if he shoots the Bad and dies if he does anything else. (The Bad will surely shoot at the Ugly if he can in round 2, since this dominates any other strategy.) So it definitely makes a difference to the Ugly in this case to shoot at the Bad.

Here's where the clincher comes in! If we "assume" the object of the game is to survive, and that there exists one unique method for survival, then all the shooters will behave in the same fashion. Obviously the above case will not hold. How do we distinguish between the good, the bad and the ugly? If the command is "Shoot" then all will shoot and somebody is going to wind up lucky (Prob that it is you is 1/3). If the command is "No Shoot", then all will fire into the ground (or just give up and go home--no sense waitin' around wastin' time, ya know)...

But wait, you cry! What if there exists more than one solution for optimal survival. Then what? Will the Good the Bad and the Ugly each randomly decide between "Shoot" and "No Shoot" with .5 probability? If this is true, then is it in your best interest to shoot someone? If it is, then we arrive back at square one: since we assume all shooters are geniouses, then all will shoot-- arriving at an optimal solution of "Shooting". If the answer is "No Shooting", we arrive at an optimal solution of "No Shooting". If there is no effect on your personal survival, then do we analyze this with another .5 probability between the chances of soemone shooting or not shooting? If the answer to this is "Shoot" then we arrive at square one: all will Shoot; if no, then all will withold. If there is no effect, then we arrive at another .5 probability... Obviously you can see the recursion of this process.

Perhaps this would be easier to discuss if we let p=1, q=1, r=0. Obviously, in terms of survival, shooting at the ugly would be wasting a shot. Thus we have

made a complex problem more simple but retaining the essence of the paradox
If there are two gunmen who shoot and think with perfect logic and are kept inside a room and are allowed to shoot at discrete time intervals without being able to "see" what your opponent will do, what will happen?

Let's say that you are one of the gunmen (the Good). You reason "My probability to survive the next round is independent on whether or not I fire at him." So you say to yourself, "Fire at the opponent! I'll get to stop playing this blasted game." But then you realize that your opponent will also think the same way...so you might think that you might as well not shoot. But if your opponent thinks that way, then you know that: 1. You can survive the next round. 2. You can shoot him if you wish on this round (if you like). So you say to yourself, "Fire at the opponent!". But you know the opponent thinks the same way so... you're dead. But really, you say. Is there a way of "knowing" what the opponent thinks? Of course not. You reason that you can know your probability of shooting your opponent (either 1 or 0). You reason that your opponent has a variable probability of shooting you. Thus from your perspective, p=1 and r<1. We already discussed this case and said "Shoot!". But wait you cry! What if the opponent figures this out too: p<1, r=1? Sorry, you're both dead. 'nuff said! This applies to the p=r=q=1 case as well.

But all this is under the assumption that no shooter can tell what the others are about to do until after all have shot.

Ay, there's the rub!

This isn't entirely unreasonable--we can certainly set up a game that plays

this way--but suppose we assume instead

All three start out with guns initially holstered. Each one is a blindingly fast shot: he can grab his holstered gun,

aim, and fire in 0.6 second.

A shooter can redirect his unholstered gun at a different target and

fire in just 0.4 second.

The reaction time of each shooter is just 0.2 second. That is, any

decision he makes to act can be based only on the actions of the other two up to 0.2 second before he initiates his own action.

The bullets travel between shooters in less than 0.1 second and

stop any further action when they hit.

Then I think the conclusion holds for p=q=r=1: The best strategy is to wait for someone else to grab for their gun, then shoot that person, therefore nobody will shoot at anyone. At least I haven't yet thought of a case in which you improve your survival by shooting at anyone. Of course this is only good if you don't mind waiting around the circle forever.

Hmmn...alternate ploy:

0.0 You begin to unholster your gun 0.2 Opponents begin unholstering guns. You aim into the ground for .2 sec. 0.4 Opponents are unholstered you are unholstered. They note you aren't

aiming at them. They haven't aimed at anyone yet.

What happens now? I'll have to think about it, but I haven't seen anything fundamentally different between this and the above case yet.

More ideas to consider:

You begin unholstering your gun but only for .1 sec (you place it by .2 ) You begin unholstering your gun but only for .09 sec (you place it by .19)

You start to aim for .1 sec and then stop aiming. You start to aim for .1 sec and then turn and aim at another. You start to aim for .09 sec and then stop aiming (or aim at another)

Looking at the answer for decision/truel, I came across the following:

Each player can obtain a positive expected payoff by shooting at one of the other players, and it is known that an infinite sequence of misses will result in a zero payoff for all players. So it is known that some player's strategy must involve shooting at another player rather than deliberately missing.

This may be true but it's not obvious to me. For example, suppose A, B, and C are passengers in a lifeboat in a storm. If they all stay aboard, the lifeboat is certain to sink eventually, taking all three to the bottom with it. If anyone jumps overboard, the two remaining in the boat are guaranteed to survive, while the person who jumped has a 1% chance of survival.

It seems to me the lifeboat satisfies the quoted conditions, in the sense that if nobody jumps then the payoff for all is zero, and the payoff for jumping is 0.01 which is positive. But it is not clear to me that the three shouldn't just all sit still until someone goes nuts and jumps overboard despite everything, for this strategy gives a 67% chance of survival (assuming everyone is equally likely to "crack" first) vs. only 1% for jumping by choice. Even if there is a wave about to swamp the boat, I'd wonder if the situation wouldn't just reduce to a game of "chicken," with each person waiting until the last minute and jumping only if it seems the other two have decided to sink with the boat if you don't jump.

On the other hand, this situation is set up so it is always worse to be the first person to jump. In the truel I don't think this is true, but only because of the asymmetry of the odds, and to determine whether anyone shoots, it is easiest to proceed directly to considering B's point of view.

Whenever it is B's turn to shoot, B can divide the possible courses of action into four possibilities (actually there are seven, but three are ruled out a priori by obvious optimizations of each individual's strategy):

Nobody ever shoots (expected value 0) A shoots first (at B, expected value <= .7) C shoots first (at B, expected value <= .5) B shoots first (at C, expected value .7)

In fact the value of "A shoots first" is strictly less than .7 because in case A misses, the same four possibilities recur, and all have expected payoff < 1.

So the value of "B shoots first" uniquely maximizes B's value function, ergo B will always shoot as soon as possible.

The rest of the analysis then follows as in the archive.

-- David Karr (karr@cs.cornell.edu)

Looking at the answer for decision/truel, I came across the following:

Each player can obtain a positive expected payoff by shooting at one of the other players, and it is known that an infinite sequence of misses will result in a zero payoff for all players. So it is known that some player's strategy must involve shooting at another player rather than deliberately missing.

This may be true but it's not obvious to me. For example, suppose A, B, and C are passengers in a lifeboat in a storm. If they all stay aboard, the lifeboat is certain to sink eventually, taking all three to the bottom with it. If anyone jumps overboard, the two remaining in the boat are guaranteed to survive, while the person who jumped has a 1% chance of survival.

It seems to me the lifeboat satisfies the quoted conditions, in the sense that if nobody jumps then the payoff for all is zero, and the payoff for jumping is 0.01 which is positive. But it is not clear to me that the three shouldn't just all sit still until someone goes nuts and jumps overboard despite everything, for this strategy gives a 67% chance of survival (assuming everyone is equally likely to "crack" first) vs. only 1% for jumping by choice. ...

Yes and no. Yes in the sense that if you treat the game as a psychological one, the best strategy is as you say. But treating it as a mathematical game, you've got to adhere to your strategy and you've got to assume that others will adhere to theirs.

I.e. as a mathematical game, "Don't jump at all" and "Don't jump unless I crack" are different strategies, and the first one is often (not always) superior - e.g. if I take "Don't jump at all" and the others take "Don't jump unless I crack", I'm certain to survive and the others each have a 50.5% chance, which is better from my point of view than a 67% chance of survival for all of us. As a psychological game, some of the mathematical strategies may simply not be available - i.e. you cannot control what you will do if you crack, and so we commonly use "Don't jump" to mean "Don't jump unless I crack", since "Don't jump at all" is not an available strategy for most real humans. But for mathematical analysis, the problem has to tell you what strategies you are not allowed to take.

What the argument above shows is that "Don't jump at all" is not a stable strategy, in the sense that if everyone takes it, it is in everyone's interest to change strategy. I.e. it shows that someone will jump eventually, even if it's only the result of someone actually having taken "Don't jump unless I crack".

Applied to the truel, the argument above does show that someone's strategy will involve shooting at another player: the strategy "Don't shoot at all" is unstable in exactly the same way as "Don't jump at all" was. But I agree it allows for a lot of leeway about how and when the deadlock gets broken, and your argument showing that it is always in B's interest to shoot is more satisfactory.

David Seal

References:

"The Truel", D. Marc Kilgour and Steven J. Brams, Mathematics Magazine 70, 5 (December 1997), p. 315

earliest reference to puzzle is in: C. Kinnaird, Encyclopedia of Puzzles and Pastimes, Citadel, Secaucus, NJ, 1946

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