# The three-way duel

The late great Martin Gardner once posed this puzzle.

Suppose you’re involved in a duel with two other people. You (Person A) shoot first, followed Person B, followed by Person C, then it goes back to you, and so on. Moreover, you know the following about everyone’s shooting skills.

• You (Person A) will hit your target with probability 1/3.
• Person B will hit her target with probability 2/3.
• Person C is a perfect marksman, will hit his target with probability 1.

You get to go first. Who would you shoot at, and why? Best solution gets a free sheep.

(Note: your options are Person B, or Person C, or neither.)

Update:  Jonathan of Unshielded Colliders has been awarded a free sheep for his solution. Here you go, Jonathan:

Soul Physics is authored by Bryan W. Roberts. Thanks for subscribing.
Want more Soul Physics? Try the Soul Physics Tweet.

## 13 thoughts on “The three-way duel”

1. Joseph Smidt

I’m torn.

If I shot and killed one I’m sure the other would kill me quickly. (As they are such good shots.) I would think I would want them to kill each other first and then I shoot the one that is left.

So neither until one is dead then shoot the remaining person. However, only having 1/3 chance to kill at this point makes me a little uneasy that this is the best solution.

2. Jonathan Livengood

I’m not sure I understand the problem being posed. Is it one against two or every man for himself? Are the probabilities like X chance that you hit your target, Y chance that you miss your target without hitting anyone else, and 0 chance of hitting something unintendedly? What assumptions can we make about the other duelists? Will they act rationally, i.e. in order to maximize their chance of living?

3. Bryan

@Joseph — I like your thinking, that was my intuitive first reaction as well.

@Jonathan — All those things matter, but I’ll let you decide. I’d set them all up to act rationally in some sense. For me, that would mean to maximize utility, suitably defined. That might mean total chance of living, or it might mean maximize the guaranteed number of turns… again, I’ll let you decide.

@Anon5:21 — Nice. But is that physically possible if you’re all standing in a triangle? I suppose if you had a boomerang…

4. Jonathan Livengood

So, if when anyone shoots, he shoots at the opponent with the greatest probability of hitting, then in the first round, you should decline to shoot. In that case, your probability of continuing to the end of any round is strictly greater than it is for the other two decisions.

The probabilities of living given you shoot at no one, p=1, or p=2/3 first go like:

R1: 1 > 7/9 > 2/3
R2: 39/81 > 23/81, and > 26/81
R3: (27/81) + (12/81)(1/3 + 2/9) > (21/81) + (2/81)(1/3 + 2/9), and > (18/81) + (8/81)(1/3 + 2/9)

After round 3, the same shape gets tacked on over and over in each condition. Distribute across the (1/3 + 2/9), take the last term and multiply again by (1/3 + 2/9).

Anyway, that’s what it looks like to me … what am I missing?

Also, Anon probably has a curved space-time in mind. ;)

5. Justin

Part 1:

Late to the party…. But I tried to answer the question before looking at the comments.

I began by thinking that A should shoot neither (as Jonathan does). But then I realized that there are two ambiguities in the question that potentially cast doubt on that answer. Beyond assuming that the shooters are rational (trying to maximize the chance that they will survive the duel)… and that each shooter takes the other shooters to be similarly rational, we are left with two other issues:

1. We aren’t told whether B or C know what A knows about their respective shooting skills.

2. We aren’t told whether A knows whether B or C know what A knows about their respective shooting skills.

Without making assumptions on those points, I’m not sure how to answer the question.

I began by assuming that B and C know what A knows and that A knows that they know. (Further, I assumed that the players play rationally and that they do not act on vengeance or employ a tit-for-tat type strategy.)

Making those assumptions, A should fire at neither:

First, with those assumptions, whether or not A shoots should not change whether or not B shoots (and whether or not A or B shoots should not change whether or not C shoots).

Second, whenever it gets down to a two person duel, each participant will shoot at the other when it is their turn to shoot.

Third, to simplify the problem, note that whatever A does we can break the probability that she dies into two parts – a “2/3” part and a “1/3” part. The “2/3” part is if it is B’s turn to shoot in the first round with both A and C alive; the “1/3” part is if A kills either B or C in the first round. Further, the “2/3” part is the same regardless of what A does in the first round: If she doesn’t shoot, shoots at B and misses, or shoots at C and misses, the rest of the scenario is the same. As such, we only need worry about comparing the “1/3” part.

The “1/3” part:

1. A shoots and kills B. C then shoots and kills B (100% probability).

On 1, the probability that A dies (for the “1/3” part) is: 1

2. A shoots and kills C. B then shoots at A. There is a 2/3 probability that B kills A. If B misses, A shoots at B. There is a 2/3 probability that A misses. If A misses, B then shoots at A. There is a 2/3 probability that B kills A… and so on.

On 2, the probability that A dies (for the “1/3” part) is:
(2/3) + (1/3)(2/3)(2/3) + (1/3)(2/3)(1/3)(2/3)(2/3) + (1/3)(2/3)(1/3)(2/3)(1/3)(2/3)(2/3) + …

3. A shoots neither. What will B do? On the assumptions made above, B will shoot at C. The reason is that B’s only chance is to shoot and kill C: If B doesn’t kill C, then C will shoot and kill B (C should kill the person who is the best shot); if B does kill C, however, B has a chance to live.

So after A shoots neither, B then shoots at C. There is a 2/3 probability B kills C. If B kills C, then A shoots at B. There is a 2/3 probability that A misses B. B would then shoot at A. There is a 2/3 probability that A is killed. If B misses, then A shoots at B. There is a 2/3 probability that A misses B. B would then shoot at A. There is a 2/3 probability that A is killed… and so on.

If B does not kill C, then C shoots and kills B. A would then shoot at C. There is a 2/3 probability that A misses C. C would then shoot and kill A (100% probability).

On 3, the probability that A dies (for the “1/3” part) is:
(1/3)(2/3) + (2/3)(2/3)(2/3) + (2/3)(2/3)(1/3)(2/3)(2/3) + (2/3)(2/3)(1/3)(2/3)(1/3)(2/3)(2/3) + …

Comparing these we find that:

P(1) > P(2) > P(3)

So A should shoot at neither.

6. Justin

Part 2:

But this answer makes multiple assumptions that are not warranted by the question (beyond the rationality bit). What if we do not assume that A knows/believes that B and C know about their respective shooting skills? That is, what if we do not assume that A knows/believes that B or C knows how good of a shot any of the three are? What should A do given that ignorance? (And shouldn’t we assume that A is ignorant in that way given the information we are given in the question?)

Continuing to assume that A assumes that each player will behave rationally (no vengeance), the problem is pretty tricky. A needs to figure out what B will do depending on what A does, then what C will do depending on what A and B do, and so on (until one person is dead – then they all just shoot at the other person).

So, what should A think that B will do if A shoots at B and misses? If A shoots at C and misses? If A shoots at neither?

This will depend on whether A thinks that B initially thinks that A knows anything more than A thinks that B knows. Let’s assume that A does think this. Will A’s decision give B reason to change her belief about what A knows? This depends on what A should do if A has no knowledge of their respective shooting skills (and assumes similar ignorance on the part of B and C). Let’s say that A assigns the probability X to any given shot hitting the target (and assign probability Y to 1 – X). Assuming that things are as A believes:

1. A shoots at B:

A kills B with probability X. C then shoots at A. C kills A with probability X. Otherwise C misses A with probability Y. A would then shoot at C. A kills C with probability X. Otherwise A misses C with probability Y. C then shoot at A… and so on.

Otherwise A misses B with Probability Y. What should B do? Well, B should do whatever A should do in this scenario (since B is now in the same position as A). I’m a little confused by this move, but I guess that we should treat B as doing what A does… that is, as shooting at the next person in line (so for B that would be to shoot at C). (This strikes me as fishy, but I’m not sure how else to avoid the lurking regress here, so I’m going for it.) So B shoots at C. B kills C with probability X. A would then shoot at B. A kills B with probability X. Otherwise A misses B with probability Y. B then shoots at A. B kills A with probability X. Otherwise B misses A with probability Y. A then shoots at B… and so on.

Otherwise B misses C with probability Y. What should C do? Same as above, I’ll assume that C should do what A does in this scenario, that is shoot at the next person in line (so for C that would be to shoot at A). So C shoots at A. C kills A with probability X. Otherwise C misses A with probability Y. A then shoots at B again… and now we repeat everything from the top.

On 1, the probability that A dies (given these assumptions) is…

Well, this has two loops. So call the probability that A dies in the first loop Z. Z is:

XX + XYYX + XYYYYX + XYYYYYYX + … +
YXYX + YXYYYX + YXYYYYYX + … +
YYX

The probability that A dies is then:

Z + (YYY)(Z) + (YYYYYY)(Z) + (YYYYYYYYY)(Z) + …

7. Justin

Part 3:

2. A shoots at C:

A kills C with probability X. B then shoots at A. B kills A with probability X. Otherwise B misses A with probability Y. Then A shoots at B. A kills B with probability A. Otherwise A misses B. Be then shoots at A… and so on.

Otherwise A misses C with probability Y. B then does what A did (i.e., shoot at the person right before her in line): B shoots at A. B kills A with probability X. Otherwise B misses A with probability Y. C then does what A did: C shoots at B. C kills B with probability X. A then shoots at C. A kills C with probability X. Otherwise A misses C with probability Y. C then shoots at A. C kills A with probability X. Otherwise C misses A with probability Y. A then shoots at C… and so on.

Otherwise C misses B with probability Y. A then shoots at C… and now we repeat everything from the top.
On 2, the probability that A dies (given these assumptions) is…

Well, this has two loops again. So call the probability that A dies in the first loop W. W is:

XX + XYYX + XYYYYX + XYYYYYYX + … +
YX + YYXYX + YYXYYYX + YYXYYYYYX + …

The probability that A dies is then:

W + (YYY)(W) + (YYYYYY)(W) + (YYYYYYYYY)(W) + …

3. A shoots at neither:

B then does what A does (shoots at neither). C then does what A does (shoots at neither). A then does what A does (shoots at neither)… and we now repeat everything.

At some point one of the participants dies of something. (Thirst if all they are doing is not shooting, or perhaps hunger, or some other natural cause, or….)

Now we have a rules issue!

When does it become the next person’s turn? Does the person have to indicate that they aren’t shooting in some way? (Waiving a hand, maybe?) If so, then the other two die waiting for a dead person to wave her hand. So the probability that A dies is 1. Call this Rule R.

Or does death count as giving up one’s turn? If so, then whoever is next in line when the first person dies will shoot at the other living person. And for all we know, it is equally probable that any given person will die first. So the probability that A dies would be 1/3. Call this Rule Q.

And I’m not sure how to decide the rules issue.

8. Justin

Part 4:

Are there any values of X and Y in which A should not shoot neither?

Well if we are using Rule R, then yes: For whatever values are assigned, A should shoot either B or C.

What if we use Rule Q?

Well, if X was 0, then P(1) < 1/3 and P(2) < 1/3; but, that devolves into A shooting neither – if they are all utterly inept shots, then it is the same as A shooting neither. What if X is 0.5? Then Z is roughly 0.54… so P(1) > 1/3
And W is roughly: 0.62… so P(2) > 1/3

What if X is 0.1?

Then Z is roughly 0.08… and P(1) would be roughly 0.28….
So shooting B would be a better option than shooting neither.

And W would be roughly 0.15… so shooting B would be the best option.

But, the point of all of this was to figure out what A should think that B would do given the assumptions being made and what A does.

But, we’ve assumed that B will do what A does in the hypothetical scenario. So what A thinks B will do will depend on what A thinks the rules are and the values… And A’s thinking about C should follow suit.

So if A shoots neither, then A should think that B will think that they are using Rule R. So if A shoots, A should expect B to shoot neither… and then for C to shoot neither.

So, knowing how good of a shot they each other, but assuming that none of the others know what A knows or that A knows what A knows, should A really shoot neither?

Well, that depends on the rules. If A shoots neither, then B will shoot neither, then C will shoot neither, and so on. And if it is Rule R they are using, then A certainly dies. So A should shoot C.

And if it is Rule Q they are using, then the probability that A dies (assuming that none of them lose their shooting skills while not shooting and waiting around for one of them to die)… then A dies with probability:

1/3 + 1/3 + (1/3)(2/3)(2/3) + (1/3)(2/3)(1/3)(2/3)(2/3) + (1/3)(2/3)(1/3)(2/3)(1/3)(2/3)(2/3) + …

Or, roughly, 83%.

That isn’t very good. And it involves a lot of waiting around for people to die… and all assumes that it is Rule Q being used anyway.

So, it seems to me, that A should think that shooting neither is a lousy option, if she thinks that B and C lack her knowledge and are rational.

So, if those assumptions are correct, then A should just shoot C.

9. Justin

Part 5:

But, maybe we shouldn’t have assumed that A will assume that B and C know nothing… rather, we might assume that A assumes that B and C know how good of a shot they themselves are, but that neither knows how good of a shot the other shooter is. Then A should assume that C at least will not shoot at neither even if A and B shoots at neither. But, then, who should A assume that C will shoot at in the different scenarios? Maybe that C will do so randomly? I’m not really sure. Will B being a 2/3 shooter change anything with respect to following A? I’m not really sure… let’s assume not.

If A shoots at B:

Kills B, then C kills A.
Doesn’t kill B, then B shoots at C; if B kills C, then A and B duke it out with A shooting first. If not, then C randomly kills A or B. If it is B, then A has a 1/3 shot to kill C.

If A shoots at C:

Kills C, then A and B duke it out with B shooting first.
Doesn’t kill C, then B shoots at A; if B doesn’t kill A, then C randomly kills A or B. If it is B, then A has a 1/3 shot to kill C.

If A shoots at neither:

Then B shoots at neither. Then C randomly kills A or B. If it is B, then A has a 1/3 shot to kill C.

So if A shoots at neither then A should expect to die with probability (1/2) + (1/2)(2/3) = 5/6

If A shoots at B, however, then A should expect to die with probability:

(1/3) + (2/3)(2/3)(2/3)(2/3) + (2/3)(2/3)(2/3)(1/3)(2/3)(2/3) + … +
(2/3)(1/3)(1/2) + (2/3)(1/3)(1/2)(2/3)

Which looks to be less than 5/6 (although I might be screwing this up).

If A shoots at C, however, then A should expect to die with probability:

(2/3) + (1/3)(2/3)(2/3) + (1/3)(1/3)(1/3)(2/3)(2/3) + … +
(1/3)(2/3) + (1/3)(1/3)(1/2) + (1/3)(1/3)(1/2)(2/3)

Which looks to be more than 5/6.

So A should shoot at B.

10. Justin

Part 6:

I’m sure I made all sorts of mistakes above… Nonetheless:

In conclusion, depending on how we fill in the assumptions that aren’t filled in in the problem, A should either shoot at B, shoot at C, or shoot at neither!

Further, we can conclude that Bryan is one sick puppy… and that I’m prone to wasting my time and mental energy on figuring out why I think that a problem seems like a trick!