I need help understanding the solution to the following puzzle:

Puzzle: There is a "triangular" duel between the three shooters. Everyone shoots one by one, can shoot once and any person he wants. Smith always hits the target (100%) and shoots last. Brown hits the target 80% of the time and shoots second. Jones is the worst, hitting the target only 50% of the time and he shoots first. Who should Jones shoot in order to increase his chance of surviving?

Answer: Jones should shoot in the air. The worst shooter has the best chance of surviving in the triangular duel, which is Jones. After him goes Smith who never misses the target. Since Smith and Brown will be shooting each other when their turn comes, Jones must shoot no one until one of his enemies dies.

After that he shoots at his enemy, again having an advantage. First, it's easier to calculate the probability of surviving for Smith. In the duel with Brown he shoots first with a probability of 1/2. In this case he kills Brown. Brown also shoots first with a probability of 1/2.

Smith survives with a probability of 1/5. Thus, Smith will live over Brown with a probability of 1/2 + 1/2 * 1/5 = 3/5. With a probability of 1/2, Smith survives the duel with Jones. All in all: the probability to survive for Smith is 3/5 * 1/2 = 3/10.

The probability for Brown to survive the duel with Smith is 2/5.

Then Jones shoots Brown. In the first round, Brown has a probability of 1/2 * 4/5 = 4/10 to win. In the second round, he has a probability of 1/2 * 1/5 * 1/2 * 4/5 = 4/100.

Thus, Brown has a chance to survive Jones: 4/10 + 4/100 + 4/1000 + 4/10000 = 0.4444(4) = 4/9.

The probability of Brown to survive both of his opponents is equal to 2/5 (over Smith) * 4/9 (over Jones) = 8/45. The probability of Jones to survive = 1 - 3/10 - 8/45 = 47/90.

In bold are the calculations I'm confused about and don't know how to arrive at. Can someone please explain it? Thank you.

UPDATE: They shoot one by one in the same order until one person survives.

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    $\begingroup$ There are definitely problems with the proposed answer. First and foremost, it doesn't explain why this strategy is the best, it just calculates a probability of one strategy. Secondly, the wording is confusing, if not incorrect. For example, the author says "Smith survives with a probability of 1/5.", but then goes on to say "the probability to survive for Smith is 3/5 * 1/2 = 3/10." $\endgroup$ Commented Aug 25, 2014 at 20:12
  • $\begingroup$ Just googled it and the calculations appear to be for a random draw of who shoots first: math.cornell.edu/~mec/Summer2009/Leung/puzzles_p3.htm $\endgroup$ Commented Aug 25, 2014 at 21:08
  • $\begingroup$ If Smith fires first at Brown, the chance of surviving is incalculable. if he kills Brown (50% chance), then Jones gets him with 100%, if he misses Brown, then Brown misses, who does Jones shoot? Jones could just fire into the air if he were so inclined. Does he stop to flip a coin? For Brown, he would try to shoot Jones, assuming he knows Jones is 100% accurate. But shooting and missing Jones might encourage Jones to target him, so should he try to miss Smith and hope Jones... Too much human influences, even with perfect probability calculation. $\endgroup$
    – Ken Y-N
    Commented Aug 26, 2014 at 4:17
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    $\begingroup$ @KenY-N I think you have to make certain assumptions for the sake of the puzzle. For example, I don't expect that emotion is involved (make them robots if necessary); I'm sure the others are trying to maximize the probability of their own survival. $\endgroup$ Commented Aug 26, 2014 at 12:51
  • $\begingroup$ Others raise an interesting point: the question seems to indicate (but isn't totally clear) that each person gets exactly one shot, but the answer assumes they repeat in the same order until one is left standing. I assume this is the second, but could you clarify this in the question? $\endgroup$ Commented Aug 26, 2014 at 12:55

7 Answers 7


The answer does not appear to apply to the question as stated. In the question it says that Jones shoots first, Brown next, and Smith last. The calculation of the chance that it is Smith that Jones faces assumes it is random which of Smith or Brown shoots first. If it is Smith (probability $\frac 12$) he survives. If it is Brown (probability $\frac 12$) Smith still lives with probability $\frac 15$ So Smith survives Brown $\frac 35$ of the time and Brown survives Smith $\frac 25$ of the time. You agree that this gives Smith a $\frac 3{10}$ overall chance of survival. To calculate Brown's chances in the duel with Jones (assuming he has already beat Smith-a $\frac 25$ chance), he can win if Jones misses and he hits, which is where the $\frac 12 \cdot \frac 45=\frac 4{10}$ comes from. He can also win if Jones misses, he misses, Jones misses, and he hits, which is the $\frac 12 \cdot \frac 15 \cdot \frac 12 \cdot \frac45=\frac 4{100}$ Continuing, we have an infinite series to sum, which results in Brown winning $\frac 49$ of the time.

  • $\begingroup$ People are only allowed one shot. $\endgroup$
    – Ken Y-N
    Commented Aug 26, 2014 at 4:18
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    $\begingroup$ @KenY-N: One shot in a given turn. The turn to shoot passes around until two are dead. $\endgroup$ Commented Aug 26, 2014 at 13:02

I don't know if this is deliberately different to an old chestnut which appeared in one of Martin Gardner's books many years ago - it appears almost verbatim, even down to the names, except that:

  • the participants keep shooting until only one remains
  • the initial order of shooting is chosen randomly

The solution you quote is straight from Gardner, and doesn't relate to the question as posed here. Smith and Brown clearly need to target their most dangerous enemy first (each other), giving Jones the option to deliberately miss until Smith v Brown is resolved and then take first crack at the survivor, guaranteeing him 50% survival chance even in the worst case when he faces sure-shot Smith.

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    $\begingroup$ I should add the assumption that noone is accidentally killed by a bullet not intended for them! $\endgroup$ Commented Aug 26, 2014 at 14:51
  • $\begingroup$ It is one of the Gardner's puzzles. I've updated the question to state that they shoot until one person survives. $\endgroup$
    – makaed
    Commented Aug 28, 2014 at 18:46

I found the explanation of the puzzle by its original author. The answer is in fact for Jones to shoot into the air until one of the opponents is dead.

The order is kept as per the question, but working with probabilities we need to consider all possible outcomes. Thus, we consider the case where Brown shoots Jones first instead of logically (as each will try to take away the strongest person) shooting Smith.

Smith: We start calculating the probability of surviving for Smith first as it is the easiest one to do. There are two ways Smith can win Brown in their duel: 1) Smith shoots first and kills Brown (since he is 100% accurate), giving him a chance of surviving of 1/2; 2) Brown shoots first and misses, then Smith kills him; this case gives Smith a chance of surviving of 1/2*1/5 (since Brown is 4/5 accurate). As a result, Smith can survive Brown with a probability of: 1/2 (case 1) + 1/2*1/5 (case 2) = 3/5. Next Smith faces Jones. Since Jones is only 50% accurate, Smith has a survival chance of 1/2 against Jones. Smith's overall chance of surviving in this triangular duel is 3/5 (case against Brown) * 1/2 (case against Jones) = 3/10.

Brown: Next comes Brown. He has a chance of surviving against Smith of 2/5 (simply deducting Smith's survival probability of 3/5 from total probability of 1). Here it gets complicated as we run into infinite series of possibilities with Brown against Jones. There is a chance of 1/2 that Jones will miss (he is 50% accurate), and Brown has a chance of 4/5 to kill Jones. At this point, Brown has a chance of surviving of 1/2*4/5 = 4/10. But! He can also miss with a probability of 1/5 (which is 1 - his accuracy of 4/5). That can go on forever, and will result in Brown's total survival chance against Jones yielding 4/10 + 4/100 + 4/1,000 + 4/10,000 + ... This can be written as 0.444444... which is the decimal expansion of 4/9. Brown's overall chance of surviving in this triangular duel is 2/5 (case against Smith) * 4/9 (case against Jones) = 8/45.

Jones: This leaves Jones with a probability of surviving of 1 (total probability) - 3/10 (Smith's chance) - 8/45 (Brown's chance) = 47/90.

Therefore, Jones has 52.22% of surviving, Smith - 30%, Brown - 17.78%; and it's true that it's better for Jones to fire into the air in order to increase his chance of surviving.

  • $\begingroup$ This cannot be true for the Problem as stated: Brown always shoots before smith and tries to kill him. He hits with 80% chance, so Smith is dead with 80% chance. So his survival rate cannot be higher than 0.2 - it can never be 3/10 $\endgroup$
    – Falco
    Commented Feb 15, 2016 at 10:57

I find all the answers here way too complicated except Julia's

Using LOGIC;

1) Smith will ALWAYS shoots Brown first, because whoever he leaves alive will automatically shoot him. He has a better chance of survival if Jones is the one alive
2) If Jones kills Brown, he is automatically dead.
3) If Brown kills Jones, he is automatically dead.

This stipulates that the other two will always fire on one another. Jones is safe as long as they are alive and has dramatically less chances of survival the moment one of them shoots at him FIRST (aka, if he kills one of them). There is literally no need for any percentages other than the initial ones already stated in the puzzle. Depending on the rules (if you are allowed to intentionally miss), either he misses intentionally, or he shoots Smith and hopes he misses.


-While Smith is alive, Jones and Brown will never shoot each other, as if they succeed, they're instantly killed after.

-Knowing this, Smith also knows he will never have the first shot when it eventually comes to a 1v1, he can discard that. His choice now comes to who he wants to take out first (as a perfect killer he is) and who he wants to make that first 1v1 shot. Obviously optimal strategy for Smith is killing Brown as fast as he can and then Jones.

-If Brown realizes this, he will use his first and maybe only shot to try to kill Smith as otherwise he is dead for sure.

These leaves us with only 4 possible scenarios, regardless the chances of each one coming to happen:

A) Jones kills Smith, then Brown shoots Jones. Long term survival chances for Jones: less than 20%.

B) Jones kills Brown, then Smith shoots Jones. Long (and short) term survival chances for Jones: 0%.

C) Jones fails first shot, then Brown kills Smith, then Jones shoots Brown. Long term survival chances for Jones: Higher than 50%.

D) Jones fails first shot, Brown too, Smith kills Brown, and Jones has one chance to kill Smith. Long term survival chances for Jones: exactly 50%.

-As you can see, it's highly interesting for Jones to fail the first shot, so if he can do that intentionally, it would be highly recommended.


The right answer to THIS question (instead of random order shooting)

I will try to consolidate some of the right answers and not only give the strategy for each participant, but also their chances of survival. - The question as stated is different from the similar puzzle by Martin Gardner, in our version they don't shoot in random order, but Jones always shoots first, then Brown and then Smith.


If someone has only one enemy left to fire at, the best option is always to shoot, because not shooting lowers your chance of survival. (Gruesome Wild West) If there are 2 possible targets, you can decide to shoot one of them or miss on purpose. Missing on purpose has the same effect as shooting and missing in this scenario, because no one holds grudges. So the only relevant part to decide if you should shoot someone is the scenario where you hit him (since you hit everyone with the same percentage).


Killing Brown will give him 50% chance to survive against Jones, killing Jones will give him 20% chance against Jones. Killing no one will only help him, if Brown or Jones shoot at each other.


Killing Smith will give him below 50% chance to survive. Killing Jones result in 0% chance to survive. Shooting no one only helps if Smith or Jones shoot at each other.


Killing Smith will get him below 20% chance and killing Brown 0% chance to survive. Shooting no one only helps if Smith or Brown shoot at each other.

Jones and Brown have no incentive to kill each other. So Smith will kill Brown and therefore Brown will try to kill Smith. Since they try to kill each other, Jones has a Chance of 50% or higher, if one of the two kills the other, since Jones gets the next shot after that. This is higher than the chance after killing one himself first, so he will not shoot as long as the other two are still standing.

Surviving Probabilities:

Smith: 10%

He has only three scenarios: a) he is killed by Brown 0.8, b) he kills Brown and is killed by Jones 0.2 * 0.5 or c) he kills both of them 0.2 * 0.5.

He only survives if Brown and Jones each miss him one time so his chance to survive is 0.1

Brown 35,56%

Brown really only has one chance to survive: He has to kill Smith with his first shot and then win the duel against Jones (with John getting the first shot). His chance to win in a round against Jones is: Jones misses him and he hits: 0.5 * 0.8 = 0.4 The chance of Jones winning a round is 0.5 the chance for the round actually ending is 0.9 so the chance of Brown winning the duel overall is 0.4 / 0.9 = 4/9 So his overall chance of survival is 0.8 * 4/9 = 16/45

Jones 54,44%

Jones has two scenarios to live: Brown kills Smith and Jones wins the duel against Brown: 0.8 * 5/9 = 4/9 or Brown misses Smith, Smith kills Brown and Jones kills Smith on the first try: 0.2 * 0.5 = 0.1 Overall: 4/9 + 0.1 = 49/90


Actually: ALL SURVIVE. No-one shoots.

Jones fires into the air, in the hope one of Brown and Smith kill the other, at which point Jones would fire at the survivor.

After Jones fires into the air, Brown knows that if he kills Smith, Jones will kill him half the time. (And of course if he kills Jones, Smith will in turn kill Brown.) Further, Brown can forsee Smith's best strategy will likewise involve not shooting. So Brown's best strategy is to fire into the air in turn.

Finally, Smith can kill either guy but if he does he stands a good chance of dying himself. His best strategy is to return initiative to Jones.

And Jones' calculus hasn't changed, so second verse same as first.

Or more briefly, they all know that if they kill one of their two opponents, they have a 50-100% chance of dying next. If they kill neither opponent, they know their opponents will face the same decision in turn and not shoot. so you're best not shooting.

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    $\begingroup$ This answer highlights that at some point mathematical rigor breaks down in all of the answers given so far - they all end up making some kind of assumption about what is the optimal move. I'd love to see a rigorous game theory proof of this problem. $\endgroup$
    – mbeckish
    Commented Aug 3, 2017 at 17:09
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    $\begingroup$ OK, the real question is, what are these three guys' utility function? What makes them happy? What is their goal? The other answers to this problem assume that the gunfighters don't mind dying, and are only want to see the other two dead. My answer here from last year assumes that goal one is to live, and goal two is to kill some people if you can do it while living. $\endgroup$ Commented Aug 4, 2017 at 5:44

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