Brain Teaser: A Duel

You find yourself in a duel with two other gunmen. you shoot with 33% accuracy, and the other two shoot with 100% and 50% accuracy, respectively. the rules of the duel are one shot per-person per-round. the shooting order is from worst shooter to best shooter, so you go first, the 50% guy goes second, and the 100% guy goes third. Where or who should you shoot at in round 1?

I wouldn’t shoot anyone the first round (you said where or who, kind of gave it away). 50% guy gets a shot at 100% guy. if 100% guy lives, then he kills 50% and you get a shot at him at 33% before he kills you. if 100% guy dies, then it’s your 33% vs. 50%.

Just curious, are we assuming that other players shoot randomly? If not, that will make a big difference…

You have to attack Mr 100…the payoffs just work out that way assuming the others pick randomly.

Yea…you’re right…not shooting is the best option.

I’ve seen this in a movie. You always shoot the chandelier causing it to fall and crush both. Bonus points if you can grab the chain and be pulled up in the air as the chandelier falls downward.


I think the solution is a bit more nuanced than that - if you consider that 50 and 100 can also choose to shoot no one.

If you map out the possible actions for each round, you get:

33’s Turn

  1. Shoot 100 = 0.33*30vs50 + 0.67*(50’s Turn)

  2. Shoot 50 = 0.33*(Die) + 0.67*(50’s Turn) ==> Clearly inferior to 1.

  3. Shoot No one = (50’s Turn)

50’s Turn

  1. Shoot 100 = 0.5*50vs30 + 0.5*(100’s Turn)

  2. Shoot 33 = 0.5*(Die) + 0.5*(100’s Turn) ==> Clearly inferior to 1.

  3. Shoot No one = (100’s Turn)

100’s Turn

  1. Shoot 50 = 0.33*(Die) + 0.67*(Win)

  2. Shoot 33 = 0.5*(Die) + 0.5*(Win) ==> Clearly inferior to 1.

  3. Shoot No one = (33’s Turn)


So, 33’s action depends on 50’s action, which depends on 100’s action, which depends on 33’s action, etc. I am unable to determine the end round based on this information.


This is clearly the only winning approach.


I would argue that the 50% guy would never not shoot at the 100% guy. If there is a chance that the 100% guy will shoot, it will be to shoot the 50% guy. Hence, the 50% guy would shoot to pre-empt it since the alternative is certain death.

Further, it doesn’t make sense to me why the 100% guy wouldn’t shoot at the 50% guy if both the 33% and 50% guys pass. Is he hoping that somehow the 33% guy will shoot the 50% guy, leaving the 100% guy the chance to kill the 33% without any comeback? That doesn’t make sense since shooting the 50% guy is a dominated strategy. Hence, the 100% guy will kill the 50% guy no matter what, if given the chance.

How do you express what you said mathematically? In this sort of question, the correct answer is not always the most intuitive answer.

Plus, you are disregarding the possibility that all three duelists shoot no one. For instance, for 100, shooting 50 dominates shooting 33. However, does it dominate shooting no one? If so, prove it.

If all three duelists shoot no one in round one, then they won’t shoot anyone in round n to infinity. I don’t find that to be a satisfactory answer. The construct of the game implies certain preferences. The players prefer to win the duel to losing the duel (they don’t particularly care if they are the first to lose or second to lose). The infinite loop of no one firing cannot go forever (eventually they must get water or something).

What I was doing was basically solve for the sub-game perfect Nash equilibruim (I’m guessing not many on the board have taken a course in Game Theory?). Basically what you have to do is go to the end and solve for what the person would do in that scenario and then work backwards. Given that there must be a winner, it is not credible for the 33% or 50% person to think that the 100% person would not shoot if they both pass. Assuming all the players are rational, the 100% person’s best chance of winning comes from shooting the 50% person. Hence, it is not credible for him to pass.

Given that if both pass the 100% person would shoot, then it follows that the 50% person would shoot at the 100% and the 33% person should pass. QED

You are on a horse, galloping at a constant speed. On your right side is a sharp drop-off.

On your left side is an elephant traveling at the same speed as you. Directly in front of you is a galloping kangaroo and your horse is unable to overtake it.

Behind you is a lion running at the same speed as you and the kangaroo. What must you do to get out of this highly dangerous situation?

If you can’t express it mathematically, your explanation fails. Stuff like “it’s just not credible” is not in the spirit of the question. The question is not about real life, where you will have to “get water”, go to the bathroom, get bored, etc. You must answer the question given the stated conditions.

Gonna have to side with jhm on this one.

There are a variety of solutions here, we’re assuming the lion is hungry and wants to eat something. Assuming the elephant is fully grown, I doubt you’d be able to jump on it, then riding it would be difficult at best. Despite what I’ve seen on cartoons, I also think jumping on the kangaroo would not be possible either. Assuming then that we’re now boxed in between the elephant, kangaroo and cliff, with the lion behind (and catching up), I would have to recommend pulling the horse down and feeding it to the lion and escaping while it rips it to shreds.

Well, the true answer is to keep running. A lion cannot keep up with a horse or kangaroo over long distances. So, the lion will eventually get tired and slow down. An elephant also cannot keep up with a horse, but presumably, we only care about the lion.

No one cares about long distances, over mid ranges lions average 5-10 mph faster than either the kangaroo or the horse. If you’re boxed in and unable to maneuver he’ll catch you for sure. Not to mention that over long distances horses aren’t nearly as fast as you think. Ultra runners have been proven capable of beating any horse over long distances (50-100 miles)

Ohai - Not all proofs need to be strictly mathematical. It is actually more clear if you draw a tree of all possible outcomes (as much of game theory is when presented like this). You’re particularly concerned about one branch of the tree (ground-ground-ground). I was showing that assuming all players are rational, then the subgame perfect Nash equilibrium rules out that tree.

Let’s say we are down to two players, for whatever reason. I’m sure you would agree that the person with the higher accuracy would want to shoot in every case. Hence, the lower probability person would shoot first.

Your problem seems to be with extending this argument to three people. It can help to think in terms of a mixed strategy. In a mixed strategy, you don’t assume that someone does something with certainty, but it’s like I shoot 33% x% of the time, 50% y% of the time, and the ground 1-x%-y% of the time.

The 100% probability person could then be under the assumption that what they have actually observed was a mixed strategy that resulted in him not getting hit. However, he has no reason to believe that they will continue to hold off in the next period. Hence, the 100% guy shooting the ground would actually open himself up to getting shot at with some probability in the next round, so he shoots the 50% guy in this round no matter what. However, all these mixed strategies then disappear as a result of him choosing to shoot the 50% guy where he doesn’t know what the other people’s probability assumptions are. Ie. once they know he will shoot the 50% guy under any circumstances if they both shoot the ground with any positive probability, you can use backward induction to solve the rest of it without mixed strategies.

You may disagree with backward induction as a technique, but what it implies makes sense.

CFAvsMBA - I actually like ohai’s reply here. Black Swan’s strategy is a bit risky, you may not be able to be fast enough to get away from the lion (compared to the horse, if it pops back up). You’d have to be an experienced rider.

To think a bit laterally, a kangaroo and a lion have different habitats (Australia vs. Africa). The only time they’ll be in the same place is a zoo or some kind of wildlife preserve. You wouldn’t be on horseback in a zoo, so you’re probably in a wildlife preserve. That means you’re probably a game warden, or the rich guy who owns it. I presume you have a tranquilizer gun or normal gun in either of those situations. Shoot the kangaroo.

Feel like the question is flawed – there is not enough information to reach a universal conclusion without making up scenarios that could be plausible if given more information.

The problem gives no clues about stamina or duration, so any solution based on that is mere conjecture (however logical sounding).

To keep running and hope one of the other animals tires out seems like the obvious conclusion, but we don’t have any information about that or other future variables either (does the lion speed up, does the elephant veer right, does the terrain change thereby changing the rates, whatever).

It’s equally likely as any of the above (logical) scenarios that the dude just busts out some Indiana Jones action and kills the lion, or that he jumps over the side of the cliff because he’s actually wearing a custom suit that allows him to “grind the crack” like Jeb Corliss with the horse in tow.

The question sucks.

Haha good one CFAvsMBA. Just googled it. Everyone’s serious discussion about the question makes it even more hilarious.