### Brain Teasers

# The Truel

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. All three competitors know one another's shooting odds. If you are Mr. Black, where should you shoot first for the highest chance of survival?

### Hint

Think from the points of view of Mr. Gray and Mr. White, not just Mr. Black.### Answer

He should shoot at the ground.If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before.

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## Comments

Ok that brain teaser is very confusing! TO many names... but it's not a bad one

The teasers are getting violent.

EZ but fun

I didn't get it right... but I enjoyed the answer!

Good one!!

Considering that this is an old classic from the literature, I'm surprised to see this here. Don't the editors know about copyright violation?

not only is it an old story, but I also think it's highly questionable.

The point is, Mr Black is hoping that Mr White is taken out of the picture. By shooting the ground, he hopes that Mr Gray will shoot at Mr White, with a 67% chance of hitting him. If Gray does kill White, then black has a 33% chance of killing Gray, followed by Gray having a 67% chance of killing Black, this goes on until one dies.

If Black shoots the ground, and Grey does not kill White, White then kills Grey, and Black has a 33% chance of killing White, if Black misses, White kills him 100% of the time.

If I'm Mr Black, I take my shot at Mr White. 33% of the time I kill him, and now it's me against Mr Gray and I have a chance. 67% of the time it reverts back to the previous scenario of Grey vs White, and nothing changes.

Give me that added chance.

The point is, Mr Black is hoping that Mr White is taken out of the picture. By shooting the ground, he hopes that Mr Gray will shoot at Mr White, with a 67% chance of hitting him. If Gray does kill White, then black has a 33% chance of killing Gray, followed by Gray having a 67% chance of killing Black, this goes on until one dies.

If Black shoots the ground, and Grey does not kill White, White then kills Grey, and Black has a 33% chance of killing White, if Black misses, White kills him 100% of the time.

If I'm Mr Black, I take my shot at Mr White. 33% of the time I kill him, and now it's me against Mr Gray and I have a chance. 67% of the time it reverts back to the previous scenario of Grey vs White, and nothing changes.

Give me that added chance.

Here's the complete solution, vikingboy.

If Mr. Black shoots the ground, there is a 1/3 chance that he will be left with Mr. White and a 2/3 chance he will be left with Mr. Gray.

If he is left with Mr. White, he has a 1/9 chance of survival. If he is left with Mr. Gray, his probability of survival is 1/3((2/9)^0+(2/9)^1+(2/9)^2+(2/9)^3...) which is an infinite series that approaches to 3/7 (I think). Therefore, If Mr. Black shoots the ground, his total chance of survival is 1/9+3/7*2/3=1/9+2/7=25/63

Now, if Mr. Black shoots Mr. Gray, there is a 1/3 chance that he will be left with Mr. White and die and a 2/3 chance that he will miss and that we will go into the same situation as if Mr. Black just shot the ground. Therefore, his chance of survival is 25/63*2/3=50/189

If Mr. Black shoots Mr. White, there is a 2/3 chance that he will miss and he will have a 25/63 chance of survival. There is also a 1/3 chance that he will kill Mr. White and we will go into the situation with just Mr. Gray and Mr. Black, except this time Mr. Gray shoots first. Therefore, Mr. Black has a 1/3*2/7 chance of survival. So his total chance of survival is 2/21+50/189=68/189

When comparing the three fractions, the chances are 75/189, 50/189, and 68/189. Therefore, Mr. Black should shoot the ground.

If Mr. Black shoots the ground, there is a 1/3 chance that he will be left with Mr. White and a 2/3 chance he will be left with Mr. Gray.

If he is left with Mr. White, he has a 1/9 chance of survival. If he is left with Mr. Gray, his probability of survival is 1/3((2/9)^0+(2/9)^1+(2/9)^2+(2/9)^3...) which is an infinite series that approaches to 3/7 (I think). Therefore, If Mr. Black shoots the ground, his total chance of survival is 1/9+3/7*2/3=1/9+2/7=25/63

Now, if Mr. Black shoots Mr. Gray, there is a 1/3 chance that he will be left with Mr. White and die and a 2/3 chance that he will miss and that we will go into the same situation as if Mr. Black just shot the ground. Therefore, his chance of survival is 25/63*2/3=50/189

If Mr. Black shoots Mr. White, there is a 2/3 chance that he will miss and he will have a 25/63 chance of survival. There is also a 1/3 chance that he will kill Mr. White and we will go into the situation with just Mr. Gray and Mr. Black, except this time Mr. Gray shoots first. Therefore, Mr. Black has a 1/3*2/7 chance of survival. So his total chance of survival is 2/21+50/189=68/189

When comparing the three fractions, the chances are 75/189, 50/189, and 68/189. Therefore, Mr. Black should shoot the ground.

Okay, I know that was kinda complicated, but think about it this way.

If Mr. Black is left with Mr. Gray and Mr. Black gets to shoot first, there is an unkown probability of survival, let us say x.

If Mr. Black shoots the ground, his chance of survival is 1/9+2/3 x.

If Mr. Black shoots Mr. Gray, his chance of survival is 2/3*(1/9+2/3 x), so he should not shoot at Mr. Gray.

If Mr. Black shoots Mr. White, his chance of survival is 2/3*(1/9+2/3 x)+1/3*1/3*x.

Now we must prove that 1/3*(1/9+2/3 x) is more than 1/3*1/3*x. First we can simplify to try to prove 1/9+2/3 x > 1/3*1/3*x. We can deduce that x is slightly larger than 1/3, and we know that 1/9+2/3*1/3 > 1/3*1/3*1/3, so we know this will always be true for any value of x greater than 1/3.

If Mr. Black is left with Mr. Gray and Mr. Black gets to shoot first, there is an unkown probability of survival, let us say x.

If Mr. Black shoots the ground, his chance of survival is 1/9+2/3 x.

If Mr. Black shoots Mr. Gray, his chance of survival is 2/3*(1/9+2/3 x), so he should not shoot at Mr. Gray.

If Mr. Black shoots Mr. White, his chance of survival is 2/3*(1/9+2/3 x)+1/3*1/3*x.

Now we must prove that 1/3*(1/9+2/3 x) is more than 1/3*1/3*x. First we can simplify to try to prove 1/9+2/3 x > 1/3*1/3*x. We can deduce that x is slightly larger than 1/3, and we know that 1/9+2/3*1/3 > 1/3*1/3*1/3, so we know this will always be true for any value of x greater than 1/3.

Slightly easier (I think) way to look at it.

You are Mr. Black. Do you want to aim at Mr. Gray? Let's say you hit him. Then you instantly die from Mr. White. Survival rate from hitting Mr. Gray = 0

Do you want, then, to aim at Mr. White? Let's say you hit him. Then it's Mr. Gray vs Mr. Black with Mr. Gray getting the first shot. Since there's a 2/3 probability in this case of Gray killing Black with the first shot, Black's chance of surviving such a duel is less than 1/3.

Let's say you miss. Then, Mr. Gray's turn to aim. If he aims at Mr. Black, and hits him, he instantly dies. Thus, aiming at Mr. Black is stupid.

Let's say Mr. Gray aims at Mr. White. If he hits Mr. White, he then is in a 2 person duel with Mr. Black.

Let's say he misses. Then Mr. White has his choice of killing Black or Gray. Either way, his opponent gets one chance to take him out. Thus, Mr. White would kill Gray. Gray, knowing that, will not shoot at the ground, knowing that White would instantly kill him.

Thus, Gray will shoot at White.

When it gets back around to Mr. Blacks turn, only one opponent will be left. Either way, Mr. Black gets AT LEAST a 1/3 chance of shooting that person, as opposed to the LESS THAN 1/3 chance that he would have if he were in a 2 person duel with Gray and Gray shooting first.

(Mr. Black has exactly 1/3 chance of winning a duel between him and White if Grey had killed White. He has a slightly better chance if he's facing Gray, since Gray's aim isn't 100% and Black gets a second turn.)

Interesting side note: logic stays the same if Gray's aim goes up to 100% and Blacks goes infinitely close to 0.

You are Mr. Black. Do you want to aim at Mr. Gray? Let's say you hit him. Then you instantly die from Mr. White. Survival rate from hitting Mr. Gray = 0

Do you want, then, to aim at Mr. White? Let's say you hit him. Then it's Mr. Gray vs Mr. Black with Mr. Gray getting the first shot. Since there's a 2/3 probability in this case of Gray killing Black with the first shot, Black's chance of surviving such a duel is less than 1/3.

Let's say you miss. Then, Mr. Gray's turn to aim. If he aims at Mr. Black, and hits him, he instantly dies. Thus, aiming at Mr. Black is stupid.

Let's say Mr. Gray aims at Mr. White. If he hits Mr. White, he then is in a 2 person duel with Mr. Black.

Let's say he misses. Then Mr. White has his choice of killing Black or Gray. Either way, his opponent gets one chance to take him out. Thus, Mr. White would kill Gray. Gray, knowing that, will not shoot at the ground, knowing that White would instantly kill him.

Thus, Gray will shoot at White.

When it gets back around to Mr. Blacks turn, only one opponent will be left. Either way, Mr. Black gets AT LEAST a 1/3 chance of shooting that person, as opposed to the LESS THAN 1/3 chance that he would have if he were in a 2 person duel with Gray and Gray shooting first.

(Mr. Black has exactly 1/3 chance of winning a duel between him and White if Grey had killed White. He has a slightly better chance if he's facing Gray, since Gray's aim isn't 100% and Black gets a second turn.)

Interesting side note: logic stays the same if Gray's aim goes up to 100% and Blacks goes infinitely close to 0.

foxxie

I guess I liked it it was okay...

alright, don't get your thongs all twisted...I understand the odds just fine, what I overlooked was that if Black does kill White, Gray gets the first shot at Black. If he shoots the ground AND the others do as expected Black gets to shoot first. Keep in mind that you are assuming that the other two have the same information you do, and that they act logically.

Still say that this one has been around forever.

Still say that this one has been around forever.

What if Mr. Gray or White decide to shoot and Mr. Black. Then Mr. White would have the better chance of surviving because Mr. Black would shoot at the ground and Mr. Gray would shoot him. leaving Mr. White to shoot Mr. Gray. Too many "if's"

But since Mr. White is better than Mr. Black, Mr. Gray would rather shoot at him for his highest chance of survival, Master Yoda.

It was confusing! What is a truel anyway?

vikingboy, lay off being too smart. You are making me look pretty stupid, I can't figure out these confusing riddles. It isn't my fault that I am not as smart as ya'll!

vikingboy, lay off being too smart. You are making me look pretty stupid, I can't figure out these confusing riddles. It isn't my fault that I am not as smart as ya'll!

I just use plain on common sense on this. I so! proud of MYSELF!!!!!!!!!!!!!!!!!!!!!!!

Cracker, a truel is supposed to be like a duel except there are three people fighting in it. However, it is not actually a real word.

SOOOOOOOOOOOOO STUUUUUPPPPIIIIIIIIDDDDDDDDDDDD.

SO DUMB

SO DUMB

I guessed the air. Ground, air, its all the same to me.

my head hurts

Someone should shoot Mr. Black for getting in a fight with two guys who are such better shots than him .. classic or not, as long as it doesn't duplicate another post .. what's the harm .. I enjoyed it.

i didnt kno ground was a option

I feel ultra stupid. I have a head ache... Way too many names... I think Mr. Black should just shoot everybody three times and not follow the rules. Y'know... just a suggestion...

Good one...made you think

my brain is pulsing toooooo much informatin to proses

You think three people has a strange outcome? I tried to figure it out for four (Mr. Red with 1/4, Mr. Yellow with 2/4, Mr. Green with 3/4, and Mr. Blue with 4/4), and there's a very strange result. Mr. Yellow has the best odds, followed by Mr. Green, Mr. Red, and Mr. Blue, in that order, if everyone follows ideal strategy. Mr. Red fires at Mr. Blue. If he misses, Mr. Yellow fires - but not at Mr. Blue or even at Mr. Green, but at Mr. Red! If he misses, Mr. Green fires at Mr. Blue. If he misses, Mr. Blue bumps off Mr. Green. In any case, once it's down to three people, it's the same strategy as the truel shown here (even if Mr. Blue was bumped off, although it was significantly harder to figure out).

That seems like a pretty interesting problem, Poker. I'll try to figure it out some day.

Note, however, that in this problem, we aren't trying to figure out who has the highest probability of staying alive, but rather where Mr. Black should shoot to have the highest probability of staying alive.

In my problem, Mr. Black has a 75/189 chance of staying alive, Mr. Gray has a 8/21 chance of staying alive, and Mr. White has a 2/9 chance of staying alive. Therefore, Mr. Black has the highest chance of survival.

Additionally, in your problem with 4 people, I think it would be better for Mr. Red to shoot at the ground too, as this would guarantee a 1 on 1 match with Mr. Red against someone else, with Mr. Red having the first shot.

Note, however, that in this problem, we aren't trying to figure out who has the highest probability of staying alive, but rather where Mr. Black should shoot to have the highest probability of staying alive.

In my problem, Mr. Black has a 75/189 chance of staying alive, Mr. Gray has a 8/21 chance of staying alive, and Mr. White has a 2/9 chance of staying alive. Therefore, Mr. Black has the highest chance of survival.

Additionally, in your problem with 4 people, I think it would be better for Mr. Red to shoot at the ground too, as this would guarantee a 1 on 1 match with Mr. Red against someone else, with Mr. Red having the first shot.

Not quite, since then it would be down to three people, not two. Red actually has better odds if he kills Blue than if he doesn't kill anyone. I've got to warn you, though, it's tricky to figure out (especially for Red, Yellow, and Green left, since they all sometimes miss - at least, that's what I thought when I first came to that point). And some of those fractions can get pretty large. For example, at the start, Blue wins 21/256 of the time, Yellow wins 11727/28840 of the time, Green wins 40257/149968 of the time, and Red wins 2914283/11997440 of the time! If this is what it's like with four, with five the fractions must be huge!

it's good and complicated, but how come so many of you people are trying to explain it when it's already a good explanation in the answer?

What I wanna know, it how do they know how often they hit or miss cause this sounds like something you can only do once.

i wanna shoot the person who rote that

Wow! How do you think of things like that? It all makes my head spin!

At first, I said that Mr. Black should shoot Mr.White because he never misses his target......then I said hit the ground.

I personally think that this teaser is awesome. It makes you think (sadly, I did not solve it correctly because I overlooked the possibility of shooting the ground).

-An Interesting Note-

I've discovered (with the necessary aid of a graphing calculator) that the limit of x^0+x^1+x^2+x^3+x^4+x^5+x^6... where 0â‰¤xâ‰¤1 is (-1)/(x-1) .

-Comments on the Above-

The boundaries for x (0 and 1) are inclusive in this case, but inserting the boundaries into the equation wields 1 and 1/0=âˆž*, respectively. The solution for x=0 exists because 0^0+0^1+0^2... =0^0=0/0, which can equal any real number (I also discovered this, but I found out that it had already been found out) (say a=0 and b is real, therefore ab=0, b=(0/a)=(0/0), thus 0/0 can equal any real number), equals, in this case, 1. As for the second boundary, you would expect 1^0+1^1+1^2...=1+1+1... to equal âˆž. If you would rather not go into the abstract realm of division by zero, just change both inequality signs to

-An Interesting Note-

I've discovered (with the necessary aid of a graphing calculator) that the limit of x^0+x^1+x^2+x^3+x^4+x^5+x^6... where 0â‰¤xâ‰¤1 is (-1)/(x-1) .

-Comments on the Above-

The boundaries for x (0 and 1) are inclusive in this case, but inserting the boundaries into the equation wields 1 and 1/0=âˆž*, respectively. The solution for x=0 exists because 0^0+0^1+0^2... =0^0=0/0, which can equal any real number (I also discovered this, but I found out that it had already been found out) (say a=0 and b is real, therefore ab=0, b=(0/a)=(0/0), thus 0/0 can equal any real number), equals, in this case, 1. As for the second boundary, you would expect 1^0+1^1+1^2...=1+1+1... to equal âˆž. If you would rather not go into the abstract realm of division by zero, just change both inequality signs to

(cont. from above)

less than signs.

* I say this tentatively, because I think that it is generally accepted. I have not experimented enough with the concept of infinity to feel certain that it is equal to 1/0. It's an interesting topic, though.

* I say this tentatively, because I think that it is generally accepted. I have not experimented enough with the concept of infinity to feel certain that it is equal to 1/0. It's an interesting topic, though.

I have to agree that Black should aim at White with his first shot. Unless he aims at and kills Gray with his first shot, he will get to shoot again. (unless either of the others is stupid enough to shoot at Black while their other opponent is still breathing) In any case, brainz, I disagree with your post of May 5 where you state that Black has a 1/9 chance of survival vs. White. His chance is 1/3. If he hits his only shot, he wins, if he misses, he's dead. This is assuming, of course, that he's facing White because White wisely took out Gray with his first opportunity.

that was a real blast I gave it my best shot!

thanks, good teaser

thanks, good teaser

An oldie but a goody. Teasers with completely unexpected answers are always more interesting. And besides, we'll all have to go and investigate the quadruel won't we? Watch this space!

ok i think that he should shoot at the guy who always hits cuz he would probly miss, and then the other guy would shoot at the guy who always hits then he wuold be dead;....

This teaser isn't racist...... right...?

Isn't is a 'dual' and not a 'truel'? I looked up 'truel' and it doesn't exist.

Isn't it a 'dual' and not a 'truel'? I looked up 'truel' and it doesn't exist.

Isn't shooting at Mr. White and missing the same as shooting the ground? Why not take the 1/3 chance that you might actually kill the most potent person first? does your probability of dying go up if you kill White on the first shot? I don't want to work it all out, just tell me.

No, shooting Mr White and hitting him gives Mr Grey only one target...YOU. You want Mr white alive because you can safely assume Mr grey will shoot at him and probably hit him. If Mr Grey misses, Mr White will shoot him first giving himself a 1/3 chance to live.

My biggest problem with this teaser is that shooting the ground was NOT an option. It clearly said that they were taking turns shooting each other, not the ground. Still, got me thinking.

My biggest problem with this teaser is that shooting the ground was NOT an option. It clearly said that they were taking turns shooting each other, not the ground. Still, got me thinking.

I meant to say in my previous comment..."giving himself a 2/3 change to live", not "1/3" sorry about that..guess I should proof read a little better.

it actually doesn't matter who or what black shoots at because if you've seen the movie, "the white, the gray and the black" you would know that mr white takes the bullets out of mr blacks gun the night before.

good one!!!

i enjoyed the question and the explanation is correct

but something is boggling me, why should Mr.white target Mr.Grey ? definitely Grey has more chance of killing him than black , but we are dealing with probability so white will choose Grey 2/3 times and black 1/3 times . but in the case of Grey , choosing black will mean instant death , so he always targets white.

i enjoyed the question and the explanation is correct

but something is boggling me, why should Mr.white target Mr.Grey ? definitely Grey has more chance of killing him than black , but we are dealing with probability so white will choose Grey 2/3 times and black 1/3 times . but in the case of Grey , choosing black will mean instant death , so he always targets white.

donga, suppose you are white. If you shoot at black and kill him, it becomes you vs. gray. If you shoot at gray and kill him, it becomes you vs black. Since gray is a better shooter, it is clear you would prefer the latter case: you vs black. Hence, white always shoots for gray.

If Mr. Black only hits what he's aiming at 1/3 of the time, does that mean he has a 66% chance of missing the ground?

Even the worst shooter should hit the ground... I think.

Shooting the ground is a guaranteed hit. Wouldn't that mean the next two shots would miss their target. Might as well try to take out white. Actually each shot has the same chance of hit or miss. There is no guarantee that in 3 shots one will be a hit.

Someone has way too much time on their hands!!! This was beyond stupid and who cares!!!

I lost interest way before getting to the end of the question.

Good teaser! But it made my head hurt! I say Messrs. Black, White, and Gray should put down their guns and go have a drink together instead.

I find this racist XD

Hey, why am I Mr. Pink?

Because you're a f*****.

Why can't we pick our own colors?

No way, no way. Tried it once, doesn't work. You got four guys all fighting over who's gonna be Mr. Black, but they don't know each other, so nobody wants to back down. No way. I pick. You're Mr. Pink. Be thankful you're not Mr. Yellow.

Because you're a f*****.

Why can't we pick our own colors?

No way, no way. Tried it once, doesn't work. You got four guys all fighting over who's gonna be Mr. Black, but they don't know each other, so nobody wants to back down. No way. I pick. You're Mr. Pink. Be thankful you're not Mr. Yellow.

This is one of the worst teasers I've ever seen on this site, as are also some of the comments. (Especially the one above.) I used to enjoy coming to this site every day, but it doesn't seem the same anymore.

I think he should have shot Mr.White first because Mr.White never misses so that would be a relief to get rid of him but i do see your point now

if i were mr. black i'd shoot myself in the foot and call it a day. after all, the question is about survival, not about winning.

Hey, if Mr. Black only hits his target 1/3 of the time, does that mean if he shoots at the ground he will probably miss? Where will the errant shot go? Could he accidentally hit one of the other two? Or himself?!!?

I've read another version of this that claims you have a greater chance of survival if you shoot in the air. Is that true and if so why?

My brain hurts.

Who cares?

MY brain hurts, too and WHO cares? OK It's an interesting puzzle, but it's either too hard, or too early! Just nice to say hi to all of you!

Too early for this one, but I'd be lying if I said I'll come back later.

Hi back to you Gayle!

Hi back to you Gayle!

Unfortuneately for Mr. Black he can only hit his target 1/3 of the time. Therefore he will probably miss the ground and hit either Mr. Gray or Mr. White. After the shooting is over, Mr. Pink runs away with all the money.

I am with phrebh! Now violence has infested our brain games. Isn't there enough in the world already??

I worked out the approximate percentages, but I was assuming that the solver is expected to assume that the other people's targets were randomly chosen when they had a choice between two. In other words, when it's Mr. Gray's turn and all three are still alive, I figured we were supposed to give a 50% chance that he shoots at me and a 50% chance that he shoots at Mr. White. It didn't really say in the teaser. Also, I didn't think about the shooting-at-the-ground option. So I got this:

Shoot at Mr. White first: 25.8% win

Shoot at Mr. Gray first: 12.3% win

And if you use the same method for shooting at the ground: 34.1% win

So yeah, shooting at the ground is best either way.

Shoot at Mr. White first: 25.8% win

Shoot at Mr. Gray first: 12.3% win

And if you use the same method for shooting at the ground: 34.1% win

So yeah, shooting at the ground is best either way.

And in defense of my reasoning, it never even says in the teaser that the three people know each other's shooting percentages.

The answer is not enough.

3 choices: shoot Gray, shoot White and no shoot (shoot ground)

If shoot Gray, is Black better off is Gray is dead? No because White will kill him => not shoot Gray

If shoot White, is Black better off if White is dead?

If White die => Gray will start first at 1vs1 combat with Black => prob Black alive = 1/7

If White alive => 2 scenarios: Gray may kill White => Black will start first => Black is better off. If Gray does not kill White, White will kill Gray => Black will start first at 1vs1 combat with White => prob Black alive=1/3 => Black is better off as well.

Therefore, Black will be better off if White is alive.

3 choices: shoot Gray, shoot White and no shoot (shoot ground)

If shoot Gray, is Black better off is Gray is dead? No because White will kill him => not shoot Gray

If shoot White, is Black better off if White is dead?

If White die => Gray will start first at 1vs1 combat with Black => prob Black alive = 1/7

If White alive => 2 scenarios: Gray may kill White => Black will start first => Black is better off. If Gray does not kill White, White will kill Gray => Black will start first at 1vs1 combat with White => prob Black alive=1/3 => Black is better off as well.

Therefore, Black will be better off if White is alive.

Side to consider:

Suppose Mr. Black has shot at the ground.

Now it is Mr. Grey's turn. Suppose Mr. Grey shoots at Mr. White. If Mr. Grey kills Mr. White, then Mr. Black and Mr. Grey obviously shoot at each other, giving Mr. Grey a 3/5 chance of survival.

Interesting. Suppose it comes to Mr. White's turn, with all 3 men alive.

Suppose Mr. White shoots at Mr. Grey. Mr. Grey dies and Mr. White now duels with Mr. Black giving Mr. White a 2/3 chance of survival.

Now we can provide insight on the best strategies for Mr. Grey and Mr. White.

In the most simplified form, let us assume any strategy must be consistent. That is to say that, given a set of circumstances, the same action is always taken. Then we have two Nash equilibria: Mr. Grey and Mr. White both choose to shoot each other when all 3 men are alive, or both men choose to also shoot the ground when all three men are alive (this game is assumed to give a result of all three men living indefinitely with no dying from other means, although does not take into account the obvious drop in quality of life).

In this simplified version, these two men have NO OPTIMAL STRATEGY. This is because the best strategy for one of the men changes if the other man's chosen strategy changes.

Let us now look at the more complex situation in which we assume the two men pick some probabilities g (for Mr. Grey) and w (for Mr. White), such that whenever it is their turn and all 3 men are alive, the chance of them deciding to shoot the other is g or w with the alternative being a ground shot.

This scenario retains the two equilibria from the previous scenario. Importantly, for any g

Suppose Mr. Black has shot at the ground.

Now it is Mr. Grey's turn. Suppose Mr. Grey shoots at Mr. White. If Mr. Grey kills Mr. White, then Mr. Black and Mr. Grey obviously shoot at each other, giving Mr. Grey a 3/5 chance of survival.

Interesting. Suppose it comes to Mr. White's turn, with all 3 men alive.

Suppose Mr. White shoots at Mr. Grey. Mr. Grey dies and Mr. White now duels with Mr. Black giving Mr. White a 2/3 chance of survival.

Now we can provide insight on the best strategies for Mr. Grey and Mr. White.

In the most simplified form, let us assume any strategy must be consistent. That is to say that, given a set of circumstances, the same action is always taken. Then we have two Nash equilibria: Mr. Grey and Mr. White both choose to shoot each other when all 3 men are alive, or both men choose to also shoot the ground when all three men are alive (this game is assumed to give a result of all three men living indefinitely with no dying from other means, although does not take into account the obvious drop in quality of life).

In this simplified version, these two men have NO OPTIMAL STRATEGY. This is because the best strategy for one of the men changes if the other man's chosen strategy changes.

Let us now look at the more complex situation in which we assume the two men pick some probabilities g (for Mr. Grey) and w (for Mr. White), such that whenever it is their turn and all 3 men are alive, the chance of them deciding to shoot the other is g or w with the alternative being a ground shot.

This scenario retains the two equilibria from the previous scenario. Importantly, for any g

for any g

My comments keep getting cut off for some reason.

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