# The Mexican Standoff

There are three cowboys in a Mexican standoff against each other, named Juan, José, and Jorge.

• Juan is the straightest shooter in the West, and can hit his target 100% of the time.
• José, who has a cataract and can't see clearly, can hit his target 70% of the time.
• Jorge, who has no aim but is very clever, can hit his target only 30% of the time.

They proceed to shoot starting from Jorge and going to José and finally Juan, continuing in the same order until only one of them is left standing. Who has the highest chance of surviving, and what strategies do each of them use to maximize their chances?

• Can they shoot anyone they want, or just the next person on the list? May 20, 2014 at 22:57
• You can assume that the hits are always fatal.
– user88
May 20, 2014 at 23:33
• @DisgruntledGoat I'm assuming they're trying to stay alive. May 20, 2014 at 23:45
• @DisgruntledGoat: They have to not be clever enough to realize that if nobody shoots anybody they all survive. Then the problem goes away. May 21, 2014 at 0:05
• Is Jorge the only one allowed to not shoot anyone? Otherwise, why isn't the solution that no one shoots anyone, guaranteeing everyone's survival? Apr 18, 2019 at 4:57

• Juan's strategy is simple: if he has to choose between opponents, then his shot will turn it into a two-man contest, and he will have to survive a shot before he can kill the second man. So of course he will shoot Jose first.

• José's strategy, similarly, is to shoot at Juan. Leaving Juan alive and Jorge dead cannot possibly benefit him.

• Jorge's strategy is the interesting case.

Aiming at José is clearly wrong, since hitting him would leave Juan with no other target.

So let's say Jorge aims at Juan and kills him. His chance of surviving is the sum of the chances that he will both survive a round and kill José in the next. Since both are 30% chances, he has a 30%×30% chance of winning on his next shot, a 30%×30%×30%×30% chance of winning on his shot after that, and so on. We can calculate his chances at $\sum_1^\infty 0.3^\left(2n\right)$ = 9.89%.

Can he do better? Yes!

Jorge's best strategy (assuming he's a good enough shot to implement it) is to intentionally miss.

Now if José misses Juan (30% chance), then Juan will kill José. Then Jorge's chances are exactly 30%: he gets one shot and must make it count.

But if José hits Juan (70% chance), then it's Jorge's turn and he has a 30% chance of winning immediately, plus the chances of surviving a duel with Jorge as calculated above. Total chance: 39.89%

This means Jorge's total chances are 70%×39.89% + 30%×30% = 36.9%. Not bad for a guy who can't shoot!

Now that we have all three strategies, we can calculate the chances for Juan and José.

José will get the first meaningful shot. He must first kill Juan (70% chance) and then survive his battle with Jorge (60.11% chance: 100% minus the 39.89% chance calculated above). Total chance: 70%×60.11% = 42.1%

Juan has a 30% chance of surviving José's shot and then killing him, then a 70% chance of beating Jorge. Total chance: 30%×70% = 21%

Putting it all together:

• Juan: 21%
• José: 42.1%
• Jorge: 36.9%

The moral of the story: never be the best gunfighter around.

• There are ways to tweak the probabilities so that Jorge actually has the highest change of winning. In the formulation I read, it was 100/80/50 and Jorge had a 52.22% chance of winning.
– user88
May 20, 2014 at 23:59
• @JoeZ. Yes, I noticed. In other words, I'd probably win a Mexican standoff - although my strategy would likely involve hiding while the other two were shooting at each other. May 21, 2014 at 0:09
• Brilliant. Great explanation. Sep 11, 2019 at 9:22
• The probabilities in this answer are incorrect. In a duel between 70% and 30% guys, with 70% guy going first, the 2nd guy has ~11.39% chance of winning - not 9.89%. You can calculate it with a recurrence equation P = 0.3*0.3 + 0.3*0.7*P Jul 24, 2020 at 15:28
• @rinspy: I've been sitting here trying to figure out if the math is wrong or not, and I'm getting confused by the fact that Jorge's chance to hit is the same as José's chance to miss. What would your formula be if José's chance were, say, 80%? In other words, which term is for Jorge and which is for José? Jul 24, 2020 at 15:38

If Juan gets a chance to shoot, he will shoot Jose (because Jose has a higher chance of killing him) and he will kill Jose because he has 100% accuracy.

Jose knows this, and he gets a chance to shoot first, so he will shoot at Juan. There is a 70% chance that he will manage to kill Juan if he gets to shoot him.

Jorge knows that the other two are going to focus on shooting each other. If he succeeds at shooting either one of them, the other will go next and shoot him instead! That is not what he wants. So Jorge will go first, and he will not shoot at either Juan or Jose. He can shoot at the ground, a passing bird, whatever. But he will not aim at either of the others.

Then it's Jose's turn. As we said, he'll aim at Juan with a 70% chance of success. If he misses, it's Juan's turn, and Juan will shoot and kill Jose. Then Jorge will have a 30% chance to kill Juan. If he fails, Juan will shoot him and he will die (Juan is the last left standing). If Jorge succeeds, he will be the last one left standing.

Going back to the top: Jorge passes, and it's Jose's turn. If he succeeds in killing Juan, it's now Jorge's turn. He has a 30% change of killing Jose. If he fails, Jose has a 70% chance to kill Jorge. They will keep exchanging shots until one of them succeeds.

Let's approach this for a purely Game-theory perspective (which means, everyone simply wants to get the biggest possible payout, to live, and no one does actions like shooting someone out of revenge for shooting at them, etc.)

We can deduce two things logically from the beginning. The first is that since Juan is such an excellent marksman, at the end of the first "round" (where everyone has had a chance to shoot at least once, assuming they haven't died), the game will always end with either one winner left alive, or a "two-man standoff".

The second thing we can deduce is that since everyone has an infinite amount of bullets (though, only three would actually be required per person until the game ends), if two people are left standing, the best thing they can do is keep shooting at each other until they or the other person is dead.

If you enjoy graphs, I have created a little ASCII diagram which illustrates how the first three turns can play out:

[Jorge]---(Shoot José)--- 0.3 HIT  ---[DEAD]-------------------------------[Juan]--(Shoot Jorge)-- # Juan WINS
\                    -- 0.7 MISS ---[José]---(Shoot Jorge)--- 0.7 HIT  --[Juan]--(Shoot José )-- # Juan WINS
\                                       \                 -- 0.3 MISS --[Juan]--(Shoot Jorge)-- # José  vs. Juan two-man standoff
\                                       \                                     -(Shoot José )-- # Jorge vs. Juan two-man standoff
\                                       --(Shoot Juan )--- 0.7 HIT  --[DEAD]----------------- # Jorge vs. José two-man standoff
\                                                      -- 0.3 MISS --[Juan]--(Shoot Jorge)-- # José  vs. Juan two-man standoff
\                                                                          -(Shoot José )-- # Jorge vs. Juan two-man standoff
--(Shoot Juan)--- 0.3 HIT  ---[José]---(Shoot Jorge)--- 0.7 HIT  --[DEAD]----------------- # José WINS
\                                      -- 0.3 MISS --[DEAD]----------------- # Jorge vs. José two-man standoff
-- 0.7 MISS ---[José]---(Shoot Jorge)--- 0.7 HIT  --[Juan]--(Shoot José )-- # Juan WINS
\                 -- 0.3 MISS --[Juan]--(Shoot Jorge)-- # José  vs. Juan two-man standoff
\                                     -(Shoot José )-- # Jorge vs. Juan two-man standoff
--(Shoot Juan )--- 0.7 HIT  --[DEAD]----------------- # Jorge vs. José two-man standoff
-- 0.3 MISS --[Juan]--(Shoot Jorge)-- # José  vs. Juan two-man standoff
-(Shoot José )-- # Jorge vs. Juan two-man standoff


## Two-man standoffs

There are three possible two-man standoffs at the end of the first round. Note that since everyone has had a chance to fire once, the person who is the poorest shot will be allowed to shoot first.

Jorge vs. Juan

Jorge is first to fire:
> If he hits Juan (30% chance), he wins.
> If he misses (70% chance), Juan fires back, and being an excellent marksman, is guaranteed to win the standoff.

José vs. Juan

Jorge is first to fire:
> If he hits Juan (70% chance), he wins.
> If he misses (30% chance), same as above, Juan guarantees a win.

Jorge vs. José

If Jorge faces off against José, since no one can hit the other person with 100% accuracy, there is an infinitesimally small probability that the standoff will go on forever with each person missing their shot until they both die of old age.

It's possible to calculate an "exact" probability out of this game using integrals, however, we are going to stick to high-school algebra and basic logic to solve this puzzle in order to make it understandable to as many people as possible. Let's just say since Jorge shoots first, he has a greater than 30% chance of winning, and José has a less than %70 chance of winning. This is actually all the information we need to get an answer.

Summary

Let's summarize the two-man standoffs. Instead of using percentages, we will be using values from 0 to 1 to make the math we will be doing later easier. The following values represent the probabilities of each person surviving the encounter:

Jorge vs. Juan: { Jorge:  0.3, José:  0.0, Juan: 0.7 }
José  vs. Juan: { Jorge:  0.0, José:  0.7, Juan: 0.3 }
Jorge vs. José: { Jorge: >0.3, José: <0.7, Juan: 0.0 }


## Juan

Let's start with Juan. If only one person remains after José and Jorge had their turns, he will obviously shoot the remaining person, hitting the target perfectly, and will win the game.

If the other two have gotten their allowed shots, and Juan is left standing, he has two options:

A. Shoot José
B. Shoot Jorge

Since he will always hit his opponent, the person he shoots will not make it to the next "round". That means, he will be left in a two-man standoff (see Deduction #2) with the person he didn't shoot.

Looking at our previous summary, we know that he would rather be in a two-man standoff with the worst shooter (Jorge), so he will always shoot José to avoid a two-man standoff with him.

We can now update our graph with this information:

[Jorge]---(Shoot José)--- 0.3 HIT  ---[DEAD]------------------------------- # Juan WINS
\                  -- 0.7 MISS ---[José]---(Shoot Jorge)--- 0.7 HIT  -- # Juan WINS
\                                      \                -- 0.3 MISS -- # Jorge vs. Juan
\                                      --(Shoot Juan )--- 0.7 HIT  -- # Jorge vs. José
\                                                     -- 0.3 MISS -- # Jorge vs. Juan
--(Shoot Juan)--- 0.3 HIT  ---[José]---(Shoot Jorge)--- 0.7 HIT  -- # José WINS
\                                      -- 0.3 MISS -- # Jorge vs. José
-- 0.7 MISS ---[José]---(Shoot Jorge)--- 0.7 HIT  -- # Juan WINS
\                -- 0.3 MISS -- # Jorge vs. Juan
--(Shoot Juan )--- 0.7 HIT  -- # Jorge vs. José
-- 0.3 MISS -- # Jorge vs. Juan


## José

Now it's José's turn. Obviously, if only one person remains, José will shoot whoever is left! But what if three people remain?

We will update the graph to include the payoffs, which will see who José is better off shooting:

   [José]---(Shoot Jorge)--- 0.7 * { Jorge:  0.0, José:  0.0, Juan: 1.0 }
\                -- 0.3 * { Jorge:  0.3, José:  0.0, Juan: 0.7 }
--(Shoot Juan )--- 0.7 * { Jorge: >0.3, José: <0.7, Juan: 0.0 }
-- 0.3 * { Jorge:  0.3, José:  0.0, Juan: 0.7 }


If we multiply the probability of José hitting the target by the probability that José will win after that round, and then get the average of the two, we get the following results:

   [José]---(Shoot Jorge)--- { Jorge:  0.09, José:  0.00, Juan: 0.91 }
--(Shoot Juan )--- { Jorge: >0.30, José: <0.49, Juan: 0.21 }


We can now clearly see that if there are three people remaining when José is ready to shoot, if he shoots Jorge he has absolutely 0% chance of surviving. On the other hand, if he shoots Juan, he has a bit less than 49% chance of winning. Not being an idiot, he will choose the latter!

Our tree now looks like this:

[Jorge]---(Shoot José)--- 0.3 HIT  ---[DEAD]------------------- { Jorge:  0.00, José:  0.00, Juan: 1.00 }
\                -- 0.7 MISS ---[José]---(Shoot Juan )--- { Jorge: >0.30, José: <0.49, Juan: 0.21 }
--(Shoot Juan)--- 0.3 HIT  ---[José]---(Shoot Jorge)--- { Jorge: >0.09, José: <0.91, Juan: 0.00 }
-- 0.7 MISS ---[José]---(Shoot Juan )--- { Jorge: >0.30, José: <0.49, Juan: 0.21 }


## Jorge

Finally, it's Jorge's turn (well, actually, he shot first, but thinking backwards logically, we finally arrive at him).

Let's take a final look at that tree, collapsing the tree down so we see what Jorge's probabilities of winning are (hiding José's move):

[Jorge]---(Shoot José)--- 0.3 * { Jorge:  0.00, José:  0.00, Juan: 1.00 }
\                -- 0.7 * { Jorge: >0.30, José: <0.49, Juan: 0.21 }
--(Shoot Juan)--- 0.3 * { Jorge: >0.09, José: <0.91, Juan: 0.00 }
-- 0.7 * { Jorge: >0.30, José: <0.49, Juan: 0.21 }


And if we multiply the payoffs by the odds of them happening:

[Jorge]---(Shoot José)--- { Jorge: >0.210, José: <0.343, Juan: 0.447 }
--(Shoot Juan)--- { Jorge: >0.237, José: <0.616, Juan: 0.147 }


It's a close tie! Jose has a slightly larger than 21.0% chance of winning if he shoots Jorge, but a very slight lead of slightly larger than 23.7% chance of surviving if he shoots Juan.

There isn't a very large chance that he will survive anyway, but Jorge hopes for the best, and aims his sights at Juan.

## Results

Assuming everyone plays rationally, here is how the game will go:

[Jorge]---(Shoot Juan)--- 0.3 HIT  ---[José]---(Shoot Jorge)--- 0.7 HIT  --[DEAD]--------------------- { Jorge:  0.0, José:  1.0, Juan: 0.0 }
\                                      -- 0.3 MISS --[DEAD]---[Jorge vs. José]-- { Jorge: >0.3, José: <0.7, Juan: 0.0 }
-- 0.7 MISS ---[José]---(Shoot Juan )--- 0.7 HIT  --[DEAD]---[Jorge vs. José]-- { Jorge: >0.3, José: <0.7, Juan: 0.0 }
-- 0.3 MISS --[Juan]---(Shoot José)------ { Jorge:  0.3, José:  0.0, Juan: 0.7 }


In the end, these are the combined chances that the respective marksmen will win:

• Jorge: >23.7%
• José: <61.6%
• Juan: 14.7%

So surprisingly, despite being a semi-poor shot, assuming Jorge acts rationally, the odds are in José's favor! Yet, one can never guarantee anything with randomness.

There is a "bonus answer" for Jorge getting a better payoff if you are willing to "think outside the box", but I'll leave that one for you to decipher. ;)

• What happens if Jorge misses deliberately? People have already revealed that trick here.
– user88
May 21, 2014 at 1:58
• @JoeZ. See the addendum. I actually plan on coming back and adding that solution to the calculations, but I feel I have written enough walls of text for one day. ;) May 21, 2014 at 2:08
• I saw the addendum. I just thought you might have made the question under the assumption that nobody else had answered yet.
– user88
May 21, 2014 at 2:14
• @JoeZ. Well, when I started writing the answer, there were no existing answers. There were three answers by the time I submitted, but I expected this, knowing it takes me a long time to write, and being familiar with the site's "Fastest Gun in the West" problem. May 21, 2014 at 2:32
• The "addendum" is the entire puzzle. I appreciate the hard work but I can't upvote without the addendum. May 21, 2014 at 3:08

There is a mistake in the selected answer! The Sum is used for the situation when both Jorge ($$70\%$$ chance to miss) and José ($$30\%$$ chance to miss) miss, so it will be $$\sum_{n=1}^\infty (0.3\cdot 0.7)^n=\frac{1}{1-0.3\cdot 0.7} =0.2658$$ by using the geometric series.

Other calculations are simply adding and multiplying, so the final results will be:

If Jorge will try to shoot Juan:

• Juan $$14.7\%$$,
• José $$57.0\%$$ (rounded),
• Jorge $$28.3\%$$ (rounded).

Better result for Jorge if Jorge will miss intentionally (until José or Juan dies):

• Juan $$21\%$$,
• José $$43.4\%$$ (rounded),
• Jorge $$35.6\%$$ (rounded).
• No, Jorge's chance of surviving is dependent on his 30% chance to hit, not his 70% chance to miss. I think my math is correct. Mar 9, 2016 at 18:36
• @MichaelMyers Your math is wrong - try running a simulation! Jul 24, 2020 at 15:23
• @MichaelMyers No, if Jorge hit second time, then game is over and infinite sum would have no sence. Also you are using wolframalpha when it is no needed, so your math is probably not as good as mine, sorry. Jul 25, 2020 at 17:44
• There is a mistake here as well. The chance for Jorge to win a dual with Jose is $(0.3)^2 \sum_{n=0}^\infty (0.3\cdot 0.7)^n = 0.1139...$.
– daw
Aug 22, 2023 at 14:15