The Hurried Duelers brainteaser This question is similar as this other one asked in the forum, but I am trying to give it a different twist. Unfortunately, I am not getting to the same answer, so there might be something wrong in my math.

The problem in this case is the following:

Duels in the town of Discretion are rarely fatal. There, each contestant comes at a random moment between 5 a.m. and 6 a.m. on the appointed day and leaves exactly 5 minutes later, honor served, unless his opponent arrives within the time interval and then they fight. What fraction of duels lead to violence?

My solution is the following. Let X and Y be uniformly distributed random variables on $[0,60]$, each corresponding to the time of arrival of each dueler. The desired probability is 
$$
P(X + 5 &lt; Y) + P(Y+5&lt;X)
$$
which, by symmetry, equals $2\,P(X+5&lt;Y)$. I have done the following:
$$
2P(X+5&lt;Y) = 2\iint_{X&lt;Y-5}f_{X,Y}(x,y)\,dx\,dy = 2\int_5^{60}\int_0^{y-5}\left(\frac{1}{60}\right)^2\,dx\,dy
$$

But this does not lead to the desired probability $\frac{1}{6}$. Is perhaps my approach wrong?
Your calculation should be $$1- \displaystyle 2\int_5^{60}\int_0^{y-5}\left(\frac{1}{60}\right)^2\,dx\,dy$$  which would be $1-\frac{55^2}{60^2} \approx 0.1597$ or slightly less than $\frac16$.  I am not sure you have multiplied by $2$ or subtracted from $1$. Perhaps this picture helps
For the violence to occur next man should come in $5$ mins after the first one so the probability of that is $5/60$. but these man can interchange the position so answer will be $2*5/60=1/6$.