Vtyjkyuk

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
$$
P(X + 5 < Y) + P(Y+5<X)
$$
which, by symmetry, equals $2,P(X+5<Y)$. I have done the following:
$$
2P(X+5<Y) = 2iint_X<Y-5f_X,Y(x,y),dx,dy = 2int_5^60int_0^y-5left(frac160right)^2,dx,dy
$$

But this does not lead to the desired probability $frac16$. Is perhaps my approach wrong?

Your calculation should be $$1- displaystyle 2int_5^60int_0^y-5left(frac160right)^2,dx,dy$$ which would be $1-frac55^260^2 approx 0.1597$ or slightly less than $frac16$. I am not sure you have multiplied by $2$ or subtracted from $1$. Perhaps this picture helps

The condition $X+5 < Y$ actually guarantees there will not be a duel.
Similarly for $Y+5<X$.

If you compute your integral correctly, you need to subtract twice the
integral from $1$ to get the probability that there will be a duel.

It looks like somewhere you lost a factor of $2$.
The integral itself should be at least $0.42$ and double it should be at least $0.84$.
The probability of a duel turns out to be less than $frac 16$.

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 $2*5/60=1/6$.