The Hurried Duelers brainteaser

Question

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: $$ 2P(X+5<Y) = 2\iint_{X<Y-5}f_{X,Y}(x,y)\,dx\,dy = 2\int_5^{60}\int_0^{y-5}\left(\frac{1}{60}\right)^2\,dx\,dy $$

But this does not lead to the desired probability $\frac{1}{6}$. Is perhaps my approach wrong?

Answer 1

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

Answer 2

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 $\frac 16$.