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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?

Community
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Iliana
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  • Check the probability you wrote down in the first equation. The given event is $(X < Y < X + 5\text{ or }Y < X < Y + 5)$, which isn't quite what you wrote. – anomaly Feb 11 '15 at 14:49
  • How many duelers come within that 1 hour? – ghosts_in_the_code Feb 11 '15 at 14:50
  • @Iliana: What you wrote corresponds to the event where $X$ leaves before $Y$ arrives or vice versa. (The events are disjoint and so their probabilities add). That works, but you should have $1 - \cdots$ term in there. – anomaly Feb 11 '15 at 15:12

3 Answers3

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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

enter image description here

Henry
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  • @ Henry why not to solve it using just simple geometry where the searched probability is the difference between the surface of the whole square - 2 * the surface of the white lower triangle $1-2 \frac{1}{2} (\frac{11}{12})^2=\frac{23}{144} \approx \frac{1}{6}$. Isn't the double integral over complicating the solution? – Michal Dec 10 '16 at 23:11
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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 $\frac 16$.

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

Gaurav
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