A man is to be executed at random time between 00:00 and 01:00. The firing squad's accuracy decreases linearly, so that at 00:00 they shoot perfectly, at 00:30 miss half the time, and at 01:00 miss always. Also, with probability 1/2, a blank round of shots will be used.

Given that the man survived, what is the probability that he faced a live round?

At first I drew a diagram with time on x-axis, a horizontal line at $y=1/2$ with "blind round" above it and "live round" below. Then I divided that lower region with a diagonal to represent the falling accuracy, and got the answer $\frac{1}{4} / (\frac{1}{4}+\frac{1}{2}) = \frac{1}{3}$ which a simulation seems to confirm.

I'm not sure why the following method gives an incorrect answer: at time $\theta$, $P_\theta (\text{survived})=\theta/2 + 1/2$ and $P_\theta(\text{live rounds and survived}) = \theta/2$. Therefore $P_\theta(\text{live rounds | survived}) = \frac{\theta}{\theta+1}$. Now integrate to get $\int_0^1 P_\theta d\theta = \int_0^1 \frac{d\theta . \theta}{\theta+1} = 1-\log 2 \approx 0.3069$.

  • 2
    $\begingroup$ Nice background story for the question. Looks taken from Games of Thrones... $\endgroup$
    – DonAntonio
    Commented Jun 8, 2019 at 23:22

2 Answers 2


This is similar to the question, "does $\displaystyle \int \frac{f(x)}{g(x)} dx= \frac{\int f(x) dx}{\int g(x) dx}?$" The answer is no, and this will resolve your issue. We know that the probability of surviving given that he faced blank rounds is the probability of both divided by the probability of surviving. So the expression we're really looking for here is $$p = \frac{\int_0^1 (\theta/2)d\theta}{\int_0^1 (\theta/2 + 1/2)d \theta}=\frac{1/2}{1 + 1/2} = \boxed{1/3}.$$

  • $\begingroup$ Yes (+1), that pinpoints the error. $\endgroup$
    – quasi
    Commented Jun 8, 2019 at 23:45
  • $\begingroup$ I understand why these integrals aren't equal, but I don't understand why "$\int_0^1 \frac{x.dx}{x+1}$" does not define the correct answer. What is this value? $\endgroup$
    – Moronic
    Commented Jun 9, 2019 at 0:02

The answer by @paulinho is excellent, but here's another way to explain why your second answer is wrong.

Let $L=$ live rounds, and $S =$ survive. I will use $P()$ for probability and $p()$ for pdf.

Your ${\theta \over \theta + 1} = P(L | \theta, S)$. So if you integrate $\int_0^1 {\theta \over \theta + 1}\; d\theta = \int_0^1 P(L | \theta, S)\; d \theta$, since $\theta$ is time, you are assuming that every moment is equally likely. However, when conditioned on survival, every moment is not equally likely!

You needed something akin to the pdf of $\theta$, conditioned on $S$, shown in red below:

  • $P(L|S) = \int_0^1 p(L, \theta | S)\; d \theta = \int_0^1 P(L | \theta, S) \color{red}{p(\theta | S)}\; d \theta$

Frankly, finding $p(\theta | S)$ requires basically solving the entire problem, so this is not a good way to proceed. Instead:

  • $p(L, \theta | S) = p(L, \theta, S) / P(S)$

  • $p(L, \theta, S) = \theta / 2$

  • So, $P(L|S) = \int_0^1 {\theta / 2 \over P(S)} \; d\theta = {\int_0^1 (\theta /2) d\theta \over P(S)}$ and you are back to the answer by @paulinho


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