Problem: $A, B, C, D$ are playing an archery game. They have equal probability of landing their arrow on any spot the target with radius of $R$. Their score is the distance to the center of the target. They take turn to shoot at the target $A, B, C, D, A, B, C, D...$. If one's score is closer to the center of the target than all previous arrows by all players, he survives, goes to the back of the queue, otherwise he is eliminated. We can assume the arrows always hit the target.

What's the probability that $D$ wins this game?

My thoughts: The distribution of the score is obviously linearly scaling from 0 to $R$. But I'm not sure if the distribution of the score matters here. Since we only care about the orders, any order of $x_1, ... x_n$ has $\frac{1}{n!}$ probability regardless of the distribution.

Other than that, I have little clue...

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    $\begingroup$ I've written a simple simulation script: pastebin.com/JQJPTffu It gives a value of about 18% (or 0.18) for D, so the actual answer is probably close to that value. $\endgroup$
    – trolley813
    Dec 30, 2021 at 16:28
  • $\begingroup$ Depends on the equal probability. If all hit inside R > 0 with probability 1, the game never ends. If the probability is 1 for R >= 0, A wins by hitting the mathematical bullseye on the first shot. If the probability is 0 for any R, everyone always misses the target and D wins by default. $\endgroup$
    – Bass
    Dec 30, 2021 at 16:35
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    $\begingroup$ @JeremyDover They have equal probability of landing their arrow on any spot the target with radius of R. so score is not uniformly distributed $\endgroup$ Dec 30, 2021 at 17:03
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    $\begingroup$ Got it. So your note in the puzzle "The distribution of the score is obviously linearly scaling from 0 to R." is not correct? $\endgroup$ Dec 30, 2021 at 17:49
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    $\begingroup$ It matters because probability is a function of area, not distance from the origin. You have a $1/2$ probability of hitting inside the circle with radius $R / \sqrt{2}$, not $R/2$. $\endgroup$ Dec 30, 2021 at 21:11

1 Answer 1


I have just answered this question on the math site.

The solution for this one is basically identical, taking into account the comments below my answer there.

In conclusion, the winning probability of $D$ is equal to $\sum_{a, b, c}\frac{a(a + b)(a + b + c)}{(a + b + c + 1)!}$ where the sum ranges through all triples of positive integers $(a, b, c)$ such that one of the following holds:

  • $a \equiv 0 \pmod 4$, $b \equiv 1 \pmod 3$, $c \equiv 1 \pmod 2$;
  • $a \equiv 0 \pmod 4$, $b \equiv 2 \pmod 3$, $c \equiv 0 \pmod 2$;
  • $a \equiv 1 \pmod 4$, $b \equiv 0 \pmod 3$, $c \equiv 1 \pmod 2$;
  • $a \equiv 1 \pmod 4$, $b \equiv 1 \pmod 3$, $c \equiv 0 \pmod 2$;
  • $a \equiv 2 \pmod 4$, $b \equiv 0 \pmod 3$, $c \equiv 0 \pmod 2$;
  • $a \equiv 2 \pmod 4$, $b \equiv 2 \pmod 3$, $c \equiv 1 \pmod 2$.

The result is approximately $0.18343765$.

  • $\begingroup$ "basically identical" does this account for the circular targets and higher chance to hit closer to the edge? $\endgroup$
    – A username
    Dec 31, 2021 at 1:41
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    $\begingroup$ Yes, this is mentioned in the comment below my linked answer: one can use any probability distribution as long as $\Bbb P(X = x) = 0$ for any $x \in \Bbb R$. This is because we have symmetry among $n$ independ random variables, hence any order of the $n$ numbers appears with probability $\frac 1{n!}$. $\endgroup$
    – WhatsUp
    Dec 31, 2021 at 1:55
  • $\begingroup$ for that particular problem you linked on the math site, I managed to come up with a close form analytical solution. Let $f_2(x)$ be if there are two people, probability of $A$ winning the game, when the lowest score so far is $x$, easy to solve that $f_2(x) = 1 - e^{-x}$, $f_2(1) = 1-1/e$. similarly define $f_3(x)$ for the case when there are 3 people, we can solve $f_3(x)$, and $f_3(1) = 1 - \frac{2\sqrt{3}}{3} e ^{-0.5} \sin (\frac{\sqrt{3}}{2})$. But this involves solving a second order ODE. for 4 persons, it got hairy... so it does seem like a close-form analytical solution exists $\endgroup$ Jan 1, 2022 at 15:30
  • $\begingroup$ I very much believe that such analytic solutions exist. I didn't work them out because the series that we currently have already converge super fast, thus there seems to be no gain in getting a closed formula. Of course it'll be interesting to know them (: $\endgroup$
    – WhatsUp
    Jan 1, 2022 at 17:30

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