You are a good sniper and there are 10 targets you need to hit. The chance of hitting any target is the same.

Strangely, with your accuracy the chance of getting 8 hits is equal to getting 7 hits in 10 hits.


What is the exact chance of hitting all targets?

Note that the chance of getting hit is greater than $0$ and not $100$% either.

  • $\begingroup$ Just realised what the question was actually asking so edited my answer to include it :) $\endgroup$ – AHKieran Oct 2 '18 at 13:35
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    $\begingroup$ Accepted answer doesn't really satisfy the condition of "good sniper" ;) $\endgroup$ – Kevin Oct 2 '18 at 15:59
  • $\begingroup$ @Kevin hahah you are right, it was misdirection :) $\endgroup$ – Oray Oct 2 '18 at 16:13
  • $\begingroup$ @Kevin : why not? The targets might be so far away and so small, that average snipers won't hit at all. $\endgroup$ – vsz Oct 3 '18 at 4:03

The answer is

$\left(\frac{8}{11}\right)^{10} \approx 0.0414$

If $p$ is the probability of hitting a target on any turn then the actual probability of hitting seven targets is

$\binom{10}{7} p^7 (1-p)^3$

and the chance of hitting eight targets is

$\binom{10}{8} p^8 (1-p)^2$

That means we want

$\binom{10}{7}(1-p) = \binom{10}{8}p$
$\Rightarrow 120(1-p) = 45 p$
$\Rightarrow 165p = 120 \Rightarrow p = \frac{8}{11}$

and the probability of hitting $10$ is

$p^{10} = \left(\frac{8}{11}\right)^{10} \approx 4.14\%$

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  • $\begingroup$ I knew I was missing something! Great answer $\endgroup$ – El-Guest Oct 2 '18 at 14:06
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    $\begingroup$ Okay this makes a lot more sense in my brain hole. And 8/11 is pretty darn close to my suggested 75% :P $\endgroup$ – AHKieran Oct 2 '18 at 14:14
  • $\begingroup$ Yeah @AHKieran, using your intuition is a good idea in this instance to get the ballpark figure. $\endgroup$ – hexomino Oct 2 '18 at 14:20

I don't know if this is allowed, but:

Either 0% or 100%.

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  • $\begingroup$ good catch, added a note :) $\endgroup$ – Oray Oct 2 '18 at 13:11

If your accuracy is:



Chances are, you'll get 7 or 8 hits, but the chances of it being either are equal.


The chance of hitting all the targets is 0.75^10 = 0.0563135147 = ~5.6%

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Let's say the chance of hitting one target is p. The chance of hitting 7 or 8 are equal so:

p^7 * (1-p)^3 * (10 * 9 * 8) = p^8 * (1-p)^2 * (10 *9)

Since 0 < p < 1, we can

divide both sides by p^7*(1-p)^2*(10*9).

(1-p) * 8 = p

p = 1/9

So the chance of hitting all targets is


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  • $\begingroup$ Unfortunately you missed some terms in your combinations. The way to choose three targets is (10*9*8)/(3*2*1) and choosing two targets is (10*9)/(2*1). If you solve from there you'll get the same answer given by hexomino $\endgroup$ – PunPun1000 Oct 2 '18 at 15:41
  • $\begingroup$ Yeah, I tried to fix my original error and made it worse :P But hexomino already explained it better anyway so it doesn't matter. $\endgroup$ – jafe Oct 2 '18 at 15:53

The answer is

$\frac{1}{2^{10}} = \frac{1}{1024}$.


$p$ be the probability of hitting. By binomial distribution, if the chance of hitting 7 or 8 is identical, then $p^7(1-p)^3 = p^8(1-p)^2 \Rightarrow p = 1-p \Rightarrow 2p=1 \Rightarrow p=\frac{1}{2}$.

Then the probability of hitting 10 is

$p^{10} = \frac{1}{2^{10}} = \frac{1}{1024}$.

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  • $\begingroup$ Augh, beat me by a minute...! $\endgroup$ – jafe Oct 2 '18 at 13:47
  • $\begingroup$ i understand how you came to your conclusion but it doesn't sit right in my head. Does this imply that the chance of hitting any number of targets out of the 10 shots is equal? $\endgroup$ – AHKieran Oct 2 '18 at 13:53
  • $\begingroup$ @jafe close one!! Great minds think alike! $\endgroup$ – El-Guest Oct 2 '18 at 14:04
  • $\begingroup$ Hexomino has the actual right answer, for what it’s worth — I rushed without considering combinations $\endgroup$ – El-Guest Oct 2 '18 at 14:06

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