2
$\begingroup$

You are a soldier in the army, and the commander is testing his soldiers for their shots. But the commander is also a mathematician and would like to test not only your shooting ability but also your intelligence. To do that, before you shot, he brings the shooting target to you which consists of 3 disc area on it:

enter image description here

and said:

Soldier, you have $9$ shots, and there are three places in this shooting target. I want you put any positive number on them. By doing so, after you shot I will go and check the sum the points you got. Tell me which numbers you need to put in these three places to increase the chance to get $44$ point in total. Do not forget, you may miss too.

Let's help our soldier and find which numbers he needs to put in those three places to increase his chance to get $44$ points in total including the fact that he may also miss.

Note: Assume that the chance to hit any disc area or miss is the same. and You are trying to hit every time, you do not want to miss at all.

$\endgroup$
  • $\begingroup$ Can you miss deliberately? I understand you're saying that there's 25% chance of the three areas and missing, but does the soldier also have the option of just aiming at the ground or whatever? $\endgroup$ – Dr Xorile Nov 11 '17 at 19:53
  • $\begingroup$ @DrXorile actually you are trying to hit, but the chances are the same and you may assume that you are not trying to miss a shot for simplicity. $\endgroup$ – Oray Nov 11 '17 at 19:57
  • 3
    $\begingroup$ @DrXorile "if you can't see you'll try to hit every time" not necessarily, put 44 in all numbers, shoot once then 8 times at the ground for a 75% success rate. If you can't do this it should be specified in the question. $\endgroup$ – ffao Nov 12 '17 at 1:55
  • 1
    $\begingroup$ @spacetyper One can still think of a spatial probability distribution for the hits. For example a Gaussian-like distribution with maximum probability density in the center would roughly make the probabilities of hitting either of the three circles or missing equal. $\endgroup$ – A. P. Nov 12 '17 at 18:39
  • 1
    $\begingroup$ are you saying p(I) = p(II) = p(III)=p(miss)=0.25 ? $\endgroup$ – Jasen Nov 14 '17 at 9:34
5
$\begingroup$

So here is a first guess to create a lower limit on the probability.

Firstly, we're taking 9 shots at 4 targets with equal probability. One of the targets has a score of 0. We need to assign a score to each of the other three to maximize our probability of getting 44.

It seems to me that we should look at divisors of 44 since there is no advantage to getting close to 44. So 2, 4 or 11 are the obvious cases. And since we are expecting to be hitting the scoring target 6 or 7 times (more likely 7, I guess - if we distribute the 8 equally, the ninth is more likely to hit than not), so we're targeting an average of around 6 or 7.

So here are a couple of possible strategies:

  1. Have one of the target areas be worth 22, and the others 0. We would need to hit that target area twice to win.
  2. Have one of the target areas be worth 4, and the others 8. I'll divide everything through by 4. So we are targeting a score of 11 now, and have scores of 2,2, and 1. We would need:
    • 5x2 and 1x1, or
    • 4x2 and 3x1, or
    • 3x2 and 5x1, or
    • 2x2 and 7x1.

The first of these has probability of $(0.25)^2(0.75)^7\binom{9}{2}\approx0.300$.

The second of these I can't see a neat way to calculate so I'll do it the painful way, which is to add up the four terms:

  • $(0.5)^5(0.25)^1(0.25)^3\left(\frac{9!}{5!1!3!}\right)\approx0.0615$
  • $(0.5)^4(0.25)^3(0.25)^2\left(\frac{9!}{4!3!2!}\right)\approx0.0769$
  • $(0.5)^3(0.25)^5(0.25)^1\left(\frac{9!}{3!5!1!}\right)\approx0.0153$
  • $(0.5)^2(0.25)^7(0.25)^0\left(\frac{9!}{2!7!0!}\right)\approx0.0005$

This is about 0.15. (Having two 1s and one 2 is worse).

I'll be interested to see if there are better strategies than this. But current leading strategy is

to have 1 with a score of 22 and no scores for anything else (equivalent strategy is to have 44/7 on all three targets).

However, this solution is no longer allowed with the new rules, so I would go for the second option:

One area worth 4 and two worth 8. This has approx 15% chance.

@Nopalaa used a computer to consider all the possibilities and confirmed that this is the optimal solution. He also confirmed the probability as being $\frac{40464}{262144}=\frac{2529}{16384}\approx15.44%$, as shown in the wolfram alpha link above.

$\endgroup$
  • $\begingroup$ Sorry for the last minute change but I forgot to add the information that the numbers that you are going to put on the target has to be positive integers. I totally forgot that, my mistake. Could you reconsider your solution with this change if possible? :( $\endgroup$ – Oray Nov 12 '17 at 17:46
  • $\begingroup$ I already had a solution for this option in my answer. I have now emphasized it. $\endgroup$ – Dr Xorile Nov 13 '17 at 0:49
  • 1
    $\begingroup$ Well, i used a computer to test all positive integers, and the 4-8-8 is the best one out there with a probability of 40464/262144. So you can edit your answer to show it is best. $\endgroup$ – Nopalaa Nov 13 '17 at 16:41
0
$\begingroup$

Put 44 in each of the target sections. Hit the target once and shoot away from the target after.

$\endgroup$
  • $\begingroup$ This is a good idea. But the OP clarified that you cannot tell until after the 9 shots whether you've hit or not. So every shot is going to be at the target. You cannot deliberately miss. See comments below the original question. $\endgroup$ – Dr Xorile Nov 12 '17 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.