11 February 2020

# Here is another puzzle

There are three archers – Hubert, Anamika and myself. Hubert is the world renowned, highly decorated Olympics champion from Belgium who is a “sure shot”. Meaning he never misses. Then there is Anamika – who is a amateur archer but has gotten her game to a point where she scores 2 out of 3 times. And then there is Rajib – yours truly – who barely knows which end of the arrow should point towards you. And I have an average starter’s chance of success – 1 in 3.

We find ourselves in a contest where we can only survive by killing the other two. (One successful shot with the arrow is enough to kill anybody). Also, we three are highly intelligent and know how to calculate probabilities of success. (With that level of intelligence, how we got ourselves into this situation is a good question but terrible for this puzzle).

The rules are the following:
Since I am the novice, I get the first shot.
Next, Anamika – being the second best shot goes. Of course, that assumes I did not aim at her and succeeded in which case Hubert goes next.
Once Hubert has had a chance (assuming he was still alive after I and Anamika had our shots), the cycle continues.
So it is me -> Anamika -> Hubert -> repeat till there is one person standing.

The question is: What should be my strategy as the first shot to maximize my chances of still living after the whole ordeal.

Posted February 11, 2020 by rajibroy in category "Puzzles

1 COMMENTS :

1. By rajibroy (Post author) on

Solution
(I need somebody to check my math here. I think I have it right. But would love somebody to review it.)

The answer turns out to be a little tricky. My best chance is to not shoot at either (deliberately miss) and let H(Hubert) and A(namika) duke it out.

To get mathematically there, let’s first look at what happens when only 2 are left.

#1 H and A are left, H shoot first. A’s chance of survival is 0, H is 100%

#2 Same as above, A goes first. A’s chance of survival is 2/3 (if she misses, next H surely gets her)

#3 H and I are left, H goes first. My chance of survival is 0

#4 Same as above, I goes first. My chance of survival is 1/3. (If I miss, I am gone with H’s first shot)

#5 A and I are left. A goes first. The probability I am dead is when she gets me in first shot OR she misses, I miss, she scores OR …. in that sequence. Which is 2/3 + (1/3)*(2/3)*(2/3)+(1/3)*(2/3)*(1/3)*(2/3)*(2/3)+….
= (2/3)*(1+2/9+2/9squared+2/9cubed+….)
= (2/3)*(1 / (1 – 2/9))
= (2/3) / (7/9)
= 6/7
So, if she goes first, my chance of survival is 1/7

#6 Same as above. I go first. The probability A is dead is when I get her in first shot OR I miss, she misses, I get her OR ….
Which is 1/3 + (2/3)*(1/3)*(1/3) + (2/3)*(1/3)*(2/3)*(1/3)*1/3 +…
= (1/3) * (1 + (2/9)squared + (2/9)cubed + ….)
=(1/3) * (1 / (1 – 2/9))
= (1/3) / (7/9)
=3/7
So, if I go first, my chance of survival is 3/7

Now back to the original case. All three are alive.

(A) If I go for A first,
A1. If I hit her (1/3), in next shot H gets me. My probability of survival is 1/3 * 0 = 0
A2a. If I miss her (2/3), she is going to aim at H. If she gets H (2/3), then we are in case #6. My chance of survival is 2/3 * 2/3 * 3/7 = 4/21
A2b. If I miss her (2/3) and she misses H (1/3), H is going to take her out with next shot. Now we are left with #4. My chances of survival is 2/3 * 1/3 * 1/3 = 2/27

The important thing to note is that if I miss, my chances of survival is 4/21 + 2/27 which is definitely higher than if I had hit A (chance of survival = 0)

(B) If I go for H first,
B1. If I hit him (1/3), we are now left with Case #5. My survival chances are 1/3 * 1/7 = 1/21
B2. If I miss him (2/3), then essentially, we are in Case A2a/b.

Again, if I miss, my chances of survival is 4/21 + 2/27 which is definitely higher than if I had hit H (chance of survival 1/21)

So, my best case scenario is not succeed in hitting H or A. Thus shooting the way way past them represents my best chance of survival.

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