# What is the probability of sinking the submarine but not the destroyer in a simplified game of battleship?

Consider a a simplified game of battleship. We are given a 4x4 board on which we can place 2 pieces. One destroyer which is a 1 × 2 squares and a submarine that is 1 × 3 squares ￼. The pieces are placed on the board and cannot overlap, and are placed either vertically or horizontally.

Your opponent picks 8 (of the 16 squares) uniformly at random and then "shoots" at these 8 squares. A ship is "sunk" if all the squares were the ship was were "shot".

What is the probability that the submarine is sunk but not the destroyer?

The answer that I got was:

$$\frac{{13 \choose 5} - {11 \choose 3}}{{16 \choose 8}}$$

This is how I got my answer:

${16 \choose 8}$ are the total number of ways that the opponent can pick where to shoot bombs. Therefore the total number of outcomes (that we could choose) that we are considering is ${16 \choose 8}$.

${13 \choose 5}$ is the total number of ways to sink the submarine (and perhaps also the destroyer). Why? Well, we can sink the submarine by choosing bombs in its 3 spots $3 \choose 3$ and then we can place our remaining bombs anywhere ${16 -3 \choose 8 - 3} = {13 \choose 5}$ yielding to a grand total of ${3 \choose 3}{13 \choose 5} = {13 \choose 5}$.

However, that could sink the destroyer too and we only want to sink the destroyer. Therefore, we subtract the "intersection". i.e. we subtract the number of outcomes were we are sinking the submarine AND the destroyer. This is given by ${11 \choose 3}$. Why? Well, given the position for all 5 submarines, we could sink all of them ${5 \choose 5}{11 \choose 3} = {11 \choose 3}$ this number of ways.

Which mostly seems to make sense to me. However, I also thought of an alternative way of doing this that does NOT seems to give the same answer (because I plugged in the numbers in wolfram and they give different probabilities, one gives 7/165 and the other 17/165 or something like that). Therefore, I doubted my answer. However, I can't tell which one is right :(.

This is my second way of doing it.

We could sink the submarine only, by first choosing the position of the submarine and then from the remaining, choose position were the destroyer doesn't get hit. i.e. ${3 \choose 3}$ for sinking the submarine and ${16 -3 -2 \choose 8 -3} = {11 \choose 5}$ for the remaining places where the destroyer survives. Thus, we could sink the submarine by ${3 \choose 3}{11 \choose 5}$ Leading to an answer of:

$$\frac{{11 \choose 5}}{{16 \choose 8}}$$

However, my two answers don't seem to match and I am not sure why.

Does someone know why?

This is what good answer should have:

1. It should say why they are different (and thus say why one is under counting or over counting)
2. It should say which one is the correct answer (and justify it of course)
3. It should say how to fix the one that is wrong by either, subtracting some number (since its over counting) or adding some number (since its over under counting) or say why its impossible to fix the wrong answer to make it correct (i.e. why isn't there way a simple way to get an alternative way to do the question).
• As I learned battleship, subs were 2 squares and destroyers were 3 squares. When you say you only want to sink the destroyer it should be the sub. Then you say 5 subs, but there is only one. The first calculation seems correct. – Ross Millikan Dec 30 '14 at 6:04
• its just a made up problem, I didn't even make it up :p lets focus on the interesting thing, the maths! :) – Charlie Parker Dec 30 '14 at 6:07

## 1 Answer

Your first calculation is correct. Given that you sink the sub, you are asking how many sets of $13$ shots will fail to sink the destroyer. You have not assessed the number of positions for the destroyer in the second calculation, you are still assessing the number of shot patterns that do not sink the destroyer. Your count is for the cases the destroyer doesn't get hit at all, not the number where it survives (and maybe gets hit once)