How to survive the Squid Games using probability theory.

There are $8\times(9-L)\times 2$ ways to place a ship of length $L$ on the board. That gives $64$, $96$, and $112$ options for $L=5$, $3$, and $2$ respectively. Because there may be possible overlaps, and because the two $L=3$ ships are exchangeable, the number of legitimate arrangements is substantially smaller than $64\times 96\times 96\times 112=66,060,288$. We have produced a list of all possible arrangements for each ship, and wrote an algorithm to retain only the non-overlapping arrangements. Not making a distinction between the two ships of size $3$, this results in $28,876,784$ distinct placements.

Arguably, placing the ships in adjacent cells makes it easier for the opponent to inadvertently find the second ship while trying to sink the first one. A more clever strategy may thus be to place ships so that no two ships are adjacent by side. The number of such “buffered” arrangements is smaller at $6,406,464$.

It is interesting to take a look at the probability that a cell will have a ship located on it for the two arrangements described above. These probabilities are shown in Figures 2 and 3.

The problem of the optimal strategy in ‘Battleships’ has been addressed before, by, for example, Rodin et al, [RCH⁺88], and more recently by [CdFG20],[ABV14]. Predictably, there have also been some forays into reinforcement learning (see [Cle13] and [Kan20]). While discussing the optimal strategy for either placing the ships or hunting is beyond the scope of this study, we are going to consider four possible strategies for finding a ship:

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  The “Diagonal” strategy involves hitting cells highlighted in Figure 4, first all the orange cells, then all the brown cells. This strategy guarantees finding the $L=5$ ship within $12$ turns, and the second ship within additional $12$ turns.

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  The “Smart Diagonal” coverage-proportionate strategy, where the cells highlighted in Figure 4 are hit in order of decreasing probability of a ship being located there as per Figures 2 and 3.

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  The “Regular” strategy, where the cells in the entire field are hit in order of decreasing probability of a ship being located there as per Figures 2 and 3.

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  The “Completely Random” strategy, where the cells are hit completely at random.

How many additional turns does it take to sink a ship, once it is hit? Keep in mind, that if the team has successfully hit a ship, it is allowed to go again. So the number of turns equates the number of misses. For a ship of length $L=2$, assuming that it is not at the edge of in the corner of the sea and assuming that no other ships are adjacent to it, the remaining square can be found immediately with probability $1/4$, or after one miss with probability $3/4\times 1/3=1/4$, or after two misses with probability $3/4\times 2/3\times 1/2=1/4$, or after three misses with the remaining probability $1/4$. For bigger ships, enumerating all the paths to destruction yields the probability distributions presented in Table 1.

We see that the average time taken to sink the ship does not substantially vary with the size of the ship. While the logic of finishing the ship appears intuitive to a human played, implementing an efficient algorithm is far from trivial, and having possibly adjacent ships only complicates it further.

After listing all possible ship deployments under the “no-buffer” and “buffer” settings, we have used simulations studies to see how long it will take to hit the first ship using each of the four strategies explained in section 2.2.2. Avoiding the issue of finishing the ship off, we then assumed that the hunt would continue until the second (different) ship was hit. The probabilities of each of the ships being hit first or within the first two are shown in Table 1. Note, that they do not sum up to one, because we have two ships of length $L=3$.

One way to determine the probability of not a single ship being sunk on the winning side, $\pi$ is to simulate a lot of games, with two games playing each other, and evaluate the frequency of the event in question. However, assuming that the strategy of one team is not in any way influenced by the strategy of the other team, one can use a one-sided simulation to obtain the distribution for the number of shots it takes to hit the first ship ($t_{1}$), and the (total) number of shots it takes to hit the first two ships ($t_{2}$). Assuming it takes $\delta_{1}$ turns to sink the first ship that was hit, and $\delta_{2}$ turns to sink the second ship that was hit, the probability of not losing any ships on the winning side is then $\pi=Pr(t_{1}+\delta_{1}>t_{2}+\delta_{1}+\delta_{2})=Pr(t_{1}>t_{2}+\delta_{2})$, and can also be obtained from simulations.

The summary statistics for time $t_{1}$ and $t_{2}$ under different strategies are shown in Table 4. As postulated, the “Smart Diagonal” strategy produces the fastest results. Figure 5

Using the information in Table 1, we have used the average estimate $\delta_{t}=1.60$ to obtain $\pi_{\text{ave}}$. We have also used an extremely conservative $\delta_{t}=0$ (assuming that ships are sunk without any misses) to provide an upper bound $\pi_{\max}$ for the probability of no sunk ship by one team while the other sinks both. The results are shown in Table 2.

Note, that for completely random shooting, the probability that the first ship is of length $L$ is $L/13$, where $13=5+3+3+2$ is the sum of the lengths of all the ships. This yields $0.154$, $0.231$, and $0.385$ for $L=2,3$, and $5$ respectively, and agrees with the estimates for random hunting whether with or without placement buffer in Table 2. Also notice that because the “Diagonal” strategies are designed to find the longest ship first, the simulated probabilities of $L=5$ being hit first are higher than that for the “Completely Random” and “Regular” strategies.

Substituting the estimates from Table 2 into Equation 3 yields the probabilities of a ship of length $L$ being eliminated, which are reported in Table 3. The probability of the ships of lengths $L=2$ and $L=3$ being eliminated are firmly below $0.5$. Thus, it is better to be on those smaller ships than to be a captain or a lieutenant. For the longest ship, $L=5$, the probability of being eliminated is higher than $0.5$, except for the Regular strategy.

Putting these numbers into Equation 3 results in the probabilities of elimination reported in Table 3.