Spoiler Warning: This blog post contains mild mechanical spoilers for the second scenario of the Innsmouth Conspiracy expansion of Arkham Horror LCG.

In The Vanishing of Elina Harper, the players need to find out who abducted agent Elina Harper and at what location she’s taken hostage. In total, there are 6 suspects and 6 locations. During setup, the players need to randomly (and secretly) choose one suspect and one location and shuffle the remaining locations and suspects together as the Leads deck. During actual gameplay, the players may then trade a certain amount of clues to look at the top (up to 3) cards of this deck, choose to draw one of the revealed cards and shuffle the remaining cards together with a replacement card from the encounter deck.

This means that at all times there are 10 cards in the Leads deck and the players know a certain amount of cards in that deck. The goal of this scenario is to see as many cards of the Leads deck as possible. Because the players have 13 rounds to do so but may only look at the top cards once per round, the question for the optimal strategy arises: Should they look at the top cards each round or should they wait until they have acquired enough clues to look at the maximum amount of (3) cards?

Related to this is the question what the expected number of draws is when looking at the top 1, 2 or 3 cards respectively. Interestingly, while the expected number of draws when looking at the top card may be computed in a brute-force manner, we desperately need to organize the computation for the other two cases. This is where Markov chains will come into play.

Because the following is math-heavy, let’s first state the results.

• If you decide to only ever look at the top card of the Leads deck, you will need over 29 rounds on average until you have seen every card.
• If you decide to look at the top 2 cards of the Leads deck, you will need over 14 rounds on average until you have seen every card.
• Finally, if you decide to look at the top 3 cards of the Leads deck, you will need over 9 rounds on average until you have seen every card.

Of course, this assumes that you are able to gather the required amount of clues each round. Here’s a table showing the probabilities of having seen each card after each draw for the three strategies.

DrawOnly drawing 1Only drawing 2Only drawing 3
4000.0149
500.00060.088
600.00950.2188
700.03880.3726
800.09390.5186
900.17130.6419
100.00040.26240.739
110.0020.35830.8124
120.00620.45190.8663
130.01430.53870.9053

• To have a decent chance at success, your group should be able to gather at least 2 clues per player per round.
• If your group is able to gather exactly 2 clues per player each round, it might be better to spend those two clues each round (you will only have 8 chances of looking at cards otherwise, totalling a probability of success at 51.86% compared to 53.87% of spending 2 clues per player each round for 13 rounds).
• This doesn’t take into account that you might want to fish for victory points on locations in the Leads deck.

## Brute-forcing the case of only looking at the top card Link to heading

Let $d$ be the number of cards we look at from the top of the Leads deck, $1 \leq d \leq 3$. Let $k$ be the number of cards we have already seen. The binomial coefficient $\binom{10}{d}$ is the number of possible outcomes to draw $d$ cards from a deck of 10 cards (disregarding order of the drawn cards). Likewise, there are $\binom{k}{d}$ ways to draw $d$ cards from the $k$ cards we have already seen. Consequently, the probability to only draw cards we have already seen when having seen $k$ cards and drawing $d$ cards at once is $q_{d,k} = \frac{\binom{k}{d}}{\binom{10}{d}} = \frac{k!(10-d)!}{(k-d)!10!}.$ In the case $d=1$ this simplifies to $q_{1,k} = k/10$ so that the probability to see a new card when already having seen $k$ cards is $p_{1,k} = 1-q_{1,k} = (10-k)/10$.

Let’s compute the expected number of rounds we have to play when only drawing one card from the Leads deck each round. Let $X_{k}$ be the random variable of rounds we have to play until we see a new card when having seen $k$ cards from the Leads deck. Then we may compute the expected value of $X_k$ as follows. $$\mathrm{E}[X_k] = 1 \cdot P(X_k = 1) + 2 \cdot P(X_k = 2) + 3 \cdot P(X_k = 3) + \dotsb = \sum_{i=1}^\infty i \cdot P(X_k = i).$$

The probability that we have to play exactly $i \geq 1$ rounds is $$P(X_k = i) = q_{1,k}^{i-1} p_{1,k} = \frac{k^{i-1} (10 - k)}{10^i}$$ because we don’t have to make any progress in the first $i-1$ rounds (which has probability $q_{1,k}^{i-1}$) and then finally see a new card in the $i$th round (which has probability $p_{1,k}$). Plugging this into our computation for the expected value, we end up with

$$E[X_k] = \sum_{i=1}^\infty i \cdot \frac{k^{i-1} (10 - k)}{10^i} = \frac{10-k}{10} \sum_{i=1}^\infty i \cdot \left(\frac{k}{10}\right)^{i-1}.$$

Why have I rewritten the result in this form? Because we see now that we’ve ended up with the first derivative of the geometric series $\sum_{i=0}^\infty (k/10)^i$. Because the geometric series has the closed form $$\sum_{i=0}^\infty r^i = \frac{1}{1-r}$$ which has the derivative $1/(1-r)^2$, we deduce $$\sum_{i=1}^\infty i \cdot \left(\frac{k}{10}\right)^{i-1} = \frac{1}{(1-k/10)^2} = \frac{10^2}{(10 - k)^2}.$$

Finally, we arrive at $$E[X_k] = \frac{10}{10 - k}.$$

Note that this formula also holds for $k=0$ and yields $E[X_0] = 1$ (if we haven’t seen any card yet, we are guaranteed to see a new card in the first round). Summing these numbers up yields the desired average number of rounds when only drawing one card:

$$\sum_{k=0}^9 E[X_k] = 10 \sum_{k=0}^9 \frac{1}{10-k} = 10 \cdot \sum_{i=1}^{10} \frac{1}{i} = 10 \cdot H_{10}.$$

where $H_{10}$ denotes the 10th harmonic number.

Likewise, we can compute the probabilities that we only take, say, 10 rounds to see all cards when only drawing one card from the Leads deck each round. In this case, we need to have $X_k = 1$ for all $0 \leq k \leq 9$ and the probability for that is $$\prod_{k=0}^9 p_{1,k} = \frac{10!}{10^{10}} = \frac{567}{1562500} = 0.00036288.$$

Doing similar computations for the case where we draw more than 1 card each round is possible but becomes unwieldy quickly. To orchestrate computations, we use Markov chains in the form of transition matrices.

First, let us think about the case where we draw 2 cards each round:

• In the first round, when we haven’t seen any cards yet, we are guaranteed to see two new cards.
• After the first round, we have three possibilities:
• We could only draw the two cards we have already seen,
• We could draw one card we have already seen and one new card,
• We could draw two cards we haven’t seen yet.
• At some point, we have either seen all cards or we end up with 9 cards seen – in the latter case, we only have two possibilities (see the last card or not).

A state diagram helps to visualize these possibilities. In our case, the state is the number of cards we have already seen: The initial state is $0$ (we haven’t seen any cards) and the terminal (or absorbing) state is $10$ (we have seen everything in the Leads deck). We draw an edge between states if it is possible to advance to that state. Here’s how it could look like:

flowchart LR 0 --> 2 2 --> 2 & 3 & 4 3 --> 3 & 4 & 5 4 --> 4 & 5 & 6 5 --> 5 & 6 & 7 6 --> 6 & 7 & 8 7 --> 7 & 8 & 9 8 --> 8 & 9 & 10 9 --> 9 & 10

From a state diagram, we may derive the transition matrix $P$ that has the probability of transitioning from state $i$ to state $j$ as its $(i,j)$-entry. This transition matrix is our tool of orchestrating our computations. We need to use two of its properties.

It can be proven (by using induction and the definition of matrix multiplication) that $P^k$ contains the probabilities of transitioning from state $i$ to $j$ in exactly $k$ steps as its $(i,j)$-entry. This is the first property we need and we’ll use it to compute the probabilities of transitioning to the absorbing state in any given number of rounds.

To explain the second property, note that our transition matrix has the canonical form

$P = \begin{pmatrix} Q & R \\ 0 & 1 \end{pmatrix},$

where $Q$ is the transition matrix of transient states, i.e. of all non-absorbing states and $R$ is the vector of probabilities of transitioning to the absorbing state.

From this, it can be shown (as a generalization of the geometric series to matrices) that $N := \sum_{i=0}^\infty Q^i = (\mathrm{id} - Q)^{-1}.$ $N$ is the fundamental matrix and contains the expected number of steps of transitioning from state $i$ to state $j$ before being absorbed as its $(i,j)$-entry. This means that $N \cdot 1$ (where we abuse notation and denote the vector that only contains ones by $1$) contains the expected number of steps before being absorbed when starting at state $i$ as its entry at position $i$. This is the property we use to compute the expected number of rounds.

We have all the theory laid down but haven’t actually defined the transition matrix in our case. Let $1 \leq d \leq 3$ be the number of cards to draw. Then, for all $0 \leq i, j \leq 10$,

$P_{ij} = \frac{\binom{i}{d - j + i} \binom{10 - i}{j - i}}{\binom{10}{d}},$ where we use the common convention that $\binom{n}{k} = 0$ whenever $k < 0$ or $k > n$ and the not-so-common convention to start the indices of our matrix at 0 because it fits the names of our states (at least in mathematics, it’s more common to start indices of matrices at 1).

Let’s try and make sense of the formula. At the beginning, in state $0$, where we haven’t seen any cards and draw $d$ cards from the Leads deck, the probability to transition to state $d$ is one (and any other probability should be $0$). Is this the case? Let’s see:

$P_{0j} = \frac{\binom{0}{d - j} \binom{10}{j}}{\binom{10}{d}} = \begin{cases} 1 & j = d, \\ 0 & \text{otherwise}. \end{cases}$

Here, we have used the convention that $\binom{0}{d-j} = 0$ if $j < d$ or $j > d$. More generally, $P_{ij} = 0$ whenever $j < i$ (cards that have been seen cannot be unseen). What about the probability to not see any new cards, i.e., about $P_{ii}$? If we have already seen $i$ cards, draw $d$ cards, and haven’t seen any new cards that means that all $d$ cards are among the already seen $i$ cards. There are $\binom{i}{d}$ ways this could happen. Dividing this by the total amount of possibilities, which is $\binom{10}{d}$, yields the desired probability. Our formula yields the same result because $\binom{10-i}{0} = 1$.

As it turns out, $P_{ij} = 0$ also holds if $j > i + d$. Indeed, this is equivalent to $d - j + i < 0$ which means that, by convention, $\binom{i}{d-j+i} = 0$.

In all of the cases we have investigated so far, our transition matrix really models drawing from the Leads deck. The only cases that remain are $i + 1 \leq j \leq i + d$. In this case, we want to make sure that the our formula really is the probability of transitioning from state $i$ to the reachable state $j$. More down to earth, we want to compute the probability of seeing exactly $j - i$ new cards assuming that we have already seen $i$ cards. First note that seeing $j-i$ new cards means that $d-(j-i) = d-j+i$ cards are among the cards we have already seen. There are $\binom{i}{d-j+i}$ possibilities for this, which explains the first factor in the numerator of our formula. Furthermore, seeing exactly $j -i$ new cards means that they must be among the $10 - i$ cards we have not seen yet. There are $\binom{10-i}{j-i}$ possibilities for this, which explains the second factor in the numerator of our formula. Dividing by the total number of possibilities to draw $d$ cards $\binom{10}{d}$ yields the desired probability.

I’ve written a small Python script that computes the first few powers of the transition matrix and the expected number of rounds. A small trick I’ve used is that I’ve only computed the coefficients (i.e., the numerator of the probabilities). To make things a little easier to understand, I’ve decided for somewhat lengthier but more readable variable names; also, I’ve decided against hard-coding the deck size of 10.

 1from scipy.special import comb as binom
2
3
4def c(
5    cards_seen: int,
6    new_cards_seen: int,
7    cards_drawn: int,
8    deck_size: int = 10,
9) -> int:
10    return binom(cards_seen, cards_drawn - new_cards_seen, exact=True) * binom(
11        deck_size - cards_seen, new_cards_seen, exact=True
12    )


The comb (for combinations) function from the scipy library computes the exact values of binomial coefficients with the same conventions for negative values (as long as, you guessed it, you supplied it with the parameter exact=True). You may verify that this really is the numerator of our formula for the entries of the transition matrix by doing the following substitutions.

• cards_seen -> $i$
• new_cards_seen -> $j - i$
• cards_drawn -> $d$
• deck_size -> $10$.

When assembling the coefficients, there’s another trick I’m using. Note that our transition matrix $P$ is determined by the transition matrix of transient states $Q$. This generally holds for Markov chains that only have a single absorbing state because all rows sum up to $1$. This means that I only need to compute the transition matrix of transient states. I’m calling this the transient coefficient matrix (or coefficient matrix of transient states).

 1import numpy as np
2
3
4def transient_coefficient_matrix(
5    cards_drawn: int, deck_size: int = 10
6) -> np.ndarray[Any, np.dtype[np.int32]]:
7    m = np.zeros(shape=(deck_size, deck_size), dtype=np.int32)
8    for i in range(deck_size):
9        m[i, i:] = [c(i, j, cards_drawn) for j in range(deck_size - i)]
10    return m


I’m first filling up a $10 \times 10$ matrix (the whole transition matrix has shape $11 \times 11$) with zeroes. We have observed that only up to $d+1$ values are possibly nonzero in each row, namely, the entries at $(i, i), (i, i+1), \dotsc, (i,i+d)$. This is what I’m using when filling in the nontrivial values.

Now we’re at the heart of the matter where we can apply Markov chain theory. To explain the following function, note that $C / \binom{10}{d} = Q$ for the coeffient matrix $C$ of transient states. Thus, $N = (\mathrm{id} - Q)^{-1} = \left(\mathrm{id} - C/\binom{10}{d}\right)^{-1} = \binom{10}{d} \cdot \left(\mathrm{id}\cdot\binom{10}{d} - C\right)^{-1}$

Info
I’m doing this because $C$ only contains integer values which produces no rounding errors for additions and subtractions.

The expected number of rounds when starting at state $0$ is the first component of the vector $N \cdot 1$. This is how the following functions computes it.

 1def expected_number_of_rounds(
2    cards_drawn: int,
3    deck_size: int = 10,
4) -> float:
5    m = transient_coefficient_matrix(cards_drawn, deck_size)
6    total_possibilities = binom(deck_size, cards_drawn, exact=True)
7    expected_rounds_vec = (
8        total_possibilities
9        * np.linalg.inv(total_possibilities * np.identity(deck_size) - m)
10        @ np.ones(deck_size)
11    )
12    return expected_rounds_vec[0]


Finally, we compute the probabilities of success by computing matrix powers of the transition matrix. To do so, we first need to compute the transition matrix from the coefficient matrix of transient states (by embedding it into an $11 \times 11$ matrix and then computing the final column by subtracting the sum of each row from $1$).

 1def probabilities_of_success(
2    cards_drawn: int,
3    deck_size: int = 10,
4    up_to_round: int = 14,
5) -> list[float]:
6    m = np.zeros((deck_size + 1, deck_size + 1))
7    m[:deck_size, :deck_size] = transient_coefficient_matrix(cards_drawn)
8    m /= binom(deck_size, cards_drawn, exact=True)
9    m[:, deck_size] = np.ones(deck_size + 1) - m.sum(axis=1)
10    probs = []
11    for power in range(1, up_to_round):
12        result_vec = (
13            np.linalg.matrix_power(m, power) @ np.eye(1, deck_size + 1, deck_size).T
14        )
15        probs.append(round(result_vec[0, 0], 4))
16    return probs


To output the results, we use Pandas to generate a nice markdown table for the probabilities of success (which is basically what I’ve pasted into this blog post).

 1def table_of_probs(up_to_round: int = 14) -> str:
2    return pd.DataFrame(
3        data={
4            f"Only drawing {i}": probabilities_of_success(
5                cards_drawn=i,
6                up_to_round=up_to_round,
7            )
8            for i in range(1, 4)
9        },
10        index=pd.RangeIndex(start=1, stop=up_to_round, name="Round"),
11    ).to_markdown()
12
13
14if __name__ == "__main__":
15    for drawn in range(1, 4):
16        print(f"d = {drawn}")
17        print(transient_coefficient_matrix(drawn))
18        print(expected_number_of_rounds(drawn))
19    print(table_of_probs())


It would be nice to obtain a closed formula for the probabilities of success or the expected number of rounds, like the one we’ve obtained via bruteforce for the case $d=1$.