Aspects of the Cooperative Card Game Hanabi

Abstract

We examine the cooperative card game Hanabi. Players can only see the cards of the other players, but not their own. Using hints partial information can be revealed. We show some combinatorial properties, and develop AI (Artificial Intelligence) players that use rule-based and Monte Carlo methods.

1 Introduction

The game of Hanabi, meaning “fire flower” or “fireworks” in Japanese, is a cooperative card game that requires the players to combine efforts in order to achieve the highest possible score. Designed by Antoine Bauza in 2011 and published by R & R Games [8] (see Fig. 1), among others, it is designated as a game in the categories “cooperative play” and “hand management” by the game analysis site BoardGameGeek [3]. The goal of the game is simple: play out several sequences of cards in the right order. The catch, however, is that the players can only see the cards in other player’s hands and not their own; information has to be gathered by a system of hints that reveal partial information.

A game in the named categories that is somewhat similar in gameplay, is The Game designed by Steffen Benndorf [3]. Though leading to interesting combinatorial and strategic questions, there is not much to be found in the literature on this kind of games. More is known about another game which bears similarity to Hanabi in view of its dealing with hints: Bridge. Yet another game which gives rise to theoretical problems comparable to those inspired by Hanabi, because of its similarity in nature, is Solitaire; as an example, NP-completeness results for Hanabi can be found in [1].

Fig. 1.

Hanabi, as sold by R & R Games [8].

In [5] the authors describe two sophisticated strategies that play the game in a near perfect way. They make clever use of the hints system, using dedicated conventions, and codings as in the well-known hats problem (cf. [6]). Note that, as in Bridge, conventions may be artificial: the natural (literal) meaning of the hints might be or is (as in [5]) even ignored. In our paper we restrict ourselves to this natural meaning only.

This paper examines some interesting properties of the game, but is far from being complete. In fact, we show complicated mathematical behavior for a one-player version without hints, and provide an exploratory examination of some artificial players (cf. [2, 7]), including rule-based and Monte Carlo versions.

We start with a comprehensive explanation of the rules of Hanabi in Sect. 2. We then deal with two main questions. In Sect. 3, we consider the playability of a game of Hanabi: given a certain start configuration, is it possible to obtain a maximal score when playing perfectly? We will address some theoretical issues. Next, in Sect. 4, we look for a good strategy for any arbitrary game. Among others, we consider Monte Carlo methods, which seem promising. In Sect. 5, we conclude with a summary of the given results as well as some interesting open questions.

2 Game Rules

The classic game of Hanabi is played with a stack of $N=50$ cards. Every card has one out of $C=5$ colors — blue ($\mathrm {B}$), red ($\mathrm {R}$), green ($\mathrm {G}$), yellow ($\mathrm {Y}$) or white ($\mathrm {W}$) — and a value between 1 and $k=5$. In the classic game, for every color there are three 1 s, two 2 s, 3 s and 4 s and one 5, hence fifty cards in total. At the start of the game, the stack is shuffled and a hand of cards is dealt to each player: $R=5$ cards are given to every player if $P=2$ or $P=3$ persons are playing, and $R=4$ cards are dealt in games with $P=4$ or $P=5$ players. Now, every player picks up his/her cards in such a way that the other players can see them, but they themselves cannot. The rest of the cards forms the face-down stack.

The goal of the game is to create C stacks of cards going from 1 through k on the table, one of each color. To do so, players take turns, choosing exactly one of the three following actions every turn:

• give a hint,

• discard a card,

• play a card.

At the start of the game, a pool of $H=8$ hint tokens is available to the players. To give a hint to another player, one token must be removed from this pool. If there are no tokens left, this action cannot be chosen. Giving a hint is done by either pointing out all cards (perhaps zero) of a certain color or all cards of a certain value in the hand of one other player. This is explained most easily by considering an example hand like $(\mathrm {R}3,\mathrm {W}1,\mathrm {B}1,\mathrm {R}4,\mathrm {B}5)$. A hint may be expended to point out the position of the $\mathrm {W}1$ and $\mathrm {B}1$, telling the player that these two cards are 1s. One might also point out the $\mathrm {R}3$ and $\mathrm {R}4$ by telling that these cards are red. Also, pointing out $\mathrm {B}5$ by saying this is a 5 is fine. However, one cannot point out $\mathrm {B}5$ by telling that this card is blue, because $\mathrm {B}1$ must then also be pointed at. One is allowed to tell the player that there are no green cards in the hand, as this hint effectively points out all (zero) green cards. This option is sometimes disallowed, as in [5].

The second possible action is to discard a card: the active player takes a card from his/her hand (without first looking at it) and announces that it will be discarded. It is then put face-up in the discard pile, which may be viewed by all players at any point in the game. Once in the discard pile, a card will never re-enter the game. A new card is now taken from the face-down stack to replenish the hand and as an added bonus, a hint token is added to the available hints pool. Therefore, if there are already H hints available, this action cannot be chosen.

Finally, one may choose to try to play a card much in the same way as discarding a card. However, the players now look whether the card can be added to one of the stacks on the table. This is again best clarified by an example. Suppose the following stacks are on the table:

It is now possible to play a $\mathrm {R}3$, which can be appended to the leftmost stack. Similarly, a $\mathrm {G}2$ or $\mathrm {W}4$ would be fine. It is also possible to play a $\mathrm {Y}1$ or $\mathrm {B}1$, starting a new stack. However, one may for example not play a $\mathrm {R}4$ (as a $\mathrm {R}3$ is first needed) or $\mathrm {R}2$ (as this is already on the table). Moreover, another $\mathrm {R}1$, $\mathrm {G}1$ or $\mathrm {W}1$ cannot be played as there may only be one stack of every color on the table at any time. If a card is successfully played, it is put in the fitting position on the table. If a card turns out not to be playable, it is moved to the discard pile and the players score one error. In either case, the active player draws a new card from the stack, but no hint token is added.

In order to show that there is more to a hint than just its plain meaning, consider the following example. If a player has only one 2, the $\mathrm {G}2$, a hint pointing at this card might in this situation be considered containing the message “come on, play this card”.

The game ends when one of three conditions is met. First, if $E=3$ errors are made by trying to play a card, the game ends with the lowest possible score of 0. Second, if the C-th stack is completed by playing a card, i.e., if there is a sequence of 1 through k of every color on the table, the game ends with the highest possible score of $C\cdot k$ (for the classic game: 25). Finally, when the last card of the face-down stack is drawn, every player gets one more turn, including the player that drew the last card. The achieved score is then determined by adding the highest number of every stack on the table. In the previous example, a score of $2 + 1 + 3 = 6$ points would be obtained. Apart from just aiming at the highest score, one might also focus on the 0–1 target of obtaining the maximal score.

Naturally, the rules for classic Hanabi can easily be generalized and extended. One can for example alter the parameters mentioned above, see Sect. 3. A more formal approach can be found in [2].

3 Playability

In this section, we will address the first main question: given an initial configuration of a game of Hanabi, is it possible to obtain the maximum score if playing perfectly? An initial configuration of a game of Hanabi (or simply a game of Hanabi) is called playable if the maximum score can be achieved. We first turn to a theoretical approach involving combinatorics; a perhaps more practical approach using dynamic programming is presented in [1] and also in [2].

As in [1], we consider the simplified situation where the players can also see their own cards (making the hints system and errors obsolete). We give a result for the one-player version, with $R=1$, called SingleHanabi: a player must immediately play or discard the newly received card. We consider only one color, but the number of cards of each value can be arbitrary.

Fix an integer $k\ge 1$. Let $x=(x_1,x_2,\ldots ,x_k)$ be a vector with k non-negative integers $x_1,x_2,\ldots ,x_k$, and let $F(x)=F(x_1,x_2,\ldots ,x_k)$ denote the number of ordered sequences of length $N = N(x) = x_1+x_2+\ldots +x_k$ with $x_1$ occurrences of the integer 1, $x_2$ occurrences of the integer 2, ..., $x_k$ occurrences of the integer k, without a subsequence 1–2–...–k: the so-called bad sequences; the others are called good. Note that a subsequence is not necessarily consecutive; e.g., the sequence 1–3–2–3 has 1–3–2 and 1–2–3 as subsequence, but not 2–3–1. So $F(x_1,x_2,\ldots ,x_k)$ is the number of unplayable SingleHanabi games, having $x_i$ cards of value i ($1\le i \le k$): there is no subsequence 1–2–...–k that would allow the player, who has “no memory”, to immediately play these k “cards” in order.

If for some i with $1\le i \le k$ we have $x_i=0$, then we know that $F(x)=\left( {\begin{array}{c}N\\ x_1,x_2,\ldots ,x_k,N-\sum _{j=1}^k x_j\end{array}}\right)$, since all sequences are bad in that case.

We also note that F is symmetric in its arguments. Indeed, we construct a bijection between good sequences with interchanged numbers of, e.g., 1 s and 2 s: $x_1 \leftrightarrow x_2$. To do this, view every good sequence in parts: the part up to but not including the first 1, the part between this 1 and the next 2, etc. Now, the required bijection is given by swapping the first and second part and changing all 1 s in the parts to 2 s and vice versa. We clarify this by an example, in which the numbers which divide the parts are shown in bold:

Note that the resulting sequence contains the proper amount of every number. Moreover, applying the proposed bijection twice results in the original sequence, which shows that it is indeed bijective.

Inspired by a discussion by anonymous contributors at StackExchange [10], our main result in this section is:

Theorem. We have

$$ F(x)=\!\!\!\!\sum _{\mathop {|y|\le k-2}\limits ^{\scriptstyle {y\prec \; x}}}\!\!\! a(N,y). $$

Here we denote $y\prec x$ if the ordered sequence y with k non-negative integers satisfies $y_i\le x_i$ for all i with $1\le i \le k$. Furthermore, |y| denotes the number of non-zero elements in y. We put $a(n,y)= (-1)^{|y|} \left( {\begin{array}{c}n\\ y\end{array}}\right) ^{\!*} (k-|y|-1)^{n-\,s(y)}$. The multinomial coefficient $\left( {\begin{array}{c}n\\ y\end{array}}\right) ^{\!*}$ is defined as follows: $\left( {\begin{array}{c}n\\ y\end{array}}\right) ^{\!*}= \left( {\begin{array}{c}n\\ y_1\ominus 1,y_2\ominus 1,\ldots ,y_k\ominus 1,n-s(y)\end{array}}\right) =n!/((y_1\ominus 1)!(y_2\ominus 1)!\ldots (y_k\ominus 1)!(n-s(y))!)$ (the bottom last term in the multinomial coefficient, here $n-s(y)$, is often omitted by convention) with $s(y)=\sum _{i=1}^k (y_i\ominus 1)$, where we used $t\ominus 1 =\max (t-1,0)$.

The equation can also be written as

$$ F(x)=\sum _{\ell =0}^{k-2} (-1)^\ell \sum _{\mathop {|y|=\ell }\limits ^{\scriptstyle {y\prec \; x}}} \left( {\begin{array}{c}N\\ y\end{array}}\right) ^{\!*} (k-\ell -1)^{N-\,s(y)}. $$

For example, with $k=3$ and $\ell =|y|=1$, in the computation of F(2, 3, 1) we encounter sequences (1, 0, 0), (2, 0, 0), (0, 1, 0), (0, 2, 0), (0, 3, 0) and (0, 1, 0); and, e.g., $\left( {\begin{array}{c}6\\ (0,2,0)\end{array}}\right) ^{\!*}=\left( {\begin{array}{c}6\\ 0,1,0,5\end{array}}\right) =\left( {\begin{array}{c}6\\ 1,5\end{array}}\right) =\left( {\begin{array}{c}6\\ 1\end{array}}\right) =6$. The term with $\ell =|y|=0$ equals $(k-1)^N$. Note that $F(x_1,x_2)$ evaluates to 1, as expected. By the way, $F(x_1)$ is 0, if $x_1>0$.

Proof. The theorem can be proven through the obvious recurrence (for $x_1>0$)

$$\begin{aligned} F(x_1,x_2,\ldots ,x_k)= & {} \left( {\begin{array}{c}N-1\\ x_1-1\end{array}}\right) F(x_2,\ldots ,x_k) + {}\\&{}+ F(x_1,x_2-1,\ldots ,x_k)+\ldots +F(x_1,x_2,\ldots ,x_k-1) \end{aligned}$$

where we interpret a term as 0 if one of its arguments is negative. The respective terms count bad sequences that start with a 1, with a 2, ..., with a k.

We first note that $F(1,\!\overbrace{0,\ldots ,0}^{k-1\ \mathrm {times}}\!)$ equals 0 if $k=1$, and equals 1 if $k>1$. Furthermore, $F(0,0,\ldots ,0)=1$. This is the basis for an inductive proof, with respect to $\langle k,N\rangle$.

Using the symmetry of F in its arguments, we may assume that $x_1>0$. If $x_2>0$ we compute (analogous for the other terms):

$$ F(x_1,x_2-1,\ldots ,x_k)= \!\!\!\!\!\!\!\sum _{\mathop {|y|\le k-2}\limits ^{\scriptstyle {y\prec \; x;\, y_2<x_2}}}\!\!\!\!\!\! a(N-1,y) = \!\!\!\!\!\!\sum _{\mathop {|y|\le k-2}\limits ^{\scriptstyle {y\prec \; x;\, y_2= 0}}}\!\!\!\!\! a(N-1,y) \,+\!\!\!\!\!\sum _{\mathop {|y|\le k-2}\limits ^{\scriptstyle {y\prec \; x;\, y_2\ne 0}}}\!\!\!\!\! a(N,y) \frac{y_2-1}{N} $$

Now we look at a fixed y, and combine all contributions of the $k-1$ terms. If $y_1=0$ we arrive at $a(N-1,y)(k-|y|-1)+a(N,y)s(y)/N=a(N,y)$. However, if y has non-zero $y_1$, we have to be more careful. We then still arrive at a(N, y), but now with an additional term $a(N-1,y) - a(N,y){(y_1-1)}/{N}$. If we let $y_1$ increase from 1 to $x_1$ (where we keep the other elements from y unchanged), these terms telescope to $a(N-1,y')$ with $y_1' = x_1$ and $y_i'=y_i$ ($1<i\le k$). We rewrite

$$ \left( {\begin{array}{c}N-1\\ x_1-1\end{array}}\right) F(x_2,\ldots ,x_k) = {}-\!\!\!\!\!\!\sum _{\mathop {|y|\le k-2}\limits ^{\scriptstyle {y\prec x};\, y_1=x_1}}\!\!\! a(N-1,y), $$

which exactly cancels the remaining terms.

If we happen to have $x_2=0$, the argument above remains valid. Indeed,

$$ \sum _{\mathop {|y|\le k-2}\limits ^{\scriptstyle {y\prec \; x}}}\!\!\!\!\! a(N-1,y) = \!\!\!\!\!\!\!\!\sum _{\mathop {|y|\le k-2}\limits ^{\scriptstyle {y\prec \; x;\, y_1<x_1}}}\!\!\!\!\! a(N-1,y) + \!\!\!\!\!\!\!\!\sum _{\mathop {|y|\le k-2}\limits ^{\scriptstyle {y\prec \; x;\, y_1=x_1}}}\!\!\!\!\! a(N-1,y) $$

is equal to 0, since the first term from the right hand side in that case equals

$$F(x_1-1,0,x_3,\ldots ,x_k)=\left( {\begin{array}{c}N-1\\ x_1-1,x_3,\ldots ,x_k\end{array}}\right) ,$$

whereas the second equals

$$-\left( {\begin{array}{c}N-1\\ x_1-1\end{array}}\right) F(0,x_3,\ldots ,x_k) = -\left( {\begin{array}{c}N-1\\ x_1-1\end{array}}\right) \left( {\begin{array}{c}N-x_1\\ x_3,\ldots ,x_k\end{array}}\right) \!=\!-\left( {\begin{array}{c}N-1\\ x_1-1,x_3,\ldots ,x_k\end{array}}\right) .$$

thereby completing the proof.   $\Box$

One consequence is that

$$ F(\overbrace{1,1,\ldots ,1}^{k\ \mathrm {times}})=k!-1=\sum _{\ell =0}^{k-2} (-1)^\ell \left( {\begin{array}{c}k\\ \ell \end{array}}\right) (k-\ell -1)^k$$

which happens to be a special case of a formula from [9]. In this same category, another special case is

$$ F(\overbrace{4,4,\ldots ,4}^{13\ \mathrm {times}}) = \sum _{\ell =0}^{11} (-1)^\ell \left( {\begin{array}{c}13\\ \ell \end{array}}\right) \sum _{i_1=0}^3 \ldots \sum _{i_\ell =0}^3 \left( {\begin{array}{c}52\\ i_1,\ldots ,i_\ell \end{array}}\right) (12-\ell )^{52-i_1-\ldots -i_\ell } $$

which evaluates to 91973270026324484565579418350194489655582912237019 (a prime number approximately equal to $9\cdot 10^{49}$) using a straightforward Python program, that takes a few seconds. Therefore, the probability that a shuffled deck of standard playing cards contains a full increasing subsequence (from ace to king, disregarding suits) turns out to be equal to $1-F(4,4,\ldots ,4)/(52!/4!^{13})\approx 0.000554$, which is a folklore result.

4 Strategies

We will now consider the second main question: what is a good strategy? In this section, a strategy is a means to determine which action to take in any given state of the game. We will look at games with $P = 3$ players in which every player has a hand size of $R = 5$ as in the classic game rules in Sect. 2. We call a card useful if it can be appended to one of the stacks on the table at the current moment. A card is called worthless if it is clear that it can never played (either because a copy of it has already been successfully played (called “dead” in [5]) or all copies of a lower numbered card of the same suit have been discarded).

In [7], strategies are considered in which the players try to estimate their hand by analyzing actions of other players. This analysis is for the two-player version, and relies heavily on online learning. The author claims an average result of 15.85 points. Here, we will try two other types of strategies, and explore their potential. The first one implements several rules of thumb that tend to come up quickly in human play: a rule-based strategy. The second is an implementation of a basic Monte Carlo strategy. Note that it is hard to determine the maximal score that can be achieved for any given game, see also [1]. However, the results from [5] suggest that a perfect score is realizable in most cases.

In contrast with [5], we will not consider conventions regarding the hints. Effectively, hints will be taken literally. In a game state, we will only use hint information on individual cards, and, for instance, not use the identity of the player who provided a particular hint. Clearly, this makes the hints system less powerful.

For the first strategy, every player acts according to the following preset rules:

• 1. If there is a card in my hand of which I am “certain enough” that it can be played, I play it.

• 2. Otherwise, if there is a card in my hand of which I am “certain enough” that it is worthless, I discard it.

• 3. Otherwise, if there is a hint token available, I give a hint.

• 4. Otherwise, I discard a card.

In this framework, there are several parameters to be determined. First, we may choose the definition of “certain enough” in steps 1 and 2: we let $\omega _{\mathrm {p}}, \omega _{\mathrm {d}} \in [0,1]$ be the thresholds above which a player knowing that the probability of a card being useful, resp. worthless, exceeds the threshold, it is played, resp. discarded, in step 1, resp. 2. Moreover, one can choose whether or not to take any risk to end the game on three errors by prohibiting to play cards which are not certainly useful after having made two errors; this is referred to as “safe play”.

In step 3, we can follow different guidelines which determine the hint to be given. We consider four of them: (#1) random; (#2) giving a hint that gives information on the largest number of cards; (#3) giving a hint on the next useful card in sight or on the largest number of cards if no useful card is seen; or (#4) giving a hint on the next useful card or on the next worthless card if no useful card is available or otherwise on the largest number of cards.

Finally, in step 4, we choose from four different rules by which the card to be discarded is chosen. There are: (#1) random; (#2) discarding the card of which it is most certain that it is worthless; (#3) discarding the card which has been stored in hand the longest; or (#4) discarding the card of which it is most certain that it is not absolutely necessary to complete all stacks (a card which is not “indispensable”, according to [5]).

In addition to these choices, we also explore whether it is profitable to sometimes swap the order of steps 3 and 4. We let $\omega _{\mathrm {h}} \in [0,1]$ be the probability that this is done during a turn.

To test the various settings of the parameters, we let three players using the same strategy play 10,000 different starting configurations. Every configuration is played ten times in order to account for the randomness in the strategies. A test run of this kind takes approximately one minute on a computer with an Intel i7 2.3 GHz core and 6 GB of RAM.

First, we take $\omega _{\mathrm {d}} = \omega _{\mathrm {h}} = 1.0$, and vary $\omega _{\mathrm {p}}$ in $\{0.5,0.6,0.7\}$. The average scores for all sixteen combinations of hint and discard rules are shown in Table 1. It is apparent that the third hint rule, giving a hint on the next useful card or otherwise on the largest number of cards, is dominant for this setting of the parameters. Surprisingly, discard rules #1 and #2 are about as effective, meaning that discarding randomly is competitive with discarding the card of which we most think it is worthless.

Table 1. Scores obtained with different rules and varying $\omega _{\mathrm {p}}$, for $\omega _{\mathrm {d}} = \omega _{\mathrm {h}} = 1.0$, playing safely.

Now, we vary the other parameters. For a full overview of the results, see [2]. It turns out that for the combination of hint rule #3 and discard rule #2, the best results can be obtained. In Fig. 2 the results for varying $\omega$-thresholds can be found, playing safely: no more risk is taken when playing cards after two errors have been made. We see (also in Table 1) that the best average score obtained is 15.4 when taking $\omega _{\mathrm {p}} = 0.6$, $\omega _{\mathrm {d}} = \omega _{\mathrm {h}} = 1.0$ and playing safely. Apparently, it is profitable to try and play a card once we are 60% sure that it will be correct (unless we have already made two errors in which case we require certainty), and only discard a card if it is certain that it is worthless and give a hint if possible otherwise.

Fig. 2.

Scores obtained using hint rule #3 and discard rule #2, for varying $\omega$-thresholds.

For the Monte Carlo strategy, the basis is well-known. In every turn, we try every action after which the game is played out by random players many times. Each of these random games is evaluated in some way, after which we choose to do the action which led to the best score.

In the implementation for Hanabi, special care has to be taken in at least two situations. First, note that by trying to play a card and seeing how this turns out, a player could illegitimately obtain information on this card. Therefore, when trying to play a card in the Monte Carlo phase, the hand of the active player is shuffled through the deck and a new hand is dealt which is consistent with all hint information obtained so far. This way, the factual information on the cards is stored without allowing the agent to cheat. However, information on the exact hints that were given and the time at which these were given is lost, which possibly results in the agent not picking up implied hints, e.g., a hint pointing out a card actually meaning that it can successfully be played. This is in sharp contrast with the inner working of the strategies as described in [7]. Still, much progress could be booked for this determinization step by better judging or even learning the probability distribution of the cards in the hand of the active player, incorporating the hints provided so far.

Second, a truly random player will end the game on three errors with high probability. To circumvent this, we choose to not end the game after three errors have been made in the play-out phase. Moreover, we prohibit the random player from playing a card of which it is certain that it cannot be played based on the hints received so far. If the random player knows that none of the cards in his/her hand can be played, he/she will always randomly discard a card or give a hint. If he/she knows that only some of his/her cards are useless, he/she may also randomly try to play one of the other cards.

Contrary to the standard implementation of Monte Carlo, we do not evaluate each of the play-outs on the final score obtained. Instead, we register for the next D turns the amount of new points obtained as well as the amount of errors made with respect to the current turn in each play-out. For each new point we administer a $+1$ and for the k-th error we administer a score of $-k$. If at least three errors were made, we administer an extra penalty of $-2$ (this value was chosen based on experimentation). Experiments showed that without this additional penalty, the average score of the Monte Carlo player improves slightly, but the number of times the game ends with a score of 0 due to three errors being made, increases as well. We then compute a weighted sum over the scores for each of the D turns, counting later turns with (linear) higher weight, after which we pick the action with the highest average score among the play-outs. Note that taking the value $D = 1$ would result in greedy play.

Furthermore, it turns out to be profitable to not let the random player choose an action uniformly at random. Indeed, a Monte Carlo player using this strategy achieves on average a score of approximately 8.5 (with representative values like 1000 play-outs and $D=5$); note that such a player is unable to differentiate between possible hint moves.

The enhanced random player first decides whether to play, discard or give a hint, with probability 1/6, 1/6 and 2/3, respectively (the chosen values may benefit from further tuning). If one or more of these action types cannot be chosen at this time, the probabilities of selecting the remaining action types keep the same proportion to each other. For example, if the player knows none of his/her cards can be played, the probability of choosing a discard and hint action will be 1/5 and 4/5 respectively. If the random player decides to play a card, he/she favors playing a card which he/she deems likely to be playable, based on the number of possible cards that still remain in play and still conform to the information gained through hints. Similarly, when discarding a card, the random player favors a card which is likely to remain useless for the rest of the game, and favors keeping those cards that are likely to be the last remaining copy of that card and need to still be played later in the game to achieve the maximum score. Finally, hints are chosen uniformly at random.

In order to test the performance of the Monte Carlo player, we let three players using this strategy play 500 different starting configurations. A test run of this kind takes approximately three hours on a computer with an Intel i7 2.3 GHz core and 6 GB of RAM, when using 1000 play-outs and a depth of 10.

In the experiments, the results of which can be seen in Fig. 3, we have varied the amount of turns D that the Monte Carlo player takes into account and the number of play-outs allowed per different action. Surprisingly, the value of D taken, when not being too small, does not fundamentally affect the obtained score. Furthermore, while raising the amount of allowed play-outs does seems to somewhat improve the average score obtained, the differences are relatively small.

In fact, we note that it is not unlikely that these differences are to be contributed for a great part to random chance. Indeed, the standard deviation in the scores obtained seems to be quite high, with scores as low as 6 (or sometimes even 0) and as high as 24 being observed with an average around 15 points; the highest average score obtained is 16, with 1000 play-outs. We also see this large range in the scores obtained by the rule-based strategy discussed earlier in this section. Apparently, as the results from [5] show, substantial improvements can be reached by using more than the literal meaning of the hints. As discussed in the previous section and in [1], it is sometimes impossible to score a perfect game with 25 points. An easy example is a game where many 1 s and/or 2 s arrive late in the deck.

A provisional conclusion could be that, with the current settings, rule-based players and Monte Carlo players show similar performance, where the latter perform a little better. The results from [7] for the two-player version, using online learning, even seem to be a little better than for the three-player version.

Fig. 3.

Scores obtained using 1000 play-outs and varying D (top), and using $D = 5$ and varying number of play-outs (bottom), both with standard deviation.

5 Conclusions and Further Research

In this paper we have examined some interesting aspects of the cooperative card game Hanabi, that has the special property that players can only see the cards in the hands of the other players — but not in their own. Hints provide partial information about the game states; in contrast with [5] we only consider the literal meaning of hints.

Even simplified versions of the game, like SingleHanabi, give rise to complicated combinatorial questions and corresponding formulas. Furthermore, we have shown that different game playing strategies offer promising results. In particular, Monte Carlo techniques require little game knowledge, but are capable of delivering high quality competitive players. One issue is the problem of the amount of information that is used during the play-outs. The simple rule-based players perform a little worse, even incorporating more game knowledge.

As further research we first mention the quest for a proof of the result from Sect. 3 using the principle of inclusion and exclusion, perhaps also providing links to more general theorems. It is also of interest to find (substantial) classes of games that are unplayable. For the Monte Carlo players, we like to combine the presented method with techniques as mentioned in [4], like UCT and/or information sets, meanwhile somehow learning knowledge. And finally, there is much potential in research regarding the hints system, as the results from [5] show; it clearly pays off to examine clever hint schemes.