# Is it possible to get the largest conceivable score in Hearts?

I should explain the necessary rules: in the four-player game of Hearts, the object is to get as few points as possible. The points you receive are determined by the cards you pick up through a hand: hearts are worth $1$ point each, and the queen of spades is worth $13$ points. Players can thus receive any score between 0 and 25 on a given hand. However, if on a hand a player obtains all 13 hearts and the queen of spades, then that player receives $0$ points and all other players receive $26$ points, an act referred to as "shooting the moon".

The game ends when a player exceeds $100$ points, at which point the player with the fewest points is the winner. In this way, the best conceivable score is

$$\begin{matrix}0&99&99&99\end{matrix} → \begin{matrix}0&125&125&125\end{matrix}$$

My question is, is this score possible? And, more generally, how can one determine whether any given score in Hearts is possible?

As best I can tell, we must determine whether $\left[\begin{matrix}99\\99\\99\end{matrix}\right]$ is a linear combination of all the possible final scores.

I would lean towards the idea that this is not possible; it is simple to produce even multiples of 13:

$$\begin{matrix}0&0&13&13 \\ 0&13&26&13 \\ 0&26&26&26\end{matrix}$$

But I cannot think of a good way to produce any given score.

• After each hand, the sum of all the scores must be a multiple of $26$. Commented Nov 2, 2014 at 20:50
• I am confused. The goal is to get as few points as possible, or not ? And if a heart counts 1 point, any score should be possible. You probably mean the highest possible score instead of the best score. Commented Nov 2, 2014 at 20:53
• @Peter, there are 26 points awarded total, between all the players, if the moon is not shot. If it is shot, then the total number of points awarded is a multiple of 26. Commented Nov 2, 2014 at 20:53
• @Peter The "best possible score" I gave is the highest possible winning margin for the player with 0. Commented Nov 2, 2014 at 20:54
• @vadim123 Well, that was trivially easy. Put that into an answer and I'll accept it. Commented Nov 2, 2014 at 20:54

After each hand, the sum of all the scores must be a multiple of 26.

• But that does not answer the question : what is the best possible performance for the winner ? Commented Nov 2, 2014 at 20:57
• I do not think that it is $0-104-104-104$ Commented Nov 2, 2014 at 20:58
• @Peter I found through some brief trial and error that the highest possible winning margin is $\begin{matrix}0&95&95&96\end{matrix} → \begin{matrix}0&121&121&122\end{matrix}$. Commented Nov 2, 2014 at 21:00
• $0-78-78-78$ $0-79-82-99$ $0-99-88-99$ $0-125-114-125$. What about that ? Commented Nov 2, 2014 at 21:06
• @Peter the sum of both our scores is still 286, so the margin is unchanged. Commented Nov 2, 2014 at 21:08

Suppose we have four desired scores $a_1, a_2, a_3, a_4$, with $a_1+a_2+a_3+a_4=26k$. Then I claim that the scores are attainable if and only if $$\lfloor a_1/13\rfloor+\lfloor a_2/13\rfloor+\lfloor a_3/13\rfloor+\lfloor a_4/13\rfloor \geq k \tag{*}$$

To see this, suppose first that nobody ever shoots the moon. Then we played $k$ hands, and so the queen of spades must have come up $k$ times. On the other hand, a player who scored $a$ points can have received the queen of spades at most $\lfloor a/13 \rfloor$ times, which shows that $(*)$ is necessary. Conversely, if $(*)$ holds, we can distribute the $k$ queens of spades among the four players in any way that doesn't cause their score to get too big, and fill in all the gaps with hearts.

Now, suppose $l$ hands involve people shooting the moon. Those hands have a total score of $3(26)=78$, while the remaining hands have a total score of $26$; thus the number of normally-scored hands is $k-3l$. Also suppose player $i$ shot the moon $s_i$ times. Then applying $(*)$ to the unshot hands gives

\begin{align} k-3l \leq &\lfloor a_1/13\rfloor-2(s_2+s_3+s_4)+\lfloor a_2/13\rfloor-2(s_1+s_3+s_4)\\&\quad+\lfloor a_3/13\rfloor-2(s_1+s_2+s_4)+\lfloor a_4/13\rfloor-2(s_1+s_2+s_3)\\=&\lfloor a_1/13\rfloor+\lfloor a_2/13\rfloor+\lfloor a_3/13\rfloor+\lfloor a_4/13\rfloor-6l \end{align} or equivalently $$\lfloor a_1/13\rfloor+\lfloor a_2/13\rfloor+\lfloor a_3/13\rfloor+\lfloor a_4/13\rfloor-3l \geq k$$ which is a strictly stronger condition than $(*)$ for $l>0$.

Note in particular that, if every player's score is either $0$ or at least $13$, then $(*)$ must hold: in that case rounding down to a multiple of $13$ cannot take away more than half of any score, and so cannot take away more than half of the total. This provides a quick way of seeing that the scores in the comments to @vadim123's answer are actually attainable.

Although the OP's rules are the standard rules, a lot of people instead play that shooting the moon loses you $26$ points. In this variant, any set of scores whose sum is a multiple of $26$ is attainable, as the last inequality becomes $$\lfloor a_1/13\rfloor+\lfloor a_2/13\rfloor+\lfloor a_3/13\rfloor+\lfloor a_4/13\rfloor+l \geq k$$ so we just have to make $l$ sufficiently large. (Intuitively, if everyone starts by shooting the moon a whole bunch, they'll have enough room to arrange their scores however they like.)