Jtdylktuy

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

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

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 $left[begin{matrix}99\99\99end{matrix}right]$ 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:

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

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.

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
$$
lfloor a_1/13rfloor+lfloor a_2/13rfloor+lfloor a_3/13rfloor+lfloor a_4/13rfloor geq k tag{*}
$$

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_1/13rfloor-2(s_2+s_3+s_4)+lfloor a_2/13rfloor-2(s_1+s_3+s_4)\&quad+lfloor a_3/13rfloor-2(s_1+s_2+s_4)+lfloor a_4/13rfloor-2(s_1+s_2+s_3)\=&lfloor a_1/13rfloor+lfloor a_2/13rfloor+lfloor a_3/13rfloor+lfloor a_4/13rfloor-6l
end{align}
or equivalently
$$
lfloor a_1/13rfloor+lfloor a_2/13rfloor+lfloor a_3/13rfloor+lfloor a_4/13rfloor-3l geq k
$$
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
$$
lfloor a_1/13rfloor+lfloor a_2/13rfloor+lfloor a_3/13rfloor+lfloor a_4/13rfloor+l geq k
$$
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.)