In Hearts, we just want to avoid getting any heart card or the Queen of Spades. But if we can get all 14 cards, we shoot the moon and everybody else gets 26 points.

So, given a fair shuffle, what's the chance that we can actually get a shot to the moon?

There must be some probability, because if in a draw, we get 1 Ace of Clubs, and the hearts from 3 through Ace, we know there is no possible configuration where we'll ever NOT shoot the moon.

Whereas, if we get ALL the diamonds, we are doomed to "lose" every hand, because all our cards will always be different in suit from the first card. There is no possible configuration where we can shoot the moon.

If nobody can give me a number, what about a process of getting to this number? I imagine it has something to do with expected values, but I'm not very sure. But, it is an interesting question. I've managed to play a "perfect" game where I never took a single point, and some games where I've shot the moon once or twice, but never a game where I shot the moon 4 times in a row to win. Is this poor strategy, or a fault of bad luck?

closed as too broad by Joe W, Toon Krijthe, Zags, np8, Forget I was ever here May 10 '15 at 19:54

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    Much depends, in real lfe, on the skill and strategy of the other players; are you assuming that they attempt to minimise your chance of shooting the moon, or to minimise their score? Also, limited information plays havoc with precise probabilities. – TimLymington Mar 6 '13 at 11:23
  • 1
    I played Microsoft Hearts obsessively at one point until I managed to shoot the moon four times consecutively for the supposed "perfect score" of 104-0-0-0, and then gave it up forever. I can't remember why I felt the need to do such a thing, and it did take quite a while! – thesunneversets Mar 6 '13 at 12:02
  • I guess, we should assume, for simplicity's sake, that everybody has perfect information and plays a perfect game -- everybody knows everybody else's cards and is working to either minimize their score OR shoot the moon. @thesunneversets, that's quite impressive! – markovchain Mar 6 '13 at 12:52
  • 2
    @markovchain with perfect information it then becomes a game of brinkmanship, I'm not going to take a heart when I could wait for you to take one. Players would rapidly discard cards which would allow them to prevent a shoot-the-moon in order to force their opponents to take hearts to avert it. – Nick Mar 6 '13 at 13:25
  • Yes, but eventually someone will have to take the hearts. With both perfect information and a perfect game, wouldnt we be able to know if a shot to the moon is possible or not, given only the initial hands, even if nobody has played a single card yet? – markovchain Mar 6 '13 at 13:45

It's certainly possible to calculate which hands can Shoot the Moon with 100% probability, but it's not going to be easy.

There are 635,013,559,600 different starting hands of 13 cards from a 52 card deck.

We know that if we are dealt all 13 cards of one suit, there's 4 hands out of 635,013,559,600 that will shoot the moon ;)

If we have 12 cards of one suit and one ace in another suit, there are 13 different ways to get 12 cards of one suit, times 3 different other ace cards in other suits, times 4 different suits so: 13*3*4 = 156 winning hands.

Of course if we have an ace and king of the other suit and the rest one suit, it's also a winning hand — wait it has to be an Ace-King in the long suit too because we can't risk an opponent having a king plus another card in our long suit.

Well, this is getting difficult soon. Thankfully, Bridge players have already figured this out for us. :) They have a concept called the Losing-Trick Count, a method of hand evaluation for counting the number of losing tricks held by a partnership. To "shoot the moon" we need a Losing-Trick Count of Zero.

According to this table I found on Wikipedia, there are 4,245,032 hands with a losing trick count of zero. As a check I summed up the number of hands in that table and it equals 635,013,559,600 exactly. Whew! Sounds right!

Therefore, your probability of Shooting the Moon with 100% accuracy is at a bare minimum .000668% percent.

HOWEVER… Trick points are scored differently in Hearts as compared to Bridge. You can lose tricks but still take all the hearts and the queen of spades. So Ace-King of Spades + AKQJT98 of Hearts plus 6 other losers is a winning hand. I expect about half of the 90,206,044 hands with a LTC of 1 are shoot the moon winners since if the losing trick is a club or diamond, at that point it doesn't matter. (Presuming the winning tricks have forced the Queen and hearts. Things get murky.)

So, I expect the number of perfect shoot the moon hands is significantly more than the .000668% percent of Bridge hands that are perfect. You can add at least .0071%, and then a bit more as you start chipping away at those losing tricks that no longer matter.

My brain is starting to hurt now.

I think I'll go lie down and hopefully this gives you an idea of how to start to get the answer. It's important to realize that the reason shooting the moon isn't as rare as you'd expect is because even a marginal hand has a decent chance of shooting the moon, for your opponents don't have perfect knowledge. If an opponent has a king-queen in one suit and a king-10 in the other, which king do they keep to stop your attempt?

And of course, at first most opponents don't expect players to go for the moon, and are trying to get rid of trick taking cards when they can. That's what makes Hearts so interesting in my view, the psychology of being wary of those greedy moon targetters. ;)

  • 3
    I don't know if this is the right answer, but it's an automatic +1 just for effort... – thesunneversets Mar 6 '13 at 18:09
  • 6
    Sadly, it's more complicated than it appears. 13 clubs is a successful (inevitable) moonshot, but 13 of anything else is automatic failure (admittedly you score no points unless someone else shoots the moon). Even your example hand could fail: ace of clubs, ace of diamonds on which somebody discards a small heart, and a heart lead will mean you score all but one point. My own opinion, more empirically based than provable, is that the number of hands on which you can guarantee to shoot the moon is vanishingly small, but the number on which you can bluff or squeeze a result is quite large. – TimLymington Mar 6 '13 at 21:12
  • 1
    @TimLymington Gah, you are correct, I forgot to account for the fact that in Hearts, the 2 of clubs is always led first. In short, any guaranteed inevitable moonshot hand requires the Ace of Clubs. Your empirical opinion matches my experience. :) – ghoppe Mar 6 '13 at 22:08
  • Further research (and a cup of coffee) reveals that the Losing Trick Count only deals with AKQ. Reasonable in Bridge terms, but a suit AKQxx (LTC 0) is not certain to take all the tricks in Hearts; somebody with Jxxx will take the fourth round. This question (with OP's conditions) seems unanswerable, since there is no way but inspection to discover whether a particular hand fulfils the criteria (or that's what my cup of coffee says, anyway.) If you removed some of the specifics, such as Microsoft's rule that nobody can play a point card on the first trick, might Maths.SE be interested? – TimLymington Mar 6 '13 at 23:42
  • 1
    Your opponents play a key part in a lot of "shoot the moon" scenarios. Canny opponents will pass you A2H, which will almost certainly cause you to win a lot of hearts and lose at least one hearts trick. Furthermore, many players will hold off on getting rid of their big hearts until two players have collected points. Hearts is a game of damage minimisation though and sometimes it's better to let a losing player shoot the moon than to give them the points. – Stephen May 7 '15 at 23:09

Some general thoughts:

1) You have to have the Ace of clubs, to win the first trick.

After this:

2) You can win every other trick with some combination of winners in the other suits, i.e. the top 12 diamonds, the top 4 cards in every other suit, etc

Notice with enough length, you don't need to have only top cards...If you have the A,K,Q,J,10,9,X of Spades, X can be anything, because the previous high cards will pull out everything in that suit.

3)If you have AKQ of spades, and A,K,Q,J,10,9,X of hearts, you can just win with the hearts and the queen of spades, and lose all the rest (sluffing the AK of spades, even), as you will have all the points.

  • 2
    Actually, since there can be no points in the first trick, you don't need to win it. Nor do you need to win any trick up to the point where a heart could be discarded (or Black Maria is played). – TimLymington Dec 22 '15 at 11:38
  • In order to guarantee shooting the moon, you do. If you lose the first trick, and someone else leads the 2 of spades the second trick, and you have no spades, that's the end. – swbarnes2 Dec 22 '15 at 17:12
  • Fair enough. But then your hand 3) won't work: you win the spades, and then are not allowed to lead a heart, so you have to let someone else win a trick, possibly garnering a heart and leading back a heart. If you want a guarantee, you have to be sure to win all 13 tricks: simpler mathematically but very unlikely. – TimLymington Dec 22 '15 at 18:00
  • 2
    Can often shoot with far less strength as other players are generally trying to avoid getting points. – Lee Sep 23 '17 at 13:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.