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Because nobody knows the game-theoretical-optimal strategy in Hearts, I am looking for an approximation based on empirical data. For example, based on logs of games between expert players from an online server, or from records of championship events.

I'm particularly curious about the J♦︎ variant, and the difference in the shoting the moon ratio w/o the J♦︎ variant. In that variant, the J♦︎ is worth -10 points.

NOTE: This is not a duplicate of this question. That question doesn't qualify in what context the game is being played. This question does, and as such, it will have an objective answer, but only if someone has access to a decent data set of games.

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    @JoeW No, 'shooting the moon' involves getting all of the hearts and the queen of spades (all the cards that would normally give you points), and instead of you getting the points everyone else does. – diego Jul 8 '16 at 0:30
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    Possible duplicate of What are the chances of shooting the moon in Hearts? – Alex Robinson May 1 '17 at 8:30
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    This is not quite a duplicate of that question - that one asks what the probability is from a theoretical standpoint, and this from an empirical one. – Benjamin Cosman May 1 '17 at 15:56
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    I got in touch with Einar. He says their site only allows playing against computer opponents. So his data is not useful for this question. – dshin Aug 9 at 17:00
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    He told me his bots are simple and never try to shoot the moon. If they do so it is always by accident. Also he says that the bots are not good at detecting and stopping others from shooting the moon. – dshin Aug 9 at 21:25
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Hearts involves the concept of (A) deciding exactly 3 cards (at least in the common 4 player variant) to change between each hand in 3/4 of the hands and (B) to keep all cards in 1/4 of them. A major enough distinction that I think they probably need to considered in these 2 different groups. Lets consider what hand distributions capable of taking all cards would look like and consider group B as more important as it is easier to site and is based off of which players will decide which cards to pass from their hands anyway.

There are a total of 39 distinct hand suit patterns, there is a large scale experiment of computer shuffled hands and results here http://playbridge.com/pb_shuffle_project.php

These are suit independent so multiplication to increase these percentages is not needed, but certain constraints need to be taken into consideration and will reduce the percentages a significant amount. Basically to get the total you would need to add up the favorable results you get from each of the 39 hand patterns to get the total % of being able to shoot the moon.

Below this point might be considered a tedious explanation on how these numbers are derived and in excess of what the question asked for, but in case of more curious readers, I thought I would include the general process as it can be used of any of the 39 hand patterns :) Edit: section added after to account for hand passing as requested.

Lets illustrate with having 2 sample hand patterns for a key in what to look for, will do 5-4-3-1 and 7-3-3-0 as examples.

  1. 5-4-3-1, about 12.93% likely to get without caring which cards are in the suits. There are 24 possible arrangements between the suits in this chosen hand pattern, lets add what restrictions we need 1 suit at a time.

For the 1 card suit we have 13 combinations of those, in 6 of the 24 suit arrangements this suit is clubs and does not need any restriction as a point card cannot be played on the first trick, for the other arrangements having the ace is safest and so would need to divide by 13. For the 3 card suit we have 13 choose 3 combinations of those (or 286) and only 1 of them realiably has us winning all tricks (akq) and so would need to divide by 286 for all other arrangements. For the 4 card suit we have 13 choose 4 combinations (or 715) and akqj would need needed a high percentage of the time so will need to divide by 715 For the 5 card suit we can finally have a little leeway; 13 choose 5 combinations (or 1287) and the suit could reliably run with just akqj, having those 4 cards accounted for we have 9 choose 1 combinations left so basically divide by 1287 and multiply by 9.

This leaves the result looking visually like this: 5-4-3-1: 12.93% A. 3.233% of which has 1 club.
B. 9.967% other arrangements

A. 3.233%/286/715/1287*9 or 1.11e-7 % or 11.1e-8 %

B. 9.967%/13/286/715/1287*9 or 2.55e-8 %

for a total of 1.365e-7 %

  1. 7-3-3-0, about 0.2652% likely without caring about which suit or which cards. There are only 12 possible suit arrangements 6 of which have three clubs for instance, again going 1 suit at a time.

For the suit with no cards in it, no restrictions as nothing can be done about it anyway

For the suits with 3 cards 13 choose 3 (287) if one of them is clubs, (edit: this also holds true if the hand has no clubs at all) that suit can have ak and any of the 11 cards, the second 3 card suit for safety should probably be akq; thus the need to divide by 287 twice.

For the 7 card suit, we have 13 choose 7 (1716) combinations, since the suit is substantially longer, we probably only need akq for a relaible run most of the time, akqj for a nearly everytime running leaving 10 choose 4 (210) or 9 choose 3 (84) open slots.

So In total

0.2652% /287/287/1716*210 or 3.94e-7 %


Hand passing section:

For the cases where 3 cards are being passed, people try to shorten their already short suits and lengthen their strong ones. It can be helpful to pair your hand with that of the person passing to you, combining their suit lengths with yours, below is a table of the 44 fit classes breifly mentioned in the comment, the first 10 are by far the most common and in cases of passing across cannot be changed by the passing, placing a logical upper bound on how uneven the suits can be distributed.

(from Thomas Andrews Deal 3.1 documentation) There are 103 fit patterns, or 65 if we consider our fit pattern and opponent's as the same (e.g., that 8-6-6-6 is the same as 7-7-7-5. If the sum of the longest fit and the shortest fit is 13, then the pattern is self-dual, for example, if our fit pattern is 9-7-6-4, then so is the opponent's.)

Normalized | Squares sum | Patterns

     0 |          170 | 7-7-6-6
     1 |          172 | 8-6-6-6,7-7-7-5
     2 |          174 | 8-7-6-5
     4 |          178 | 9-6-6-5,8-8-5-5,8-7-7-4
     5 |          180 | 9-7-5-5,8-8-6-4
     6 |          182 | 9-7-6-4
     8 |          186 | 10-6-5-5,9-8-5-4,8-8-7-3
     9 |          188 | 10-6-6-4,9-7-7-3
    10 |          190 | 10-7-5-4,9-8-6-3
    12 |          194 | 10-7-6-3,9-9-4-4
    13 |          196 | 11-5-5-5,10-8-4-4,9-9-5-3,8-8-8-2
    14 |          198 | 11-6-5-4,10-8-5-3,9-8-7-2
    16 |          202 | 11-7-4-4,11-6-6-3,10-7-7-2,9-9-6-2
    17 |          204 | 11-7-5-3,10-8-6-2
    18 |          206 | 10-9-4-3
    20 |          210 | 12-5-5-4,11-8-4-3,11-7-6-2,10-9-5-2,9-8-8-1
    21 |          212 | 12-6-4-4,9-9-7-1
    22 |          214 | 12-6-5-3,11-8-5-2,10-8-7-1
    24 |          218 | 12-7-4-3,10-10-3-3,10-9-6-1
    25 |          220 | 12-6-6-2,11-9-3-3,11-7-7-1,10-10-4-2
    26 |          222 | 12-7-5-2,11-9-4-2,11-8-6-1
    28 |          226 | 13-5-4-4,12-8-3-3,10-10-5-1,9-9-8-0
    29 |          228 | 13-5-5-3,12-8-4-2,11-9-5-1,10-8-8-0
    30 |          230 | 13-6-4-3,12-7-6-1,10-9-7-0
    32 |          234 | 13-6-5-2,12-8-5-1,11-10-3-2,11-8-7-0
    33 |          236 | 13-7-3-3,10-10-6-0
    34 |          238 | 13-7-4-2,12-9-3-2,11-10-4-1,11-9-6-0
    36 |          242 | 13-6-6-1,12-9-4-1,12-7-7-0
    37 |          244 | 13-7-5-1,12-8-6-0
    38 |          246 | 13-8-3-2,11-10-5-0
    40 |          250 | 13-8-4-1,12-9-5-0,11-11-2-2
    41 |          252 | 12-10-2-2,11-11-3-1
    42 |          254 | 13-7-6-0,12-10-3-1
    44 |          258 | 13-9-2-2,13-8-5-0,11-11-4-0
    45 |          260 | 13-9-3-1,12-10-4-0
    48 |          266 | 13-9-4-0
    50 |          270 | 12-11-2-1
    52 |          274 | 13-10-2-1,12-11-3-0
    54 |          278 | 13-10-3-0
    60 |          290 | 12-12-1-1
    61 |          292 | 13-11-1-1,12-12-2-0
    62 |          294 | 13-11-2-0
    72 |          314 | 13-12-1-0
    84 |          338 | 13-13-0-0
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    While this is a good lower mathematical bound, there are several major flaws with this answer. First, this is only an analysis of keep hands, which are only a quarter of hands and are typically the hardest to shoot on. Second, this assumes you must win all the tricks beyond the first to shoot the moon, which is not the case; you only need win all the tricks on which points are played. Finally, it assumes you need to have the 4 highest cards in each suit, which is also not true; other people may get rid of high cards on other tricks or be holding just the king which they must play to your ace. – Zags Oct 20 at 17:00
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    Also, it is important to keep in mind that that there is a difference between being mathematically guaranteed to shoot the moon and actually shooting the moon. When shooting the moon, you frequently get several tricks of leeway where everyone is avoiding taking points because they don't know you are trying to shoot yet. And, if you hold the queen of spades in your hand until the last trick, people may not try to take the lead from you as they still risk 13 points for spoiling your moon shoot (this depends a lot on the current scores of the players). – Zags Oct 20 at 17:03
  • Yes, it's only an analysis of keep hands, even hands where you eventually pass 3 cards to another player first start out being dealt according to the table. One would then have to look at about which cards are statistically chosen by each player to be passed given the hand that was actually originally dealt to them, direction of passing being dependent in some cases. Using qualities such was favoring short suits high cards, clubs or not favoring low spades. Certainly you can add some more free cards or use second highest, etc. The original question asked for empirical data and is a good start – Toni Stack Oct 20 at 17:57
  • It might also be insightful that in the passing form of the game you can pair up your hand with the hand passing cards to you, combined they make up a fit pattern, there are 103 possible fit patterns divided into 44 different classes based on standard deviation, they are symmetric so that if you two have one fit class, the other 2 players have the exact same fit class. Example: you have 4 spades 2 hearts, 3 diamonds 4 clubs, the player on your left dealt 4 spades, 4 hearts, 3 diamonds 2 clubs with passing to right. together you two have 8 spades 6 hearts 6 Diamonds 6 Clubs or 8-6-6-6. – Toni Stack Oct 20 at 18:33
  • The opponents fit pattern would then be 7-7-7-5, which is in the same fit class as you. This is also the most likely of the 44 fit classes to be dealt out and passing strategies often use these to advantage. – Toni Stack Oct 20 at 18:38
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Undefined.

You have made the question a contradiction: There are no "perfectly playing agents", and thus there can be no empirical data on their results.

Reason: Hearts, like Contract Bridge, is a stochastic game of incomplete information: There is incomplete data throughout most of the play of each hand, meaning statistical likelihoods need to be estimated by each player for several rounds. There is no theoretic framework for such, and no evidence that even AI can do so in a foreseeable future.

Even in Contract Bridge, where additional information about the hidden hands is available from the bidding and one hand is face down on the table, there is o such thing as a perfectly playing agent, even considering just the play of the cards. The closest that is available is termed Double Dummy Analysis, which presents an assessment of makeable contracts when both opponent hands are assumed visible for the play.

So, until such time as a perfectly playing agent has at least been constructed for Contract Bridge, a game with one hand on the table for all players to see, there can be no such thing in the much more challenging conditions of a Hearts game, with all opponents always hidden.


Addendum

In card games such as Hearts and Contract Bridge, one of the most powerful analysis techniques is that of:

  • drawing correct inferences from what other players have not done;

  • then from an assessment of their habits and presumed skill making an assessment of holdings they don't have; and

  • from this finally creating an assessment of what they might have by removing the eliminated possibilities.

This is an extremely challenging skill to master, even in Contract Bridge where at least one partnership, and often both, have participated in the auction and described their hands in considerable detail to both partnerships - as there are no secret bidding or play agreements in Bridge.

If this cannot even be approached perfectly in Bridge, imagine how much harder it will be in Hearts, where only the three cards you received on the Pass is available information about the other hands when the play starts.

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    John Nash proved the existence of a perfect strategy in hearts in 1950. You can read the proof here: en.m.wikipedia.org/wiki/Nash_equilibrium – dshin Aug 10 at 2:15
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    Also, in 2019, being a stochastic game of imperfect information is no longer a reason the assume a game is intractable for AI. Poker shares these same characteristics and much progress has been made recently in that area leading to superhuman agents. – dshin Aug 10 at 2:22
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    Also, I did not ask for empirical data between games played by perfect agents. Of course no such data can exist, as such agents do not exist, and probably never will. I’m asking for an educated guess based on the assumption that current world class players might shoot the moon at similar frequencies as perfect agents (which again are known to be theoretically possible due to John Nash’s 1950 proof). – dshin Aug 10 at 2:25
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    @JoeW Imperfect information is a rigorously defined mathematically concept. Poker is an imperfect information game according to this definition. – dshin Aug 10 at 2:28
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    @dshin but you have a a massive amount of information in poker compared to hearts. – Joe W Aug 10 at 2:34

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