<p><strong>Update:</strong> here used to be my solution for 16 points in 16 turns, but since then I've managed to find an even better solution with <a href="https://github.com/monk-time/splendor-fastest-win" rel="nofollow noreferrer">my bruteforce tool</a>. Still 15 turns only, but this feels so efficient that I doubt a 14 turn one exists. This also exploits the green/blue imbalance that I mentioned below.</p>
<p>16 points in 15 turns:</p>
<pre><code> 1. Take    blue,  red, ⚫ black
 2. Take    blue,  red, ⚫ black
 3. Take    blue,  red, ⚫ black
 4. Claim   blue card  (3 ⚫ black) + 0 pt
 5. Take    blue,  green,  red
 6. Claim   blue card  (4  red) + 1 pt
 7. Claim   blue card  (6  blue) + 3 pt
 8. Take    blue,  green,  red
 9. Take    blue,  green,  red
10. Take    blue,  green,  red
11. Claim   blue card  (2  blue, 2  green, 3  red) + 1 pt
12. Claim   blue card  (5  blue) + 2 pt
13. Claim   green card (7  blue) + 4 pt
14. Take    blue,  blue
15. Claim   green card (7  blue, 3  green) + 5 pt
</code></pre>
<p>One thing I've noticed is that there's an interesting disbalance among the 2 pt cards for 5+3 gems: only 2/5 cards (blue and green) give the bonus of the same color as one of the requirements. Feels like something that might yield an even better solution.</p>

CC BY-SA 4.0
2019-02-05T20:56:43.643Z
@John: Yes, I think I have the same version as the ones I have linked in the comment to the main post: https://i.imgur.com/12DIzjz.jpg
monk-time
CC BY-SA 4.0
2019-02-06T03:32:04.033Z
I am amazed that there is an imbalance on the 2 pt 5+3 cards. That is very unexpected.  Nice catch.  Can you exploit it?  Your above solution is very clever.
John
CC BY-SA 4.0
2019-02-06T15:16:24.680Z
I think you have a calc error in there ... I step through it a few times .. but keep getting different #'s at your step #8:  #1 (Stash: W,G,B)
#2 (Stash: WW,GG,BB)
#3 (Stash: WW,GG,BBB,U,R)
#4 (Stash: W,GG,B,U,R) (Perm: U)
#5 (Stash: G,U) (Perm: U)
#6 (Stash: G,UUU) (Perm: U)
#7 (Stash: G,UUUUU) (Perm: U)
#8 (Stash: G,U) (Perm: UU) (but you claim you have G,UU in stash .. which makes step #10 impossible) :(  Did I make a mistake? or are you off there?
Ditto
CC BY-SA 4.0
2019-02-06T18:31:03.697Z
@Ditto: You missed that I have not one but two blue cards after step #5, which allows me to save 2 blue chips out of 5 that I have on step #8. That leaves me with 2 blue chips.
monk-time
CC BY-SA 4.0
2019-02-06T21:06:59.043Z
@monk-time:  you're right!  Gotcha!  :)
Ditto
CC BY-SA 4.0
2022-10-25T16:25:05.217Z
@Acccumulation All current best solutions accumulate at most 7 cards in 15 turns, and you need minimum 8 cards to get a noble. It's very tight. My non-exhaustive brute-force checker found a 16 turns/19 pts solution with a blue/green noble when I ran it on only b/g cards: 0B3-0B12-0B113-0G113-2B5-2G35-3B6-4G7-5G37
monk-time
CC BY-SA 4.0
2021-08-12T03:52:46.313Z
The nobles don't get you to 15 points faster?
Acccumulation
CC BY-SA 4.0
2019-02-05T16:50:26.587Z
2023-07-31T15:29:48.290Z
2023-07-31T15:29:48.290Z
monk-time
<p>EDIT: fixed violation of limit on taking 2 of same gems (Forbidden when &lt;4 gems remaining).</p>

<p>First of all, @Arcanist Lupus has provided good answer with simple yet efficient approach and also provided link to cards table in the comments, so I want to thank them. But looking at the table I noticed that 4 pts. cards have better conversion rates than 5 pts. ones, and a lot of gems are lost when going straight for 5 pts. cards.</p>

<p>So, assuming all required cards are available, no nobles and no interference from other players, here is my solution:</p>

<p>Cards I'm going for: White 4 for 7 blacks; Blue 4 for 7 whites; Green 4 for 7 blues; Green 3 for 6 greens.</p>

<p>Steps:</p>

<hr>

<ol>
<li>+2 black</li>
<li>+2 black</li>
<li>+1 black, 1 green, 1 {any}</li>
<li>+1 black, 2 {any}</li>
<li>+1 black, 1 white, 1 green. Return 3 {any}</li>
<li>Claim White 4 card for 7 blacks. (1 white and 2 green remain in the stash)</li>
<li>+2 white</li>
<li>+2 white</li>
<li>+1 white, 1 blue, 1 green</li>
<li>Claim Blue 4 card for 6 whites + card. (1 blue and 3 green remain in the stash)</li>
<li>+2 blue</li>
<li>+2 blue</li>
<li>+1 blue, 1 green, 1 {any}. Return 1 {any}</li>
<li>Claim Green 4 card for 6 blues + card. (4 green remain in the stash)</li>
<li>+1 green, 2 {any}</li>
<li>Claim Green 3 card for 5 greens + card</li>
</ol>

<p>Done. 15 points in 16 turns.</p>

<ul>
<li>steps 3 and 4 are just about getting 2 blacks, since it's forbidden to get 2 at once at that point. Alternatively you can take golden tokens and secure couple of cards for yourself, in that case in step 15 just take 2 greens (you will have 1 less, so it would be legal move).</li>
</ul>

CC BY-SA 4.0
2019-02-01T18:27:28.133Z
This is good, but you've forgotten that you can only take 2 gems of the same color if there are 4 or more left, so taking 2 of the same 3 times in a row is impossible
Arcanist Lupus
CC BY-SA 4.0
2019-02-02T18:23:51.610Z
@ArcanistLupus Gotcha, forgot that rule. Only played the game once. Looking at the steps, I think only step 3 violates the rule, so it can be split into 2 "take 1 of each" to get same 2 blacks, which bumps total steps to 16. I'll edit after more thorough check.
Deo
CC BY-SA 4.0
2019-02-01T16:04:01.697Z
2019-02-05T16:52:21.857Z
2019-02-05T16:52:21.857Z
Deo
<p>The top answer by Marcin_smu does not work in two player games. On move 10, he picks up 2 white chips but has one in his hand, in 2 player this is not possible. In 3 player or 4 player, it would be possible.</p>
<p>I have 15 points in 15 turns for 2 player:</p>
<ol>
<li>1 blue + 1 black + 1 red (0 points, 1u1b1r, 0 cards)</li>
<li>1 blue + 1 black + 1 red (0 points, 2u2b2r, 0 cards)</li>
<li>1 blue + 1 black + 1 red (0 points, 3u3b3r, 0 cards)</li>
<li>Claim the 3 black for a blue card (0 points, 3u3r, 1 blue)</li>
<li>1 blue + 1 green + 1 red (0 points, 4u1g4r, 1 blue)</li>
<li>Claim the 4 red for a blue card plus 1 point (1 points, 4u1g, 2 blue)</li>
<li>Claim the 6 blue for a blue card plus 3 points card (4 points, 1g, 3 blue)</li>
<li>1 blue + 1 blue (4 points, 2u1g, 3 blue)</li>
<li>Claim the 5 blue for a blue card plus 2 points card (6 points, 1g, 4 blue)</li>
<li>1 blue + 1 blue (6 points, 2u1g, 4 blue)</li>
<li>1 blue + 1 green + 1 white (6 points, 3u1g1w, 4 blue)</li>
<li>Claim 7 blue for a green card plus 4 points (10 points, 1g1w, 4 blue 1 green)</li>
<li>1 blue + 1 blue (10 points, 2u1g1w, 4 blue 1 green)</li>
<li>1 blue + 1 green + 1 white (10 points, 3u2g2w, 4 blue 1 green)</li>
<li>Claim 7 blue 3 green for a green card plus 5 points (15 points, 2w, 4 blue 2 green)</li>
</ol>
<p>note: a duplicate version of this could be done with green but not red, black or white because the 2 point 5 chip card is not monochrome</p>

CC BY-SA 4.0
2022-03-01T15:22:15.450Z
As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer).
Community
CC BY-SA 4.0
2022-03-01T14:57:10.257Z
2022-03-01T15:26:24.517Z
2022-03-01T15:26:24.517Z
Mark A
<p>Each color has a single card worth 5 points, which can be taken by 7 gems of one color and 3 of another.  Conveniently, the maximum number of gems you can hold at once is 10, and in a 4 player game there are 7 gems of each color available, which means that you can go after these cards immediately.</p>

<h2>For example:</h2>

<ol>
<li>2 red  </li>
<li>2 red  </li>
<li>1 red, 1 black, 1 {any}  </li>
<li>1 red, 1 black, 1 {any}  </li>
<li>1 red, 1 black, 1 {any}, return 3 {any}  </li>
<li>Claim the 5 pt black card  </li>
<li>2 black</li>
<li>2 black</li>
<li>1 black, 1 white, 1 {any}</li>
<li>1 black, 1 white, 1 {any}</li>
<li>1 white, 1 <strong>blue</strong>, 1 {any}, return 3 {any}</li>
<li>Claim the 5 pt white card (keep the blue)</li>
<li>2 blue</li>
<li>2 blue</li>
<li>1 blue, 1 white, 1 {any}</li>
<li>1 blue, 1 white, 1 {any}</li>
<li>Claim the 5 pt blue card.</li>
</ol>

<p>Congratulations!  You've won!</p>

<p>It's possible that there's a faster Nobles strategy, but I'm not sure.  3 Nobles requires at least 12 turns of taking cards, which means you have a total of 4 spare turns for gem collecting (5 to tie my score, 4 to win outright), and it takes three gem turns to get your first 3 cards.  Plus you have to stay in cards that produce only 3 different colors.</p>

CC BY-SA 4.0
2019-01-31T07:05:15.720Z
Actually, I think you can shave off a turn by saving a gem in the second round, but it's too late tonight to lay it out completely.
Arcanist Lupus
CC BY-SA 4.0
2019-01-31T13:04:25.963Z
nice upper bound, I think you can do better than 21
Cohensius
CC BY-SA 4.0
2019-01-31T15:07:06.877Z
@Cohensius You are indeed correct!  Especially since I couldn't math, and it turns out that 6*3 is 18, not 21
Arcanist Lupus
CC BY-SA 4.0
2019-01-31T19:01:03.303Z
"Conveniently, the maximum number of cards you can hold in your hand is 10" Cite? Are you confusing cards with tokens?
Acccumulation
CC BY-SA 4.0
2019-01-31T20:09:58.357Z
@Acccumulation fixed
Arcanist Lupus
CC BY-SA 4.0
2019-02-02T00:45:30.713Z
@Cohensius Fixed
Arcanist Lupus
CC BY-SA 4.0
2019-02-04T16:52:56.380Z
@Cophensius 4 Green + 4 Blue + 4 Red card gems, is 12 cards.  What am I missing?
John
CC BY-SA 4.0
2019-02-08T20:14:09.837Z
must been very late when I wrote the comment, indeed 4 Green + 4 Blue + 4 Red card gems, is 12 cards.
Cohensius
CC BY-SA 4.0
2019-01-31T06:32:22.823Z
2019-02-07T20:27:25.060Z
2019-02-07T20:27:25.060Z
Arcanist Lupus
<p>As a partial solution, here are some of the qualities I suspect will need to be part of such a "perfect game":</p>

<ol>
<li><p>All Noble tiles claimed (since they represent a large source of points for no additional action taken).</p></li>
<li><p>Very few turns where chips are taken, and a turning point somewhere in the middle of the game after which <em>no</em> chips are taken (since every turn taking chips is a turn you aren't taking cards which gain you points).</p></li>
</ol>

<p>Additionally, I suspect that the structure of the game will be something like (assuming a 2-player setup, so there are 9 points available in nobles):</p>

<ol>
<li><p>Spend a few turns taking chips and acquiring cheap level 1 cards.</p></li>
<li><p>Get enough level 1 cards that all future purchases are free, probably earning the first noble near the end of this step.</p></li>
<li><p>Get 1 or 2 level 2 cards worth 2-3 points, earning the second noble.</p></li>
<li><p>Get a level 3 card and the third noble.</p></li>
</ol>

<p>It might be possible to shortcut step 3 if you get a couple of the 1 point level 1 cards in step 2, but I don't think that there's a lot to be gained going that way (the 1 point cards are not particularly cost-effective).</p>

CC BY-SA 4.0
2019-01-31T12:46:56.373Z
Maybe I am missing this but this does not answer the questions of how fast can the game be won but is just a general strategy of how to win faster
Joe W
CC BY-SA 4.0
2019-01-31T22:32:09.110Z
It's me trying to theory-craft what the fastest game is likely to look like, since I haven't had the chance to actually sit down and turn the above "shape" of a game into actual turns. The fact that it might also be decent advice on winning faster is a neat corollary.
ConMan
CC BY-SA 4.0
2019-01-31T05:40:28.533Z
2019-01-31T05:40:28.533Z
ConMan
<p>I have found 15 points in 15 turns, which works even for 3 player rules:</p>
<ol>
<li>Take: white, blue, green</li>
<li>Take: white, blue, green</li>
<li>Take: white, blue, red</li>
<li>Claim: white card (3 blue)</li>
<li>Take: green, red, black</li>
<li>Claim: white card (2 red, 1 black)</li>
<li>Take: white, blue, green</li>
<li>Claim: white card (4 green) +1 pt</li>
<li>Claim: white card (6 white) +3 pt</li>
<li>Take: 2 white</li>
<li>Claim: blue card (7 white) +4 pt</li>
<li>Take: white, blue, black</li>
<li>Take: 2 white</li>
<li>Claim: blue card (7 white, 3 blue) +5 pt</li>
<li>Claim: green card (4 white, 2 blue, 1 black) +2 pt</li>
</ol>
<p>Although, I don't have proof that 14 turns game is not possible.</p>

CC BY-SA 4.0
2021-11-02T07:35:02.107Z
14 turns looks *really* close to possible but I haven't been able to quite get there. [This game](https://pastebin.com/gza5BVRt) would be a turn 14 win in 4p except it holds 11 gems one turn.
xnor
CC BY-SA 4.0
2022-01-22T09:16:33.570Z
Great work on shaving off 1 turn! I've been trying to bruteforce all shorter solutions to find the shortest path but can't quite optimize the code enough.
monk-time
CC BY-SA 4.0
2021-08-16T23:12:14.697Z
Given that it's been a year and a half to get a one turn improvement, I don't think you should feel bad that you don't have a 14 turn proof....
John
CC BY-SA 4.0
2021-08-11T14:21:51.030Z
2021-08-12T13:09:48.287Z
2021-08-12T13:09:48.287Z
Marcin_smu
<p>In the game of Splendor, what is the fewest number of turns you could take to get 15 points?</p>
<p><a href="https://i.sstatic.net/BobKC.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BobKC.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2019-01-31T05:23:24.890Z
I disagree. You can definitely figure out some kind of "perfect setup" and find the minimum number of turns based on that. It will never happen in practice (especially since you would have to assume either a solo game or a really stupid opponent), but it's a valid theoretical question.
ConMan
CC BY-SA 4.0
2019-01-31T06:14:34.163Z
BGG has a [complete decklist](https://boardgamegeek.com/filepage/136731/full-splendor-card-list-wpics) if anyone wants to use it.
Arcanist Lupus
CC BY-SA 4.0
2019-02-05T17:32:40.250Z
On BGG there's also [a google sheet with the same data by luless](https://www.boardgamegeek.com/filepage/101565/all-cards#comment4655548) (make sure to scroll down, the one on the top of the page isn't correct) and [a group photo of all cards](https://www.boardgamegeek.com/image/2414647/splendor) in case anyone wants to double-check.
monk-time
<splendor>
What is the fastest possible game in Splendor?
John
# What is the fastest possible game in Splendor?

(Asked by: username_0 on 2019-01-31)

In the game of Splendor, what is the fewest number of turns you could take to get 15 points?

[![enter image description here](https://i.sstatic.net/BobKC.png)](https://i.sstatic.net/BobKC.png)

**Tags:** splendor

> **Comments:**
> **username_1**: I disagree. You can definitely figure out some kind of "perfect setup" and find the minimum number of turns based on that. It will never happen in practice (especially since you would have to assume either a solo game or a really stupid opponent), but it's a valid theoretical question.  
> **username_2**: BGG has a [complete decklist](https://boardgamegeek.com/filepage/136731/full-splendor-card-list-wpics) if anyone wants to use it.  
> **username_3**: On BGG there's also [a google sheet with the same data by luless](https://www.boardgamegeek.com/filepage/101565/all-cards#comment4655548) (make sure to scroll down, the one on the top of the page isn't correct) and [a group photo of all cards](https://www.boardgamegeek.com/image/2414647/splendor) in case anyone wants to double-check.  

---

## 6 answers

### Answer by username_11 on 2021-08-11. Score: 8 | Accepted answer

I have found 15 points in 15 turns, which works even for 3 player rules:

1. Take: white, blue, green
2. Take: white, blue, green
3. Take: white, blue, red
4. Claim: white card (3 blue)
5. Take: green, red, black
6. Claim: white card (2 red, 1 black)
7. Take: white, blue, green
8. Claim: white card (4 green) +1 pt
9. Claim: white card (6 white) +3 pt
10. Take: 2 white
11. Claim: blue card (7 white) +4 pt
12. Take: white, blue, black
13. Take: 2 white
14. Claim: blue card (7 white, 3 blue) +5 pt
15. Claim: green card (4 white, 2 blue, 1 black) +2 pt

Although, I don't have proof that 14 turns game is not possible.

> **Comments:**
> **username_0**: Given that it's been a year and a half to get a one turn improvement, I don't think you should feel bad that you don't have a 14 turn proof....  
> **username_12**: 14 turns looks *really* close to possible but I haven't been able to quite get there. [This game](https://pastebin.com/gza5BVRt) would be a turn 14 win in 4p except it holds 11 gems one turn.  
> **username_3**: Great work on shaving off 1 turn! I've been trying to bruteforce all shorter solutions to find the shortest path but can't quite optimize the code enough.  

---

### Answer by username_3 on 2019-02-05. Score: 6

Update: here used to be my solution for 16 points in 16 turns, but since then I've managed to find an even better solution with my bruteforce tool. Still 15 turns only, but this feels so efficient that I doubt a 14 turn one exists. This also exploits the green/blue imbalance that I mentioned below.
16 points in 15 turns:
 1. Take    blue,  red, ⚫ black
 2. Take    blue,  red, ⚫ black
 3. Take    blue,  red, ⚫ black
 4. Claim   blue card  (3 ⚫ black) + 0 pt
 5. Take    blue,  green,  red
 6. Claim   blue card  (4  red) + 1 pt
 7. Claim   blue card  (6  blue) + 3 pt
 8. Take    blue,  green,  red
 9. Take    blue,  green,  red
10. Take    blue,  green,  red
11. Claim   blue card  (2  blue, 2  green, 3  red) + 1 pt
12. Claim   blue card  (5  blue) + 2 pt
13. Claim   green card (7  blue) + 4 pt
14. Take    blue,  blue
15. Claim   green card (7  blue, 3  green) + 5 pt

One thing I've noticed is that there's an interesting disbalance among the 2 pt cards for 5+3 gems: only 2/5 cards (blue and green) give the bonus of the same color as one of the requirements. Feels like something that might yield an even better solution.

> **Comments:**
> **username_3**: @username_0: Yes, I think I have the same version as the ones I have linked in the comment to the main post: https://i.imgur.com/12DIzjz.jpg  
> **username_0**: I am amazed that there is an imbalance on the 2 pt 5+3 cards. That is very unexpected.  Nice catch.  Can you exploit it?  Your above solution is very clever.  
> **username_4**: I think you have a calc error in there ... I step through it a few times .. but keep getting different #'s at your step #8:  #1 (Stash: W,G,B) #2 (Stash: WW,GG,BB) #3 (Stash: WW,GG,BBB,U,R) #4 (Stash: W,GG,B,U,R) (Perm: U) #5 (Stash: G,U) (Perm: U) #6 (Stash: G,UUU) (Perm: U) #7 (Stash: G,UUUUU) (Perm: U) #8 (Stash: G,U) (Perm: UU) (but you claim you have G,UU in stash .. which makes step #10 impossible) :(  Did I make a mistake? or are you off there?  
> **username_3**: @username_4: You missed that I have not one but two blue cards after step #5, which allows me to save 2 blue chips out of 5 that I have on step #8. That leaves me with 2 blue chips.  
> **username_4**: @username_3:  you're right!  Gotcha!  :)  
> **username_5**: The nobles don't get you to 15 points faster?  
> **username_3**: @username_5 All current best solutions accumulate at most 7 cards in 15 turns, and you need minimum 8 cards to get a noble. It's very tight. My non-exhaustive brute-force checker found a 16 turns/19 pts solution with a blue/green noble when I ran it on only b/g cards: 0B3-0B12-0B113-0G113-2B5-2G35-3B6-4G7-5G37  

---

### Answer by username_6 on 2019-02-01. Score: 5

EDIT: fixed violation of limit on taking 2 of same gems (Forbidden when <4 gems remaining).

First of all, @username_2 has provided good answer with simple yet efficient approach and also provided link to cards table in the comments, so I want to thank them. But looking at the table I noticed that 4 pts. cards have better conversion rates than 5 pts. ones, and a lot of gems are lost when going straight for 5 pts. cards.

So, assuming all required cards are available, no nobles and no interference from other players, here is my solution:

Cards I'm going for: White 4 for 7 blacks; Blue 4 for 7 whites; Green 4 for 7 blues; Green 3 for 6 greens.

Steps:

---

1. +2 black
2. +2 black
3. +1 black, 1 green, 1 {any}
4. +1 black, 2 {any}
5. +1 black, 1 white, 1 green. Return 3 {any}
6. Claim White 4 card for 7 blacks. (1 white and 2 green remain in the stash)
7. +2 white
8. +2 white
9. +1 white, 1 blue, 1 green
10. Claim Blue 4 card for 6 whites + card. (1 blue and 3 green remain in the stash)
11. +2 blue
12. +2 blue
13. +1 blue, 1 green, 1 {any}. Return 1 {any}
14. Claim Green 4 card for 6 blues + card. (4 green remain in the stash)
15. +1 green, 2 {any}
16. Claim Green 3 card for 5 greens + card

Done. 15 points in 16 turns.

* steps 3 and 4 are just about getting 2 blacks, since it's forbidden to get 2 at once at that point. Alternatively you can take golden tokens and secure couple of cards for yourself, in that case in step 15 just take 2 greens (you will have 1 less, so it would be legal move).

> **Comments:**
> **username_2**: This is good, but you've forgotten that you can only take 2 gems of the same color if there are 4 or more left, so taking 2 of the same 3 times in a row is impossible  
> **username_6**: @ArcanistLupus Gotcha, forgot that rule. Only played the game once. Looking at the steps, I think only step 3 violates the rule, so it can be split into 2 "take 1 of each" to get same 2 blacks, which bumps total steps to 16. I'll edit after more thorough check.  

---

### Answer by username_7 on 2022-03-01. Score: 1

The top answer by Marcin\_smu does not work in two player games. On move 10, he picks up 2 white chips but has one in his hand, in 2 player this is not possible. In 3 player or 4 player, it would be possible.

I have 15 points in 15 turns for 2 player:

1. 1 blue + 1 black + 1 red (0 points, 1u1b1r, 0 cards)
2. 1 blue + 1 black + 1 red (0 points, 2u2b2r, 0 cards)
3. 1 blue + 1 black + 1 red (0 points, 3u3b3r, 0 cards)
4. Claim the 3 black for a blue card (0 points, 3u3r, 1 blue)
5. 1 blue + 1 green + 1 red (0 points, 4u1g4r, 1 blue)
6. Claim the 4 red for a blue card plus 1 point (1 points, 4u1g, 2 blue)
7. Claim the 6 blue for a blue card plus 3 points card (4 points, 1g, 3 blue)
8. 1 blue + 1 blue (4 points, 2u1g, 3 blue)
9. Claim the 5 blue for a blue card plus 2 points card (6 points, 1g, 4 blue)
10. 1 blue + 1 blue (6 points, 2u1g, 4 blue)
11. 1 blue + 1 green + 1 white (6 points, 3u1g1w, 4 blue)
12. Claim 7 blue for a green card plus 4 points (10 points, 1g1w, 4 blue 1 green)
13. 1 blue + 1 blue (10 points, 2u1g1w, 4 blue 1 green)
14. 1 blue + 1 green + 1 white (10 points, 3u2g2w, 4 blue 1 green)
15. Claim 7 blue 3 green for a green card plus 5 points (15 points, 2w, 4 blue 2 green)

note: a duplicate version of this could be done with green but not red, black or white because the 2 point 5 chip card is not monochrome

> **Comments:**
> **username_8**: As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer).  

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### Answer by username_2 on 2019-01-31. Score: 0

Each color has a single card worth 5 points, which can be taken by 7 gems of one color and 3 of another. Conveniently, the maximum number of gems you can hold at once is 10, and in a 4 player game there are 7 gems of each color available, which means that you can go after these cards immediately.

## For example:

1. 2 red
2. 2 red
3. 1 red, 1 black, 1 {any}
4. 1 red, 1 black, 1 {any}
5. 1 red, 1 black, 1 {any}, return 3 {any}
6. Claim the 5 pt black card
7. 2 black
8. 2 black
9. 1 black, 1 white, 1 {any}
10. 1 black, 1 white, 1 {any}
11. 1 white, 1 **blue**, 1 {any}, return 3 {any}
12. Claim the 5 pt white card (keep the blue)
13. 2 blue
14. 2 blue
15. 1 blue, 1 white, 1 {any}
16. 1 blue, 1 white, 1 {any}
17. Claim the 5 pt blue card.

Congratulations! You've won!

It's possible that there's a faster Nobles strategy, but I'm not sure. 3 Nobles requires at least 12 turns of taking cards, which means you have a total of 4 spare turns for gem collecting (5 to tie my score, 4 to win outright), and it takes three gem turns to get your first 3 cards. Plus you have to stay in cards that produce only 3 different colors.

> **Comments:**
> **username_2**: Actually, I think you can shave off a turn by saving a gem in the second round, but it's too late tonight to lay it out completely.  
> **username_9**: nice upper bound, I think you can do better than 21  
> **username_2**: @username_9 You are indeed correct!  Especially since I couldn't math, and it turns out that 6*3 is 18, not 21  
> **username_5**: "Conveniently, the maximum number of cards you can hold in your hand is 10" Cite? Are you confusing cards with tokens?  
> **username_2**: @username_5 fixed  
> **username_2**: @username_9 Fixed  
> **username_0**: @Cophensius 4 Green + 4 Blue + 4 Red card gems, is 12 cards.  What am I missing?  
> **username_9**: must been very late when I wrote the comment, indeed 4 Green + 4 Blue + 4 Red card gems, is 12 cards.  

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### Answer by username_1 on 2019-01-31. Score: -2

As a partial solution, here are some of the qualities I suspect will need to be part of such a "perfect game":

1. All Noble tiles claimed (since they represent a large source of points for no additional action taken).
2. Very few turns where chips are taken, and a turning point somewhere in the middle of the game after which *no* chips are taken (since every turn taking chips is a turn you aren't taking cards which gain you points).

Additionally, I suspect that the structure of the game will be something like (assuming a 2-player setup, so there are 9 points available in nobles):

1. Spend a few turns taking chips and acquiring cheap level 1 cards.
2. Get enough level 1 cards that all future purchases are free, probably earning the first noble near the end of this step.
3. Get 1 or 2 level 2 cards worth 2-3 points, earning the second noble.
4. Get a level 3 card and the third noble.

It might be possible to shortcut step 3 if you get a couple of the 1 point level 1 cards in step 2, but I don't think that there's a lot to be gained going that way (the 1 point cards are not particularly cost-effective).

> **Comments:**
> **username_10**: Maybe I am missing this but this does not answer the questions of how fast can the game be won but is just a general strategy of how to win faster  
> **username_1**: It's me trying to theory-craft what the fastest game is likely to look like, since I haven't had the chance to actually sit down and turn the above "shape" of a game into actual turns. The fact that it might also be decent advice on winning faster is a neat corollary.  

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