I want to reduce the level of chance associated with the game Greed (also known as Farkle and Zilch).

Has anyone looked into this or have a source that can give some good stats to answer the following questions. At what score should I stop rolling:

  • When I have 1 die left?
  • When I have 2 dice left?

Edit: ... to maximise my score and minimise risk. Assume that I've entered the game already (scored 500).

In other words: when is it worth rolling a single die to try to get a re-roll.


2 Answers 2


Greed (a.k.a. Farkle) has an extensive amount of statistical analysis done on the game.

Now that you have updated your question, here is my updated answer.

If you want to maximize your score, what you need to figure out is what is your expected score if you risk rolling, compared to your score if you choose not to roll. So, the first thing you need to know, is what is your expected number of points you will earn for rolling 6-dice. Below is a chart of the different possible Patterns (6-of-a-Kind, 5-of-a-Kind, etc.), the possible ways of ORD 'ering those particular patterns (which are all the same combination), the number of DIF 'erent patterns that can be made with other numbers from 1-6, and the Total permutations (which should add up to 6^6=46656).

|Pattern    |ORD |DIF|Total|
|a,a,a,a,a,a|   1|  6|    6| 6-Kind
|a,a,a,a,a,b|   6| 30|  180| 5-Kind (possible '1' or '5' score singles)
|a,a,a,a,b,b|  15| 30|  450| 4-Kind (possible '1' or '5' score singles) 
|a,a,a,a,b,c|  30| 60| 1800| 4-Kind (possible '1' or '5' score singles)
|a,a,a,b,b,b|  20| 15|  300| 2 * 3-Kind
|a,a,a,b,b,c|  60|120| 7200| 3-Kind (possible '1' or '5' score singles)
|a,a,a,b,c,d| 120| 60| 7200| 3-Kind (possible '1' or '5' score singles)
|a,a,b,b,c,c|  90| 20| 1800| 3 Pair
|a,a,b,b,c,d| 180| 90|16200| (possible farkle, possible '1' or '5' score)
|a,a,b,c,d,e| 360| 30|10800| (possible '1' or '5' score)
|a,b,c,d,e,f| 720|  1|  720| straight
|      Total|1602|462|46656|

I didn't separate these individual patterns by how much they actually score, because it would be way to big of a table. The table would have to differentiate from each different kind of 3-of-a-Kind, and possible '1' and '5' single die scores. I did separate them when running a calculation of the average (total points/permutations). The only important information from that computer analysis is:

  • 1080/46656 : The odds that when you roll 6 dice, you cannot score anything. (Farkle)

  • 388.033 : The average number of points that you make on your first roll of 6 dice, if you score.

You can use the above information, you can examine all possible outcomes of rolling a single die, and look at the risk (how many points you will lose if you fail to score) versus the amount of points that you will gain. You can think of this as you would any sort of gambling question.

  • Question: Imagine a game where you pay $X, and you roll a die. If you roll a '1' you win $100 plus your original bet ($X), if you roll a '5' you win $50+$X, if you roll anything else you lose your bet ($X). How much would you pay to play this game?

  • Answer: Anything up to $37.50. Until that point, you expect to win more money than you lose. The sum of your losses multiplied by their probability of happening (-37.5 * (4/6) = 25 = ((100*(1/6)+(50*(1/6)) is equal to the sum of your winnings multiplied by their probability of happening.

    1) +$100

    2) -$37.5

    3) -$37.5

    4) -$37.5

    5) +$50

    6) -$37.5

Now, lets add up all the possibilities for Greed assuming you have X points already, and you intend to roll six dice if you win the single die roll.

|Outcome              |Odds the Outcome Occurs | Points *Odds |
|Lose rolling one die |4/6                     | -4/6*(X)     |
|Win  50 +Avg six dice|(1/6)*(46656-1080)/46656|(AVG+50)*Odds |
|Win 100 +Avg six dice|(1/6)*(46656-1080)/46656|(AVG+100)*Odds|
|Lose rolling six dice|(2/6)*(1080/46656)      | -(5/648)*(X) |
                Total | 1.0                    |              |

Time for Some Algebra

Solve for the sum of all the above outcomes equals zero.

  • (add each line above together)

0 = -(4/6) * X - (5/648) * X + (2*AVG+50+100)*(1/6) * (46656-1080)/46656)

  • (multiply the first X term by 108 to get the same denominator for both X terms)

0 = -(432/648) * X - (5/648) * X + 150.771547

  • (Add X term to both sides, simplify two added terms)

(437/648) * X = 150.771547

  • (Solve for X)

    X = (648/437) * (150.771547) = 223.6 (approximately)

Since scores are always a multiple of 50, you should stop rolling when you hit 250 points. I leave the the 2 dice question as an exercise for the OP. You should have all the tools necessary to calculate it (I will check your work if you attempt to answer it. If you do attempt to answer it, post the question in a new thread linking back to this thread, and explain your work).

Optimal Strategy for Multiplayer

Another more difficult question to answer is what is the optimal strategy to win. The calculations for this are more complicated, because you have to take into account how likely the opponent is to win before you (their likely score before your next turn, etc.) I have no doubt that a nearly optimal Greed computer program can be created for (two players). As you increase the number of players though, the problem gets exponentially more difficult. You might want to check the links in this post that talks about how optimal Yathzee for one player was created, and how the two player version is pretty close to optimal.


The answers to your questions depend on the details of how you score the game. For example, Facebook Farkle scores three pairs as 750 points, Gaby Vanhegan's Zilch implementation scores it as 1500 points, and Wikipedia's entry on Greed scores it as 800. These and other scoring differences affect the expected value of rolls and ultimately the optimal play strategy.

To correctly answer your question for the case of the one die roll, you must not only consider the potential gains and losses of that immediate one die roll, but also the weighted value of ALL subsequent rolls. For example, if you roll the sixth die and get a 1 (for hot-dice) and then roll six dice and score three-pair (for another set of hot-dice), are you going to stop and bank? No! you're going to roll again! So the true value of rolling that sixth die by itself is more than the weighted value of that one roll; and is more than the weighted value of that roll plus the subsequent 6-die roll. To get it's true value, you must calculate the expected points for the remainder of the entire turn.

I've analyzed exactly this problem at my website. The analysis is straightforward, but too long to repeat here. I wrote a web-app called the Farkle Strategy Generator (FSG) which allows you to generate farkle strategies that provide exacting answers to your questions. And here they are...

1 and 2 die rolling rules for Farkle

Setting the FSG for Farkle scoring and overriding the banking theshold to zero you find:

  • 2 dice: bank when you have 250 or more points.

  • 1 die: bank when you have 300 or more points.

But the min-bank threshold is actually 300 points, so the above rules boil down to just: Never roll one or two dice unless the minimum banking threshold forces you to.

1 and 2 die rolling rules for Zilch

  • 2 dice: bank when you have 250 or more points.

  • 1 die: bank when you have 350 or more points.

Again, the 300 point banking threshold will prevent you from banking 250 points with 2 dice to roll, but the 1 die rule is more interesting as it suggests you roll when you have 300 points (which is enough to bank).

I suggest you run the FSG yourself to generate the strategy customized to your own scoring rules.

Why my answer is different

user1873's analysis says to bank when you have 250 points and one die to roll, but the strategy produced by my FSG says to roll in this situation. Why the difference? While the analysis presented by user1873 is both instructive and a good estimator of how to play to maximize expected scores; it does not consider the possibility of rolls subsequent to the 6-die roll and neglects the weighted contribution of these positive events. If you correctly calculate these higher-order effects, you find that it is actually worthwhile to roll one die when you have 250 points for the turn.

  • I agree, I made an error. The 100 or 50 points from the previous roll should have been multiplied by the odds of the event occurring, not added in. I will correct it soon. +1 for catching my error.
    – user1873
    Commented Jul 24, 2012 at 12:19
  • Cool. If you update your post, I'll update mine to remove the needless noise. BTW, I double checked the 388.56. I think that's right. Let me know if you want to compare notes. Commented Jul 26, 2012 at 4:59
  • I will correct my post regarding adding in the 50/100 points instead of multiplying those points by the odds, which will get the correct risk total of 223. I haven't updated it yet, because I was still checking my math regarding that calculation and others used in my analysis. The scoring rules I used were the ones listed on Wikipedia for Greed linked in the OP, I will recheck those numbers as well, but I agree that the difference between 388.56 and 388.033 is inconsequential when the grandularity of score is by divisions of 50 points.
    – user1873
    Commented Jul 26, 2012 at 13:58

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