Optimal Strategy for Farkle Dice!

In this post, I will discuss my findings in terms of an optimal strategy for the game of Farkle.  All of my findings are based on sound mathematical methods and a computer program was used to determine the optimal strategy.  Before I begin, let me first state the Rules of the game that I was using while developing the strategy
Each 1: 100 points
Each 5: 50 points
3 1's: 1000 points
3 2's: 200 points
...
3 6's: 600 points
4 1's: 2000 points
...
4 6's 1200 points
5 1's 3000 points
...
5 6's 1800 points
Straight 1-6 1500 points
3 Pairs (E.G. 22 33 44): 750 points
Everything else is considered to be a Farkle (0 points).

I calculated these figures under the assumption that we are trying to maximize expected value, however that is not necessarily the case.  To employ a true optimal strategy, we want to maximize the probability of winning which depends on both your current score and your opponents score.  However, to make the calculations easier, I ignored those two variables altogether and simply found how to maximize the expected value for any single turn.

Here is a small table that shows my findings.
Each entry in the table shows the expected value in the given situation if you were to employ this optimal strategy.
For example, lets say I have 250 points and I am rolling 3 dice, it can be seen that the best option for me is to roll again (yielding an expected value of 293.4 which is better than 250).  However, if I had 500 points rolling 3, banking the 500 points is the best play because the risk of Farkling is too high to wager the 500 points.
Note that not all entries are filled in in the table.  This is because they are impossible states.  (I.E. you can't have 50 points rolling 3).
Now let's consider a more complex situation.  Let's say I just rolled the following 6 dice:
1 2 2 2 5 4
Clearly I have a lot of options, but which one is best?  Let's use the table!
50 rolling 5 (5) - EV = 363.834
100 rolling 5 (1) - EV = 399.911
150 rolling 4 (1 5) - EV = 307.673
200 rolling 3 (2 2 2) - EV = 257.584
250 rolling 2 (2 2 2 5) - EV = 250
300 rolling 2 (2 2 2 1) - EV = 300
350 rolling 1 (2 2 2 1 5) - EV = 350

So the best option is to bank the 1 and roll the other 5 dice, resulting in an average of about 400 points. Using this same methodology, you can take any situation and determine the best possible play.

	Number of Dice
Points	1	2	3	4	5	6
0						548.858
50					363.834
100				274.776	399.911
150			227.968	307.673
200		221.706	257.584	343.8
250	274.669	250	293.4
300	300	300	329.228			802
350	350	350			596.22	846.014
400	400	400	400.947	498.2	638.291	890.036
450	450	450	450	540.176	680.368	934.065
500	500	500	500	582.157	722.453	978.099
550	550	550	550	624.144	764.545	1022.666
600	600	600	600	666.139	806.642	1067.891
650	650	650	650	708.139	849.953	1113.281
700	700	700	700	750.142	893.894	1158.674
750	750	750	750	792.147	937.847	1204.068
800	800	800	800	834.154	981.801	1249.464
850	850	850	850	876.162	1025.756	1295.174
900	900	900	900	918.17	1069.711	1341.411
950	950	950	950	960.178	1114.429	1387.917
1000	1000	1000	1000	1002.188	1160.198	1434.468
1050	1050	1050	1050	1050	1206.254	1481.147

If you're interested in seeing the extended table (up to 10000 points), or would like to understand the methods that I used to come up with this strategy, feel free to send me an email at RMcKenna21@gmail.com.