I'm playing Farkle a lot on Facebook. It's a mindless but somehow interesting game. The idea is that you throw 6 dice. If any of those dice make one of the scoring combinations, you can pick them up and roll the remaining dice again, hoping to score some more. If you run out of dice, you get to continue with 6 more. If you reach at least 300 points, you can "bank" the points you've made this turn and start with 6 fresh dice.
The trick is that if, at any time, you roll no scoring combination, you "Farkle". You lose all the points you made this turn. Farkle three times in a row, and you lose 500 points, to boot.
Being the mathematically inclined person I am, I've been trying to find the most advantageous method of picking up the dice.
It all started when I decided to shoot for the high score. That was 10,500 points. You only get 10 turns to score; naturally, this means you should roll more than 1,000 points before banking and moving on to your next turn. I started calculating how much I "needed this turn", and refusing to bank my points unless I got it... even if the odds seemed insurmountable.
This led me to wonder, "What are the odds?" Well, the scoring system isn't too tough. A die with a 1 scores 100 points. A die with a 5 scores 50 points. Triples score 100 times their face value (so triple 2s are 200, triple 5s are 500), with triple 1s getting a bonus: 1000! Finally, there are two six-dice scores: a straight (1500 points) and three pairs (750 points).
My first problem came up when rolling three dice and getting two 1s and a trash value. Should I pick up both 1s and roll the single die, or pick up only one die and roll the remaining two?
When rolling a single die, the only way to score is to roll a 1 or a 5. There are six possible values, and only two winners, so your chance of surviving to roll again is 1/3.
I figured that rolling two dice gives me that same 1/3 chance, PLUS the possibility that I could pick up both dice; so I only picked up one and rolled two. That's seat-of-the-pants, though; let's think about the actual probability.
There are 36 possible combination when you roll two dice. Let's think of them as red and green. 12 of the combinations have a red die you can pick up, and 12 have a green die you can pick up. You might jump to the conclusion that your chances are 24/36, or 2/3 -- but you'd be wrong. Four of those combinations (1+1, 1+5, 5+1, and 5+5) are already covered by picking up the red die! So that's a total of 20 combinations out of 36, or 5/9. That's still a little better than 1/3, of course.
Of course, with three dice to roll you can hit the big time! Roll triples, and you get a high score. But what's your actual chance of not Farkling? Well, if we imagine adding a blue die to the previously-considered red and green, you get 36 red+green combinations * 6 blue combinations, or 216 total combinations. We already know 20 of the red+green combinations allow you to continue; if the blue didn't matter at all, that would be 20 * 6 = 120 chances to continue out of 216, which is 5/9 (the common factor is 24). (Good; so far the math is still holding up.)
But the blue die does matter! When it's a 1 or 5, we still get to continue. There are 36 combinations of red+green where the blue is a 1 or 5, so that's 72 combinations where you get to pick up the blue die. But it's not 72 new combinations. Of the 36 for each red+green, you already get to pick up either the red or green for 20 combinations, leaving 16 losers. The blue die changes those 16 losers to winners in two cases, giving you 32 new winning combinations. So, all told: that's 20 red+green winners times 6 blue possibilities PLUS 16 red+green losers times 2 blue winners. (20 * 6) + (16 * 2) = 120 + 32 = 152 winners out of 216. That's 19/27. That's better than a half-and-half chance!
We can apply this same logic to four dice. 216 * 6 total combinations; 125 winners * 6 don't-care on the extra die, plus (216 - 152) losers times 2 new winners. 1296 total, 750 old winners + 64 * 2 (128) conversions = 878/1296 chances to continue.
If you are also of a mathematical bent, you might notice that there are 4 ^ (number of dice) losers for each combination. That is, for 1 die, there are 4 losers; for 2 dice, 16 losers; for 3 dice, 64 losers. If I decide to write (number of dice) as d, that makes the chance of losing (4^d)/(6^d). That's going to make the calculations much easier.
So, here are your chances of not Farkling with any number of dice:
|Number of dice||Fraction not Farkling||Percent not Farkling|
Let the mass Farkling begin!