I'm addicted to a game called "Farkle". If you are on Facebook, you can play it for yourself at http://apps.facebook.com/hotdices.
It goes by a lot of other names: Zonk, Zilch, 5000, 10000, Wimp Out, Hot Dice, Buzzball, Oh Crap, Greed, and Squelch. But I like calling it "Farkle", since that's what I say when I lose my turn (and all the points that go with it!)
You can read the rules at http://en.wikipedia.org/wiki/Farkle. But basically, you roll dice and score points for 1s and 5s. You can stop at any time, or keep rolling. But if you roll the dice and don't get a 1 or a 5, you lose all the points you've accumulated that round!
Similar games I've enjoyed like this include "Liar's Dice" (or "Perudo", the dice game featured in the last Pirates of the Caribbean movie), "Cosmic Wimpout", and the classic Sid Sackson boardgame "Can't Stop". It's a great game to play against people with bad impulse control who always seem to want to push their luck.
Even though there are some great statistics pages on Farkle already on the internet (I love the one at http://graciesdad.wordpress.com/2009/08/30/farkle-odds/) I wanted to crunch some numbers for myself.
The probability of getting a "farkle" with one die is:
4/6 = 0.67 (rounded to two decimal points)
Since a 2, 3, 4, or 6 is a "farkle", while a 1 or 5 keeps you alive
The probability of getting "farkle" with two dice remaining is:
16/36 = 0.44 (Note: I did this by hand, writing down all combinations of 11, then 12, then 13, etc.)
Getting a "farkle" with three dice is trickier, because you can also get three-of-a-kind, which is a good thing!
There are 64/216 combinations of three dice that have no 1s or 5s. I need to subtract the four three-of-a-kind-combinations (222, 333, 444, and 666) to get the probability of a three-dice "farkle":
64/216 - 4/216 = 60/216 = 0.28
Summary so far
What have we learned? If you have one die remaining, you have a 67% chance to farkle. Say goodbye to all of your points 2 out of every 3 times you find yourself in that position.
If you have two dice remaining, you will "farkle" 44% of the time. That's almost half, minus a little bit. And if you have three dice remaining, you have a 28% chance of losing all your points, or 1 in every 3 times. Keep that in mind when you want to go "one turn more"!
It gets complicated with more dice. There are 1296 possible dice combinations of rolling 4 dice, and 240 of them are "bad" with no 1s or 5s at all. However, there are now 16 ways to get three-of-a-kind, which is nice.
I've forgotten how to talk about "combinations". When I'm saying that there are now 16 ways to make a three-of-a-kind with 4 dice, I think I'm saying "4 choose three". Is that right? You can see the math for yourself at http://en.wikipedia.org/wiki/Combination
C(4,3) = C(n,k) = n!/(k!(n-k)!) = 4!/(3!(4-3)!) = 4
Since there are four three-of-a-kind combinations that will avoid a "farkle" (222, 333, 444, and 666), I'm saying that there are 16 dice combinations I need to subtract (4*C(4,3)). So, the total probability for a "farkle" with four dice is:
240/1296 - 16/1296 = 224/1296 = 0.17
Please note that I don't need to subtract the probability of getting a four-of-a-kind, since any 2222 already has a three-of-a-kind 222 embedded inside of it.
Now, 896 out of 7776 dice combinations are "bad". But a whopping 40 of them (i.e. 4*C(5,3)) are possible three-of-a-kind combinations. So, I think the probability for a "farkle" with five dice is:
896/7776 - 40/7776 = 856/7776 = 0.11
I couldn't finish this one. I can prove that there are 46656 possible dice combinations, and 3424 of them contain no 1s or 5s. In addition, there are (4*C(6,3)) = 80 ways of getting three-of-a-kind.
So, there are very few ways of rolling six dice and getting an instant "farkle":
3424/46656 - 80/46656 = 3344/46656 = 0.07
However, I think I also need to subtract the ways you can roll six dice and get three pairs. So, the probability will be even lower, though only slightly.
If you have four dice in your hand, there is only a 17% chance of getting a "farkle". For five dice, that goes down to a 11% chance. And the probability of getting a "farkle" on your very first throw of six dice is less than 7%, so go ahead and throw!
My results disagree with the web page at http://graciesdad.wordpress.com/2009/08/30/farkle-odds/, so I think there's something wrong with my math.
For me, the whole purpose of calculating the odds of getting a "farkle" or so I can figure out if I should stop rolling and take my points, or try for one more turn!
The math term for this is "expected return":
E(R)= Sum: probability (in scenario i) * the return (in scenario i)
If we were playing with only one die and one roll, and I told you that I'd give you five dollars if you win, where you pay me two dollars if you "farkle"... would you play?
The answer should be NO! You are paying $2.00 for each roll, but you only have an expected return of E(one_roll) = 0.33 * ($5.00) = $1.67. That is, you should "expect" to lose 33 cents every time you play the game.
Now, I admit it's kind of stupid to "expect" to lose thirty-three cents every turn, since either you win $5 or lose $2. But over time, that's the amount you will lose. In other words, if we played a hundred times, you should expect to lose $33.00 to me. Thanks for playing!
But how does this fit Farkle? Well, if you knew the "expected return" from throwing one die, you could figure out if it was worth it. I know that there is a two-in-three chance to getting a "farkle" with one die, but how much can I expect to win if I get a 1 or 1 5 and get to throw all 6 dice *again*?
People who are smarter than me figured out that the "average" farkle score with one throw of six dice is 302 to 446.6, depending on the house rules you follow. I'll use 400 as a "guesstimate". That means the expected return of one dice is:
E(one_die_left) = 0.33 * (400 plus the 100 you get for rolling a 1) = 167 points
This doesn't sound right to me. That's saying if you have 200 points and have only 1 die remaining, you should cash in your points and avoid re-rolling, since you only expect to win 167 points on your 200 point "wager".
Someday, I'll figure out the optimal Farkle strategy. I'd love to program a computer to play the game for me. They call this a "Monte Carlo" strategy... estimating probabilities by playing millions and millions of games. I did something similar for the card game War... you can see my online Java applet playing the game at: http://patrickkellogg.com/school/papers/war/cgi-bin/WarApplet.htm
I'd love to figure out:
1. The expected return per turn given various Farkle strategies
2. The problem I didn't solve above for the probability of getting a "farkle" with six dice
3. How to modify a strategy given different "goals" (i.e. how do you play differently against conservative players, or if you are trying to beat a friend's high score of 10,000 points) See also http://notaboutapples.wordpress.com/2009/08/22/farkle-expectation-and-knowing-what-you-want/
In the meantime, please feel free to try the game for yourself. And if you're one of my facebook friends (I'm "Patrick Kellogg" on there), you can challenge me head-to-head!