Happy Thanksgiving to all my fellow Canadian readers!
While I was up at my sister’s house yesterday for an early Thanksgiving dinner, she brought out a dice game that a friend had introduced her to, called ‘Greed$’. The game comes with six dice, each labeled with all six of the characters in the name, (in different colors, which are important with the Es – one E on each die is black and one is green,) and a score card for various results – a single G is worth 50 points, D is 100 points, but combinations range from 3 Es of the same color being worth 300 to six of a kind worth 5000.
Each player rolls in turn, starting with rolling all six dice. If you get any results that are worth points, you can ‘bank’ any or all of them, and then either keep your banked points or re-roll all of the dice that you didn’t bank. But you have to bank at least one die from each roll, and if you don’t roll anything that’s worth points, then you ‘zilch out’ and lose everything you’ve banked for that turn. There are a few other rules about needing to get to 500 points the first time you score, or being able to re-roll all the dice again if you bank them all, but we actually didn’t play with those the first few games.
It was a fun diversion for all 3 generations we had in the house, and as the name implies, often games were won or lost based on a decision of ‘do I take the points I have now, or do I risk them to try for more?’ Since I’m a big math geek, I was fascinated by the possibility of doing a full probability analysis of the game, and I’ve started on that, but haven’t cracked it, especially since the combinatorics with this many dice get complicated.
I’ve solved some of the simpler cases, though:
Re-rolling one or two dice, as you might imagine, is a sucker’s game. You can’t even score any of the big combinations, which are all 3 or more, (and you have to roll those all at the same time or they don’t count – you can’t bank two Gs and then add a third G to get the GGG bonus.) And there’s a really high chance of zilching out. Theoretically, if you only had 112 points or less, it’d be worth taking a chance on re-rolling two dice – but since the lowest number of points you can get per banked die is 50, if you’re re-rolling two, you’ve got four banked and at least 200 to risk.
With 3 to re-roll, things get much more interesting. Adding up all the possible combinations and other scores you could make with 3 dice, I get an expected return of just over 84 points. Based on the results for re-rolling 1 or 2, I don’t need to worry about the possible added value of re-rolls at this stage: if you get anything of value when re-rolling 3, you stick with it.
On the other hand, when re-rolling 3 dice, you have just under a 28% chance of zilching out – 60 out of 216. There are 216 ways that the 3 dice can come up, and to zilch, they need to be all black Es, green Es, Rs, or dollar signs – so those are 64 possible combinations. However, if you get all three the same, you get a bonus score, so that takes out 4 of the 64 combinations, leaving 60.
So, if you have 300 points when re-rolling 3; say, if you’ve banked 3 Ds, or 3 Es of the same color, you stand to gain an estimated 84 points by re-rolling, but you have a 27.8% chance of losing 300, which comes out to slightly less than 84, so you’re better off re-rolling. But it’s close enough that if you’re feeling unlucky I wouldn’t blame you for playing it safe.
The next step is to work out all the probabilities and combinations for 4 dice, which is going to be tricky, especially considering that you might roll a combination of 3 and extra points on the fourth die. I’ll also need to allow for the fact that if you only score a single point die with 4, you can re-roll the other 3 as above. Whether it makes sense to do that is going to depend on what you rolled and how many points you had banked before you rolled the 4 dice (which could only be 100 to 200 points, I guess.) And then I’ll need to tackle the 5-dice case, and then the all-6.
But I’m looking forward to it. 🙂 Math wonkery!
By the way, if there’s any other probability nerds out there, can you check my figures on one case? Rolling the six dice to start, how likely is it that you’d roll 1 of each symbol? This is a starting combination worth 1000 points, and as far as I can tell, it should come up better than once in 65 rolls: 6!/(6^6) = 1 / 64.8 Â That didn’t sound right to any of my family, and we probably did that many rounds of the game, and it didn’t come up. Did I make a mistake? It’s the same thing as asking how likely you are to roll 6 normal craps dice and get 1, 2, 3, 4, 5, and 6 come up at the same time.
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