Help with Dice Probabilities (small runs and sets)

Hello everyone!

I am a card and board game designer, working on a new concept for a press-your-luck style game, and I have a probability problem that is more complex than I can figure out.

In the game I'm designing, players choose a target card which bears an unlucky configuration of dice: 3 of a kind, run of 3, or 3 odd numbers, for example. They start with a pool of three 6-sided dice. If they do not roll that combination, they earn a number of points and have the option to roll again next turn by adding a die to their pool. If they roll at least that combination, they lose all the points they've accumulated for that card and the card is discarded. (For example, their unlucky combination is three of a kind and they roll four 4s, they still lose.) If they choose not to roll again, they bank the points they've accumulated. They may attempt up to 5 turns on a given card, for a total of 7 dice on their 5th turn. Other cards alter gameplay, but that is outside the scope of my question.

I would like to make the rewards for each roll relative to their risks. I can't seem to crunch the numbers correctly, especially trying to figure out rolling at least a specific result. Here are the possible patterns:

A Pair
Three of a Kind
Run of 3 (three sequential numbers)
3 Odds/Evens
3 Low (1-2) Medium (3-4) or High (5-6) Numbers

I've been trying to figure out a scaling solution for each of these (so I can use one equation for any size die pool) but have run into too many problems. If you can, please post the math behind the odds. I can punch the equations into excel so I can manipulate the payouts. If you come up with any other possible results, let me know!

Thank you in advance!

Nick

I'm beginning to think I'm making this out to be much harder than it needs to be. For example, getting exactly a pair on 5d6 is 5C2*6*3!*5C3:

- 5C2 for the positions of the pair (pull the size of the set from the number of dice rolled)
- 6 for the value of the pair (number of possible outcomes)
- 3! for the order of the singletons (number of dice minus the size of the set)
- 5C3 for the values of the singletons (combinations of the remaining values for the number of singletons)

However, the value of the singletons can be anything. If I change the singletons 3!*6C3, wouldn't that fix the problem?

Here are the calculations for 5d6 to get a pair:

6C1 : pick the value of the pair (of course, this is 6, but I think it helps keep things clear)
5C2 : pick the position of the pair
5P3 : pick the values and the order of the singleton (this is the same as 3!*5C3)

So, your original calculation is correct.

Quote: darquehope

However, the value of the singletons can be anything. If I change the singletons 3!*6C3, wouldn't that fix the problem?

The singletons cannot be any value. If one of them matches the pair, then it is no longer a pair, but a 3 of a kind. This is why you will pick the remaining dice values out of 5 instead of 6.

2 pair:

6C2 : pick the values of both pairs
5C2 : pick the position of the first pair
3C2 : out of the remaining spots, pick the position of the second pair
4C1 : pick the value of the singleton

full house:

6C1 : pick the value of the 3 of a kind
5C3 : pick the position of the 3 of a kind
5C1 : pick the value of the 2 of a kind

Hopefully, this can get you on the right track.

Thanks for the reply!

Quote: CrystalMath

The singletons cannot be any value. If one of them matches the pair, then it is no longer a pair, but a 3 of a kind. This is why you will pick the remaining dice values out of 5 instead of 6.

But they can. If the card states "a pair" or "2 of a kind" (I think I might adopt that latter wording) then any size set will qualify. So a pair, three of a kind, four of a kind, five of a kind, and so on will meet the criteria, because they all contain at least a pair. Similarly, a "run of 3" is met by three or more sequential numbers. So by extension, any dice in excess of the set/run/etc can be of any value. The more dice you roll, the more likely you are to roll the qualifying pattern. This is the why subsequent rolls become riskier, and garner higher payoffs. Some rolls become impossible as the die pools increase in size. (It's impossible not to get a pair on 7d6, without the use of a special card, but that's another story.)

I'm gonna crunch some more numbers with some of your insights and see if I can suss some of it out. Thanks again!