For 6, once dice needs to be 2, so there are five combinations. That part, I'm not having issues with. But as you get into the midrange, the number of possible combinations which could result in a given total increases exponentially. There must be some formula which can let me calculate this. I've been attempting to work it out from wikipedia ...

2021年11月26日 · In using $\frac{_{12}C_2}{2}$ to calculate the number of possible combinations, you assume that each possible dice roll appears twice in the two-element subsets of the 12-element set containing each face twice. This is true for when you roll two different numbers (one die has the higher number, other die has the higher number).

For this you would want to use combinations not permutations, since the dice are indistinguishable. $6^3$ gives the number of possible rolls under the assumption that $(3,3,2)$ is different than $(3,2,3)$ but since this is a casino and casinos don't have numbered or different colored die, $(3,3,2)$ should be the same as $(3,2,3)$ since there is no way to tell these two …

It there any formula for calculating dice combinations? For example, for hex-dice, when I roll one die there are six possible results. If I do the same with two dice there are 21 results instead possible 36 because “ $1$ ” with “ $2$ ” is the same as “ $2$ ” with “ $1$ ”. For three dice, there are $6^3 = 216$ combinations, but ...

Where, for example: (1,3,1,4,6) is considered the same outcome as (1,1,3,6,4) How many total outcomes are there?

2015年4月3日 · This topic is called 'convolutions' in probability and computer science. Because of the intensive and repetitive computation necessary, finding exact probabilities of sums on n > 2 dice is usually done by computer algorithm and several examples are available by googling 'convolutions discrete probability dice'; some have nice pictures, even if you ignore the code.

When you roll $2$ dice, there are $36$ possibilities. However, there are only $21$ combinations, if order does not matter. Rolling a $(4,2)$ = rolling a $(2,4)$. Let's say in a game, rolling a $(1,1)$ makes you lose. The odds of rolling this is a $1/36$. But why can't you say the probability is a $1/21$, assuming you roll both dice at the same ...

2024年5月17日 · If we have {0, 8} on the first dice, the second dice must have {1, 4}. Each dice must contain at least one of 6 or 9, so the dice combinations would be {0, 8, 6-9} for the first one and {1, 4, 6-9} for the second one. We suppose that the first dice has a 2, so the second one must have a 5. Finally, we can have any number as the two remaining ...

2018年11月25日 · The $21$ possible outcomes you counted consist of $6$ outcomes in which both six-sided dice display the same number and $15$ outcomes in which the two six-sided dice display different numbers. $$\binom{6}{1} + \binom{6}{2} = 21$$ This is the number of distinguishable outcomes you would see if you had two indistinguishable six-sided dice, for ...

2018年9月22日 · Specifically I want to know about the case where Two dice showing one number, two other dice showing a different number, and the fifth die showing a third number. The number of possible outcomes for this case, that is 2-2-1, is $\binom62\binom41\times \frac{5!}{2!2!} = …