Question regarding combinatorics formula I was solving this math problem:
If we roll three standard dices what is the possibility that the sum of eyes will be equal to 7?
So I tried to approach this problem in two different ways:
1) The order in every sequence IS IMPORTANT (a,a,b) and (a,b,a) sequences are counted as different. In this case all the possible variants are counted using formula A*(6,3)=6^3=216. When we look for the favorable endings there are 4 different possible sums: (5,1,1) (3,2,2) (4,2,1) (3,3,1) . All that is left is to count the different variants in which these sums can be displayed and we will have all possible outcomes. This is done using formula P(3,2) = 3! / 2! = 3 and for sequence (4,2,1) formula 3! = 6 . We get 3+3+3+6 favorable sequence and the possibility is 15/216 = 5/72. 5/72 is also the answer according to the book.
2) However, It is way more logical to think that sequences (3,2,2) and (2,2,3) are same and should be counted as one because all that matters is the sum of eyes on the three dices. However, when I make the calculations here, the answer seems to be different. All the possible variants can be counted using formula C*(6,3)=C(6+3-1,C)=(6+3-1)!/(3!*(6-1)!) = 56. We know that the desired results can be achieved with 4 possible sequences (5,1,1) (3,2,2) (4,2,1) (3,3,1) and since they are all the same we do not need to count all the possible displays, therefore, we have 4 favorable endings. In this case the possiblity is 4/36 = 1/9. Obviously, the possibilities in both 1) and 2) cases should have matched, so, could You please tell me where have I made a mistake? Thank you in advance :)
P.S the suggested answer is in the first 1) variant, but the way it is being calculated does not seem the best approach since order in sequences should really not matter. I am particularly interested in this formula where you have to count all the possible endings where order does not matter but the same element can repeat. I fail to use it every single time, maybe there is something wrong with the way I use it?
 A: You made a mistake in the second example by assuming that cases like $(2,2,3)=(3,2,2)$. The issue with this assumption is that it makes $(3,3,3)$ equally likely with something like $(1,2,3)$, which is clearly false.
A: You have a barrel containing a million jellybeans, $999,999$ black and $1$ pink.  You reach in and pick out two jellybeans at random.  What is the probability  you get two black jellybeans?
Method 1: There is a first jellybean and a second jellybean.  There are $1000000\cdot999999$ ways the selection could occur, out of which $999999\cdot999998$ result in two black jellybeans.  Hence the probability is $\frac{999999\cdot999998}{1000000\cdot999999}=\frac{999998}{1000000}=0.999998$.
Method 2: Well, order of jellybeans doesn't matter, even if, in reality, I will always touch one of the jellybeans I pick at least slightly earlier than the other.  So, ignoring order, there are $\binom{1000000}{2}$ possible pairs of jellybeans I could pick, out of which $\binom{999999}{2}$ pairs have the desired property.  Dividing gives
$$
\frac{\frac{999999\cdot999998}{2\cdot1}}{\frac{1000000\cdot999999}{2\cdot1}}=\frac{999998}{1000000}=0.999998.
$$
Method 3: Well, really all I care about is color, not particular identity of jellybeans.  So there are two possible color selections, represented by the multisets {black, black} and {black, pink}, of which one has the desired composition.  Hence the probability is $1/2=0.5$.
You can see that Methods 1 and 2 agree with each other, but are in violent disagreement with Method 3.  The reason for this is that, while it is true that order doesn't matter, it is false that physical identity of jellybeans doesn't matter.  Methods 1 and 2 are right; Method 3 is wrong.  What we learn from this is that when modeling a physical selection process using mathematics, physics matters.  The outcome {black, black} is tied to a very much larger set of physical outcomes than is the outcome {black, pink}.  (There's another big physical assumption here, namely that we can mix up the jellybeans sufficiently well so as to make all selections equally likely.)
Applying this insight to your problem, we see that the outcome "dice sum to $7$" is tied to a larger set of physical outcomes than is the outcome "dice sum to $3$".  Even analyzing in finer detail, the outcome "the roll includes a $2$, a $3$, and a $5$" is tied to a larger set of physical outcomes than is the outcome "roll includes two $2$s and a $5$", which in turn is tied to a larger set of physical outcomes than is the outcome "roll contains three $2$s".  It is generally physically reasonable to assume that the dice are fair and independent of each other.  It is not physically reasonable to assume that the dice do not have distinct physical identities.
To expand a bit on the analogy between the two problems: in computing probability in the jellybean problem it matters, not just which colors you got, but which particular jellybeans of that color you got.  In the dice problem, it matters, not just which numbers you got, but which particular dice had those numbers.
A: I'll just put in my own insight.
Let's say the die were rolled in order. Then we can look at the entire thing in terms of the first die.
By the way the total probability space is $6^3=216$. These are permutations that allow for repetition.
So the first die can never be $6$. Unless there's an option of $0$ on one of the other die you'd always add up to something greater than $7$.
If the first die has a $5$, then the number remaining to be added is $2$. You can't roll a $2$ on the die left because that would ruin our target sum of $7$. So the second die can either be $1$ or $1$ and the last can keep the remaining $1$. That gives us only one permutation.
If the first die has a $4$, then the second die can only possess values from $1$ to $2$, not $3$, because we need to leave something for the third die. That gives us $2$ permutations.
Repeating this for the other values of the first die:
$3 \implies 3 \text{ permutations}$
$2 \implies 4 \text{ permutations}$
$1 \implies 5 \text{ permutations}$
Combining this with:
$4 \implies 2 \text{ permutations}$
$5 \implies 1 \text{ permutations}$
$6 \implies 0 \text{ permutations}$
So the probability is:
$=\dfrac{0+1+2+3+4+5}{216}=\dfrac{5}{72}$
And the thing you were proposing. This system is full of permutations and not combinations. Think about it like this: would it be fair to serve the guy behind you in a line just because mathematically there is one combination of your arrangement on line? That's how the die feel. The sequences $1,5,1$ and $5,1,5$ exist on different die so they can't be equated as one. Changing the order changes which die have the number. And that's definitely a permutation.
