6 Dice Arrangement

We shall attempt to solve this question systematically by listing cases according to the distribution of the dice.

Firstly, note that each 6-sided die can be reduced to a 3-sided one by taking $\pmod 3$ of each roll. For example, 1 and 4 are treated as identical because they both contribute $1\pmod3$ to the cumulative sum, which is what matters in this question.

There are 7 cases of distribution: $(6,0,0), (5,1,0), (4,2,0), (3,3,0), (4,1,1), (3,2,1), (2,2,2)$

Case 1: Distribution of $(6,0,0)$ (ie. 6 of a certain mod value and 0 of the other 2 values). In this case, the sum of values of the 6 dies is a multiple of 3, so when k = 6, case 1 fails.

Cases 4,5,7 : Distribution of $(3,3,0), (4,1,1), (2,2,2)$ All fail due to the same reason as case 1.

Case 2: Distribution of $5,1,0$ There are 6 possible sets here, of the form $(x,x,x,x,x,y)$ . When x =0, $(y,x,x,x,x,x)$ is a possible derangement that fulfils the requirements of the question. It works when y = 1 or 2. In each set of values for $(x,y)$ , there are $\frac {6!}{5!1!} = 6$ possible ways to choose 6 dice of that configuration, thus this contributes $6*2 = 12$ solutions to the question. When x = 1 or 2, by pigeonhole principle there exists a block of 3 numbers such that all 3 numbers are x. Thus the cumulative sum will cycle through all values of mod 3 including $0\pmod3$ . Thus this subcase fails. Therefore case 2 contributes 12 cases.

Case 3: Distribution of $(4,2,0)$ . There are 6 possible sets here, of the form $(x,x,x,x,y,y)$ . When x =0, $(y,y,x,x,x,x)$ is a possible derangement that fulfils the requirements of the question. It works when y = 1 or 2. In each set of values for $(x,y)$ , there are $\frac {6!}{4!2!} = 15$ possible ways to choose 6 dice of that configuration, thus this contributes $15*2 = 30$ solutions to the question. When x = 1, y = 2, and x = 2, y =1, there are solutions in $(1,1,2,1,2,1)$ and $(2,2,1,2,1,2)$ respectively, each contributing $\frac {6!}{4!2!} = 15$ solutions for a total of 30 solutions to the question. Lastly, when y = 0, x = 1 or 2, there will be a value of k, $1 \leq k \leq 5$ such that the cumulative sum is $0\pmod3$ , so this subcase fails. Therefore case 3 contributes a total of 60 solutions to the question.

Case 6: Distribution of $(3,2,1)$ There are 6 possible sets here, of the form $(x,x,x,y,y,z)$ . There are solutions for every set of values of $(x,y,z)$ , namely $(1,1,2,0,0,0), (2,2,1,0,0,0), (1,1,2,1,0,0), (1,1,2,1,2,0), (2,2,1,2,0,0), (2,2,1,2,1,0)$ for the cases where $(x,y,z)$ are $(0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), (2,1,0)$ respectively. Each solution here contributes $\frac {6!}{3!2!1!} = 60$ solutions to the question for a total of 360 solutions in case 6.

Thus, the final probability $p = \frac {12+60+360}{3^6} = \frac {432}{729} = \frac {16}{27}$ Thus $a+b = 43$

We will instead count the probability that the dice can't be arranged with the above's property. An arrangement will be presented in the form of string of length 6 containing 3 types of numbers : '0','1','2' which represent the numbers from 1-6 which is 0,1,2 mod 3, respectively (and so each slot has probability of 1/3).

Let $a,b,c$ be the number of $1s,2s,3s$ in an arrangement of a dice, then the sum of the numbers is $a+2b+3c \equiv (a-b) \mod 3$

Now, divide into cases :

1) The sum of the number of the dices is a multiple of 3, means this will always fail the property for $k = 6$ . Regardless of the sum of numbers in the last 5 dices, only 1 from $\{1,2,3\}$ if used as the number in the 1st dice, will get you a sum which is divisible by 3, since they are pairwise distinct in mod 3. Same for $\{4,5,6\}$ . Hence the probability is $\frac{1}{3}$

2) The sum of the number is not a multiple of 3. It means that $3 \not| (a-b)$ and this implies that at least one of $a,b$ is positive (the arrangement can't solely contain $3s$ ), and $a \neq b$ . Divide into subcases :

2.1) Case where there are exactly 1 other type besides '3', which we can re-arrange into $1..13...3 (b=0)$ or $2..23..3 (a=0)$ (not necessarily contains a '3'). WLOG, we take the case $b=0$ (later multiplied by 2). Firstly, we can't have $a = 0,3,6$ as the sum will be divisible by 3.

For $a = 1$ , arrange as follows : $133333$ .

For $a = 2$ , arrange as follows : $113333$ .

If $a \geq 3$ , then we can't rearrange it to satisfy the property since no matter what, eventually we will have 3 of 1s,as we start counting the sum from the first to the last.

The probability for all possible rearrangement for $a =4 (111166)$ is $\left(\frac{1}{3} \right)^6 \dbinom{6}{4} = \frac{15}{3^6}$

The probability for all possible rearrangement for $a =5 (111116)$ is $\left(\frac{1}{3} \right)^6 \dbinom{6}{5} = \frac{6}{3^6}$

Hence, for this case, the probability is $2 \times \left(\frac{15}{3^6} + \frac{6}{3^6} \right) = \frac{14}{243}$

2.2) Case where there are exactly 2 other types besides '3', which we can re-arrange into $1..12..23..3$ (not necessarily contains a '3'), Since $a \neq b$ , WLOG assume $a > b$ , then later we multiplied by 2 as we can swap the $1s$ and $2s$ in the arguments. Clearly $a-b = 4$ as $a-b \geq 5$ implies that $a \geq b+5 \geq 6$ which is impossible.

For $a-b = 1$ , then arrange as $112333$ for $a=2,b=1$ and $112123$ for $a=3,b=2$

For $a-b = 2$ , then arrange as $112133$ for $a=3,b=1$ and $112121$ for $a=4,b=2$

For $a-b=3$ , then this will fail the condition $3 \not| (a-b)$

For $a-b = 4 \Rightarrow a=5,b=1$ we will always fail to get a rearrangement satisfying the propery, as will be shown here :

-) If we start by '2', then it must be followed by another '2' (since otherwise the sum is 3) which we don't have.

-) If we start by '1', then it must be followed with '1' (sum = 2). And then a '2' (sum = 4). And then a '1' (sum = 5). At the 5th position, we must add another '2', which we don't have.

Hence for this case, the total is just twice the probability of all rearrangements of $111112$ which is $2 \times \left(\frac{1}{3} \right)^6 \dbinom{6}{5} = \frac{4}{243}$

Giving us the grand total of $\frac{1}{3} + \frac{14}{243} + \frac{4}{243} = \frac{11}{27}$ for the probability that we can't arrange the dice.

Hence, the desired probability is just $1 - \frac{11}{27} = \frac{16}{27}$ which gives $16+27 = 43$ as the answer,

let $d_n$ be the value of the nth die. We want to avoid:

$\displaystyle \sum_{n=1}^k d_n \equiv 0 \pmod{3}$ for $1 \leq k \leq 6$

In base 3, our dice can take values of 0,1 or 2. We notice a total of $3^6$ or $729$ possible outcomes.

$0+0 = 0$

$0+1 = 1, 1 + 1 = 2$

$0+2 = 2 ,1 + 2 = 0, 2 + 2 = 1$

Next, we use the base 3 ternary addition rules above to create a method of rearranging the dice. The conditions below are sufficient (but not necessary) to avoid our beginning equivalence.

• Dice with value 0 are moved to the last positions where they do not change the value of the sum. These dice make our equivalence true if placed in position 1.
• The first two die must be equal.
• Subsequent die alternate between (1 and 2) or (2 and 1)

With our standard format we can consider all possible cases in terms of the zero valued dice.

• Case 1: no zero valued dice. $(1,1,2,1,2,1)$ or $(2,2,1,2,1,2)$ has $2 \times {6 \choose 4} = 30$ orderings
• Case 2: 1 zero valued die. $(1,1,2,1,2,0)$ or $(2,2,1,2,1,0)$ has $2 \times {5 \choose 3} \times {6 \choose 1} = 120$ orderings.
• Case 3: 2 Zero valued dice. $(1,1,2,1,0,0)$ or $(2,2,1,2,0,0)$ has $2 \times {4 \choose 3} \times {6 \choose 2} = 120$ orderings.
• Case 4: 3 zero valued dice. $(1,1,2,0,0,0)$ or $(2,2,1,0,0,0)$ has $2 \times {3 \choose 2} \times {6 \choose 3} = 120$ orderings.
• Case 5: 4 zero valued dice. $(1,1,0,0,0,0)$ or $(2,2,0,0,0,0)$ has $2 \times {6 \choose 4} = 30$ orderings.
• Case 6: 5 zero valued dice. $(1,0,0,0,0,0)$ or $(2,0,0,0,0,0)$ has $2 \times {6 \choose 5} = 30$ orderings Case 7: 6 zero valued dice. No possible solutions regardless of ordering.

Summing the useful number of orderings divided by the total number of orderings gives

$\frac{a}{b} = \frac{30+120+120+120+30+12}{729} = \frac{432}{729} = \frac{16}{27}$

$16 + 27 = \boxed{43}$

We are interested in multiples of 3, so let's consider the possible numbers mod 3: 0,1, and -1. We'll find all permutations consisting of 6 or less 1 and 2's. If such a permutation exists, we can add on threes to the end, leaving the sum unchanged mod 3. Make a table with columns labelled 1, 2, and 3. Starting with 0 threes, experiment with possible values for 1 and 2. Here are the possibilities: 4/2/0 2/4/0 3/2/1 2/3/1 1/3/2 3/1/2 1/2/3 2/1/3 2/0/4 0/2/4 1/0/5 0/1/5 For each case, calculate the number of ways to arrange the numbers in a line. For example, the expression for 4/2/0 would look like 6!/(4!2!), as the 4 one's, as well as the 2 two's are indistinguishable. The number of legitimate permutations is 432. The total number of permutations is 3^6=729. 432/729=16/27. 16+27=43, and we are done.

Firstly, we change the problem into an easier problem by treating the 6 sided dice as 3 sided dice. We have not changed the problem as the 5 can be replaced as a 2, the 6 as a 3 and the 4 as a 1.

We are left with 6 dice with 1, 2 and 3 on it. When rearranging, note that we can move all the 3s to the back of the sequence. Suppose the rearrangement started with a 1, then the full sequence without 3s would be 1, 1, 2, 1, 2, 1. If the rearrangement started with a 2, the full sequence would be 2, 2, 1, 2, 1, 2.

Replacing the ends of the sequence with 3s, it can be seen that the only possibilities to (#1s, #2s #3s) (# means number of) are:

(4, 2, 0) and (2, 4, 0)

{3, 2, 1} (All combinations of 3, 2, 1)

(2, 0, 4) and (0, 2, 4)

(1, 0, 5) and (0, 1, 5)

Counting the number of ways to roll for these cases gives us:

$4 \times \frac{6!}{4! \times 2!} + 2 \times \frac{6!}{5! \times 1!} + 6 \times \frac{6!}{3! \times 2! \times 1!}=60+12+360=432$

Counting the number of possible rolls gives $3^6=729$

Thus, the probability of an arrangement is $\frac{432}{729}=\frac{16}{27}$

Dealing with multiples of 3, we may choose to approach the problem modulo 3. That is represent 1 and 4 as 1 (mod 3), 2 and 5 as -1(mod 3), 3 and 6 as 0(mod3). We can either start with a 1 or -1. Set the number of 1's as a and the number of -1's as b. For a string of 6 numbers composed of 1,0,-1, for the first k terms to be indivisible by 3, $|a-b|$ is either 1 or 2, and $a+b \leq 6$ . The ordered pairs a and b satisfying this are: (0,1), (0,2), (1,2), (1,3), (2,3),(2,4) and interchanging a and b to get 6 more. The number of ways of doing this is $2\times (\frac{6!}{0!1!5!} +\frac{6!}{0!2!4!}+\frac{6!}{1!2!3!}+\frac{6!}{1!3!2!}+\frac{6!}{2!3!1!}+\frac{6!}{2!4!0!})=432$ . The total number of ways to get a string of 6 numbers is $3^6$ [not $6^6$ ]. Hence, the probability desired is $\frac{16}{27}$ , giving the answer of $43$ .

In this problem, it is sufficient to consider only the remainders of the dice numbers on a division by $3$ (that is, $0,1$ and $2$ ). This view facilitates the cases we need to analyse, since we are interested in the divisibility of the number, not in the number at all. Hence, we dont need to look for the sum of the dice numbers. Instead, we add their remainders on a division by $3$ .

The key of the problem is to note that we can't have a row where the sum of its respective remainders on a division by $3$ is a multiple of $3$ . That would mean the sum would be a multiple of $3$ .

Therefore, let us list the favorables arrangements in function of their remainders, not by the dices itself. The feasible configurations and their permutations are:

$(1,0,1,2,1,2): 6!/3!2! = 60$ ;

$(1,1,2,1,2,1): 6!/4!2! = 15$ ;

$(1,0,0,0,0,1): 6!/4!2! = 15$ ;

$(1,0,0,0,1,2): 6!/3!2! = 60$ ;

$(1,1,0,0,2,1): 6!/3!2! = 60$ ;

$(0,0,0,0,0,1): 6!/5! = 6$ .

suming up to $216$ (the permutations were considered in the count because the enunciation allows us to choose the order of the dice). But note that we can also interchange the $2$ 's with $1$ 's, which gives us more $216$ favorable combinations.

Therefore, we have a total of $216+216=432$ possible arrangements that satisfies the condition of the problem.

And as the total of possible permutations is $3^6 = 729$ , our propability p is $p = \frac {432}{729} = \frac {16}{27}$ .

Therefore, $a=16$ and $b=27$ , which gives the desired answer $a+b=43$

First assume that only 1,2 and 3 are spun. It makes it simpler.Then replace 3 with 0, and 2 with -1. Now, any 6-tuple with the same number of +1s and -1s must fail for k=6 any solution where the difference of +1s and -1s is more than 2 or less than -2 cannot work either, (think of a random walk, but you can switch the moves order).So we find the number of 6-tuples of ${-1,0,1}$ that have a difference of either 1 or 2 in their -1s and 1s. Then we divide the cases by number of 0s in the tuple.

For no zero, there must be a 2-4 split. There are 2 ways to choose which side has the 2, and 15 to choose the 2,so that makes 30

For 1 zero, , there must be a 2-3 split. There are 2 ways to choose which side has the 2, and 10 to choose the 2, and 6 ways to choose the 0,so that makes 120

For 2 zeros, , there must be a 1-3 split. There are 2 ways to choose which side has the 1, and 4 to choose the 1,and 15 to choose the zeros,so that makes 120

For 3 zeros, , there must be a 1-2 split. There are 2 ways to choose which side has the 2, and 3 to choose the 1,and 20 to choose the zeros,so that makes 120

For 4 zeros, , there must be a 0-2 split. There are 2 ways to choose which side has the 2, 15 to choose the zeros,so that makes 30

For 5 zeros, , there must be a 1-0 split. There are 2 ways to choose which side has the 1,6 to choose the zeros,so that makes 12

These sum to 432 out of 729, which simplifies to 16/27