Let The Mouse Go

There is a total of 6^6 distinct paths (6 steps, 6 choices each).

To be back at the starting cell, we need all of the following:

• left moves = right moves
• front moves = back moves
• up moves = down moves

Using 3-D coordinates, the starting point would be (x,y,z) = (0,0,0)

Let

• A = (+1, 0, 0)
• B = (-1, 0, 0)
• C = (0, +1, 0)
• D = (0, -1, 0)
• E = (0, 0, +1)
• F = (0, 0, -1)

The possible ways of moving back to the starting point are (then considering ordered ways):

• AAABBB: 6!/(3!*3!) = 20 ways
• CCCDDD: 6!/(3!*3!) = 20 ways
• EEEFFF: 6!/(3!*3!) = 20 ways
• AABBCD: 6!/(2!*2!) = 180 ways
• AABBEF: 6!/(2!*2!) = 180 ways
• CCDDAB: 6!/(2!*2!) = 180 ways
• CCDDEF: 6!/(2!*2!) = 180 ways
• EEFFAB: 6!/(2!*2!) = 180 ways
• EEFFCD: 6!/(2!*2!) = 180 ways
• ABCDEF: 6! = 720 ways

Total number of possible ways to get back to starting cell = 1860

Total number of distinct paths after 6 steps = 6^6 = 46656

Probability of getting back = 1860/46656 = 0.039866

Denote by $P_n(i,j,k)$ the probability that the walker is at position $(i,j,k)$ in the lattice after $n$ moves. Then the rule governing the walker's motion implies that:

$\begin{aligned} P_{n+1}(i,j,k) &= \frac{1}{6}(P_n(i+1,j,k)+P_n(i-1,j,k)\\ &+P_n(i,j+1,k)+P_n(i,j-1,k)\\ &+P_n(i,j,k+1)+P_n(i,j,k-1))\\ \end{aligned}$

Define $Q_n(x,y,z)=\sum_{i,j,k=-n}^{n} P_n(i,j,k)x^iy^jz^k$

with $Q_0=1$ . Consequently, $Q_n(x,y,z)=\frac{1}{6^n}((x+x^{-1})+(y+y^{-1})+(z+z^{-1}))^n$

$P_6(0,0,0)$ represents the probability that the walker returns to the origin on the sixth step, and is the constant term in $Q_6(x,y,z)$ . To find that term, expand $Q_6$ using the trinomial theorem as

$Q_6(x,y,z)=\frac{1}{6^6} \sum_{n_1+n_2+n_3=6}^{ }\frac{6!}{n_1!n_2!n_3!}(x+x^{-1})^{n_1}(y+y^{-1})^{n_2}(z+z^{-1})^{n_3}$

where $n_1,n_2,n_3$ are all even, in which case the binomial theorem implies that the constant term is $\binom{n_1}{n_1/2}\binom{n_2}{n_2/2}\binom{n_3}{n_3/2}$ . Equivalently,

$\begin{aligned} \\ Q_6(x,y,z) &= \frac{6!}{6^6} \sum_{m_1+m_2+m_3=3} \frac{(x+x^{-1})^{2m_1}(y+y^{-1})^{2m_2}(z+z^{-1})^{2m_3}}{(2m_1)!(2m_2)!(2m_3)!} \\ P_6(0,0,0) &= \frac{6!}{6^6} \sum_{m_1+m_2+m_3=3} \frac{1}{(2m_1)!(2m_2)!(2m_3)!}\binom{2m_1}{m_1}\binom{2m_2}{m_2}\binom{2m_3}{m_3} \\ &= \frac{6!}{6^6} \sum_{m_1+m_2+m_3=3} \frac{1}{(m_1!)^2(m_2!)^2(m_3!)^2}. \end{aligned}$

Calculating this sum shows

$P_6(0,0,0)=\frac{6!}{6^6}\cdot \frac{31}{12}=\frac{155}{3888}\approx \boxed{0.03987}.$

Six moves, choices are u, d, l, r, f, b. All moves in every direction must cancel out. Thus:

Move in 1 dimension: choose 3 x a and 3 x c out of six moves: ${6\choose {3} }=20$ , where a, c is u, d or l, r or f, b. This makes $20\times 3=10\times 6$ possible trajectories.

Move in two dimensions: choose 2 x a and 2 x c in one dimension and 1 x e and 1 x g in another dimension, out of 6 moves: ${6\choose { 2,2,1,1}}=6!/2!/2!/1!/1!=180$ where there are 6 combinations of dimensions to take for a, c, e, g. This makes $180\times 6$ possible trajectories.

Move in three dimensions: choose 1 of each u, d, l, r, f and b moves: $6!=120\times 6$ possible trajectories.

Totalling, dividing by the total nr of moves $6^6$ gives $\frac {(10+180+120) 6}{6^6}=\frac {310}{6^5}$ which gives $\approx\boxed {0.0399}$

Let Up, Right, and Forward be the "positive" directions, and let the others be negative. We need to count the number of sequences of 6 moves that get us back to the start. Such a sequence must contain three positive moves and three negative moves, so there are first of all $6 \choose 3$ ways to assign positives and negatives.

Now, consider the three positive moves. If $x,y,z$ are the three directions, the three positive moves will have one of the following forms: $xxx$ , $xxy$ , or $xyz$ .

• For the pattern $xxx$ , there are 3 ways to choose which direction "x" represents, but all orders are the same. The three negative moves must be the opposite of the three positive moves, so there are 3 ways total in this case.

• For the pattern $xxy$ , there are 3 directions to choose for what "x" represents, then 2 directions remaining to choose for what "y" represents. Then, there are $3$ ways of ordering the positive moves ( $xxy, xyx, yxx$ ), and similarly $3$ orders for the negative moves, for a total of $3 \cdot 2 \cdot 3 \cdot 3 = 54$ possibilities.

• For the pattern $xyz$ , we use all three directions, so we don't need to choose which directions are represented by which letter. However, there are $6$ ways of ordering the three letters, and $6$ ways of ordering the negative moves also, for $6 \cdot 6 = 36$ total possibilities.

There are $6^6$ moves altogether, so the answer is $\frac{{6 \choose 3} (3 + 54 + 36)}{6^6} = \frac{20 \cdot 91}{6^6} = \frac{310}{6^5} \approx \boxed{0.039866}.$

Starting from (0, 0, 0) the mouse moves 1 unit in the positive or negative direction of x, y, or z axis at each turn.

The probability = {Constant i.e. the coefficient of (x^0) (y^0) (z^0) in (x + y + z + 1/x + 1/y + 1/z)^6}/(6^6) = 1860/46656 = 155/3888 = 0.03986625514403292181069958847737

Consider the generating function $f (x, y, z)=(x+\frac {1}{x}+y+\frac {1}{y}+z+\frac {1}{z})^{6}$ . The constant term is 1860 after tedious expansion. Hence the probability is $\frac {1860}{6^{6}}=0.039866...$ hence the answer.

Firstly we must realise that there are 3 different axes along which the mouse can move. Whichever way the mouse moves, the positive movement along an axis has to equal the negative movement along the axis.

Hence, let's label any positive movement along the x axis as $x$ , and any negative movement as $-x$ , and do the same for y and z.

There are three distinct cases to address - if the mouse moves along 1 axis only, if he moves along two, and if he moves along all three.

If he moves along just one, then there is only one way for him to do it, $x + x + x - x - x - x$ . There are $\frac{6!}{3! \times 3!} = 20$ ways of doing this. $20 x 3 = 60$ (to take into account the other 2 axes).

So there are 60 ways of doing it this way.

The second case is if the mouse moves along just two axes. Using the same logic as above, we realise that there is also only one way in which he can do this $x + x - x - x + y - y$ . There are 6 ways of arranging $x$ and $y$ and $z$ in the above form.

The total ways in which any version of $x + x - x - x + y - y$ can be arranged is $\frac{6!}{2! \times 2! } = 180$ .

So there are $180 \times 6 = 1080$ ways of doing it with movement along two dimensions.

The third distinct case is movement along all three dimensions. Again, there is only one way in which to do this $x - x + y - y + z - z$ , which can be arranged in $6! = 720$ ways.

So the total ways of moving back to the original spot are $720 + 1080 + 60 = 1860$ .

The total number of possible moves are $6^6$ .

So the probability of that happening is $\frac{1860}{6^6} = \boxed{0.03986625514}$

mouse moves in only +/-x, +/-y, +/-z directions. to come back, it has to do equal number of +ve or -ve moves in all axes. May be all +/- 3 in same axes, or +/-2 in one and +/-1 in another or +/-1 in all 3 axes first case in any axes gives 6!/(3!3!) = 20 for all 3 axes 3*20 = 60 in 2nd case 6 *(6!/2!2!) 3rd case 6! Total 1860 out of 46656 cases

This answer might be a little too big, but I like to consider it comprehensive.

---

Notation used for the solution:

• $+x$ denotes moving one unit along the positive $x-axis$
• $-x$ denotes moving one unit along the negative $x-axis$
• similar moves can be defined for the positive and negative $y$ and $z\ \ axes$ .
• one particular solution sequence of moves using the above notation looks like $+x +y -x +z -y -z$

---

Basic Ideas/Formulae used for the actual solution:

1. In order to return to the starting position any move along any positive axis must be balanced out by another move along the same axes in the negative direction. In other words, a $+x$ must always be followed by a $-x$ in the sequence.
2. Number of possible ways of arranging $n$ objects of which $r$ are alike, another $s$ are alike and the remaining $t$ are all distinct is given by $\large\frac{n\,!}{r\,!s\,!}$
3. The total number of moves, 6 here, plays an important role in the solution and solving the same problem for any other number might prove to be difficult.

---

Solution

The total possible movements possible in 6 moves with 6 choices in each move is $\boxed{6 \times 6 \times ... 6 \times 6 = 6^6\quad\dots\quad (I)}$

The required number of moves is calculated below:

Any solution sequence can be classified into 3 types:

• Moving along just one axes, throughout.
• Moving along two axes.
• Moving along all the three axes

We will calculate the number of ways in which the above movements can be made while returning to the original position using "Idea no. 1" (seen above)

1.Moving along just one axis

An axes can be chosen in $\large 3 \choose 1$ ways. After choosing an axes (say $x-axis$ )we have to arrange $+x$ and $-x$ in such a way that they balance out meaning 3 $\ +x's$ and 3 $\ -x's$ . Now number of ways of arranging 6 $\ x's$ of which 3 are alike (the 3 $\ +x's$ ) and another 3 are alike is given by $\large\frac{6!}{3!3!}$ (using "Idea no. 2" (seen above))

So total number of ways of returning to starting point while moving along just one axis is \mathrm\large\large\boxed{{3 \choose 2} \times \frac{6!}{3!3!}\quad\dots\quad (II)}

2.Moving along two axes

Two axes can be chosen in $\large 3 \choose 2$ ways. After choosing two axes (say $x-axis$ and $y-axis$ ) they can be arranged in the manner $+x -x +y -y +y -y$ where one of the two selected axes repeats twice(here $y\ axes$ ), so this axis can be chosen in $\large 2 \choose 1$ ways. Now the number of ways of chosing axes is same as number of ways of arranging 6 objects of which 2 are alike(the 2 $\ +y's$ ) another 2 are alike(the 2 $\ -y's$ ) and the remaining 2 are distinct( $+x$ and $-x$ ) which is $\large{\frac{6!}{2!2!}}$

So total number of ways of returning to starting point while moving along only two axes is \mathrm\large\large\boxed{{3 \choose 2}\times{2 \choose 1}\times {\frac{6!}{2!2!}}\quad\dots\quad (III)}

3.Moving along all the three axes

Three axes can be chosen in $\large 3 \choose 3$ or 1 way. Further any sequence in this category will be a permutation of $+x-x+y-y+z-z$ . So number of ways of arranging these 6 objects is $\large 6!$ .

So total number of ways of returning to starting point while moving along all the three axes is \mathrm\large\large\boxed{6!\quad\dots\quad (IV)}

Finally the probability of returning to the starting point is

\mathrm\large\large{\frac{number\ of\ ways\ of\ returning\ to\ starting\ point}{total\ number\ of\ ways\ possible\ in\ 6\ moves}\ = \frac{II\ +\ III\ +\ IV}{I}\ = \frac{1860}{6^6}}

$\boxed{= 0.039866\dots}$