# Probability of ending at $0$ after $3$ times moving $6$-sided-dice-rolled number of steps in a coin-flip direction?

I just wrote a question and I'm not sure if my solution is correct. Could someone please help me check it?

Question: Mary starts at 0 on the number line. She first tosses a coin, and then rolls a fair six-sided dice. If the coin lands on heads, then she moves to the positive side; if the coin lands on tails, then she moves to the negative side. The number of steps she takes depends on the number that appears on the dice she rolled. For example, if the coin lands on heads and 4 is rolled, then Mary would go 4 steps to the positive side. What is the probability that Mary will end at 0 after three rounds of tossing and rolling?

My solution: In order for Mary to end at 0, one of the dice must be the sum of the other two dice and be the opposite sign.

P (sum = 2) = ⅙ x ⅙ x ⅙ x 3

P (sum = 3) = ⅙ x ⅙ x ⅙ x 6

P (sum = 4) = ⅙ x ⅙ x ⅙ x (3 + 6)

P (sum = 5) = ⅙ x ⅙ x ⅙ x (6 + 6)

P (sum = 6) = ⅙ x ⅙ x ⅙ x (3 + 6 + 6)

P total = ⅙ x ⅙ x ⅙ x 45 = 5/24.

There is 1/2 probability that the two dice are the same sign, and we multiply it by another 1/2 to ensure that the other dice is the opposite sign. Therefore my answer is 5/96.

This is correct, and your reasoning is good. For a more general and mechanical way to get the result, note that the movements are additive, with possible values $$\{-6,-5,-4,-3,-2,-1,1,2,3,4,5,6\}$$ occurring with equal probability ($$1/12$$); so the probability of landing on $$0$$ after $$n$$ rolls is the coefficient of $$x^0$$ (i.e., the constant term) in the expansion of $$\left(\frac{1}{12}\left(x^{-6}+x^{-5}+x^{-4}+x^{-3}+x^{-2}+x^{-1}+x+x^2+x^3+x^4+x^5+x^6\right)\right)^n.$$ WolframAlpha gives this probability as $$\frac{1}{12}$$ for $$n=2$$ (clearly correct), $$\frac{5}{96}$$ for $$n=3$$ (agreeing with OP's result), $$\frac{259}{5184}$$ for $$n=4$$, $$\frac{1825}{41472}$$ for $$n=5$$, and so forth.

Assume for the moment that Die-1 is the sum of the other two dice. Then, the other two dice must sum to $$2,3,4,5,$$ or $$6$$. This can happen in $$1 + 2 + 3 + 4 + 5 = 15$$ ways.

Once Die-2,Die-3 are set, Die-1 is forced to be their sum.

So, the probability that Die-1 is the sum of the other two is $$~\displaystyle \frac{15}{216} = \frac{5}{72}.~$$

In order for this to lead to returning to $$0$$, the coin flips have to specifically be HTT or THH. So the probability of Die-1 being the leveraging die is $$\displaystyle ~\frac{5}{72} \times \frac{1}{4} = \frac{5}{288}.$$

Further, anyone of the three dice can be the leveraging die. Therefore, the overall probability of returning to $$0$$ is

$$3 \times \frac{5}{288} = \frac{5}{96},$$

A symmetrical shortcut to the first portion of my answer is that the probability of two dice summing to less than $$(7)$$ must, by symmetry, be the same as the probability of the two dice summing to greater than $$(7)$$. Further, the probability of two dice summing to exactly $$(7)$$ is $$~\displaystyle \frac{1}{6}.$$
Therefore, the probability of two dice summing to less than $$(7)$$ must be
$$\frac{1}{2} \times \left[1 - \frac{1}{6}\right] = \frac{5}{12} = \frac{15}{36}.$$