Toss two dice, A starts from $1$, At each turn moves forward with number of steps equal to the sum of numbers on the two dice that are face up. What is the probability that A reaches $440$?
I thought that the expectation of every step is $7$, then A can reach $435$ with $62$ steps, and the last step is $5$. Therefore, the answer is the probability of two dice sum when sum is $5$, which is $\frac{4}{36}$.
I'm not sure whether the answer is correct or not, need some help, thanks.
To see that your method is not correct, change $440$ to $9$. If you apply your reasoning, you would end up with a probability of zero, which is clearly incorrect.
For each integer $n$, let $p(n)$ be the probability that a sum of $n$ can be achieved in some number of steps.
Then $p$ satisfies the recursion \begin{align*} p(n) = & \frac{1}{36}p(n-2) + \frac{2}{36}p(n-3) + \frac{3}{36}p(n-4) + \frac{4}{36}p(n-5) \\[4pt] & + \frac{5}{36}p(n-6) + \frac{6}{36}p(n-7) + \frac{5}{36}p(n-8) + \frac{4}{36}p(n-9) \\[4pt] & + \frac{3}{36}p(n-10) + \frac{2}{36}p(n-11) + \frac{1}{36}p(n-12) \end{align*} for all $n > 0$, together with the initial conditions $p(0)=1$ and $p(n)=0$ for all $n < 0$.
Here are the values of $p(n)$ for $0\le n\le 12$ . . . \begin{array}{c|c} n&p(n)\\ \hline 0&1\\ \hline 1&0\\ \hline 2&{\large{\frac{1}{36}}}\approx .02777777778\\ \hline 3&{\large{\frac{1}{18}}}\approx .05555555556\\ \hline 4&{\large{\frac{109}{1296}}}\approx .08410493827\\ \hline 5&{\large{\frac{37}{324}}}\approx .1141975309\\ \hline 6&{\large{\frac{6841}{46656}}}\approx .1466263717\\ \hline 7&{\large{\frac{1417}{7776}}}\approx .1822273663\\ \hline 8&{\large{\frac{279397}{1679616}}}\approx .1663457600\\ \hline 9&{\large{\frac{32653}{209952}}}\approx .1555260250\\ \hline 10&{\large{\frac{8935921}{60466176}}}\approx .1477837957\\ \hline 11&{\large{\frac{4271189}{30233088}}}\approx .1412753140\\ \hline 12&{\large{\frac{292122973}{2176782336}}}\approx .1341994412\\ \hline \end{array} Given that $A$ starts at position $1$, the probability that $A$ reaches position $440$ is equal to $p(439)$. Applying the recursion using a CAS such as Maple or Mathematica, $p(439)$ evaluates to a rational number with huge numerator and denominator, but it is approximately $1/7$.