Probability of going only right in a random walk

Question

A guy walks a 2-dimensional random walk in the following way: He has $100$ notes in his pocket, $70$ of which says go right ($+1$), $20$ of which says go left ($-1$), and 10 of which says stay put ($+0$). In each steps, he picks randomly one of these notes without returning it to his pocket.

Let $X_n$ be his position after n steps.

What is the probability that there exists an $n$ such that $X_n=70$?

I understand that all the "go right" steps should be before the "go left", so I can count using total probability the number of times he goes only right in the first 70 steps + the number of times he goes right and one time stays put in the first 71 steps + the number of times he goes right and two times he stays put in 72 steps and so on, but I assume there is more elegant way.

I possess a final solution which is - $\frac{1}{\binom{90}{20}}$ but I don't understand why is it true?

Any help or explanation would be appreciated.

Answer 1

Only the $90$ notes that mention a direction are relevant.

$X_n=70$ will arise if and only if the first $70$ selected notes with direction (on a total of $90$ notes with direction) are all notes that send the guy to the right.

The probability that this occurs is:$$\frac{\binom{70}{70}\binom{20}0}{\binom{90}{70}}=\frac1{\binom{90}{70}}=\frac1{\binom{90}{20}}$$

(hypergeometric distribution).

Answer 2

First, forget about the notes that say "stay put", they can be shuffled in at will. So, you are left with 90 notes of which you know 70 should go before the other 20. Actually, there is only one way to choose 70 out of 90 in a way to satisfy this condition.