On the real line place an object at 1. After every flip of a fair coin move the object to the right by 1 unit if the outcome is head . [on hold]

On the real line place an object at 1. After every flip of a fair coin move the object to the right by 1 unit if the outcome is head and to the left by 1 unit if the outcome is tail.

Let $$N$$ be a fixed positive integer.Game ends when the object reaches either $$0$$ or $$N$$.

What is the probability of the object reaching at $$N$$, i.e $$P(N)$$. Give me some hint..

put on hold as off-topic by Saad, Arnaud D., Xander Henderson, Did, user21820Jan 14 at 15:05

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• This is a case of the famous "gambler's ruin" problem. For this version with a fair game, looking at the probability of reaching one state before another, the easiest approach is to track expected values. OK, where's a good version of this to link to on MSE? – jmerry Jan 12 at 8:08

Here is a short proof using induction and symmetry. We use induction on $$N$$ to show that $$P(N) = \frac{1}{N}$$. For $$N = 1$$, this is obvious.
Now, suppose that $$P(k) = \frac{1}{k}$$ for $$k \leq n$$. We shall show that the statement holds for $$N = n+1$$. Note that, in order to reach $$n+1$$, we must reach $$n$$. The probability of this is $$P(n) = \frac{1}{n}$$. Now, the probability of reaching $$n+1$$ when standing on $$n$$ is the same as the probability of reaching $$0$$ when standing on $$1$$, by symmetry. So we get $$P(n+1) = P(n)(1 - P(n+1)) \Leftrightarrow P(n+1) = \frac{P(n)}{1+P(n)} = \frac{\frac{1}{n}}{\frac{n+1}{n}} = \frac{1}{n+1}$$ So by the principle of induction, $$P(N) = \frac{1}{N}$$ for all positive integers $$N$$
• Why is the probability of reaching 0 standing on 1 equal to $1 - P(n+1)$? Did the "standing on 1" have any relevance to that value? – Pedro A Jan 12 at 13:13
• The game ends when we reach either $n+1$ or $0$, and the probability of reaching $n+1$ standing on $1$ is (by definition) $P(n+1)$ so the probability of reaching $0$ becomes $1 - P(n+1)$. – nesHan Jan 12 at 14:01