# Probability of choosing the correct missing digit > 0.5?

The question goes as follows:

A person has forgotten the last digit of a telephone number, so he dials the number with the last digit randomly chosen. How many times does he have to dial (not counting repetitions) in order that the probability of dialing the correct number is more than 0.5.

The way I answered it was in the following format:

Denote try 1 success as $T_1$, and subsequently follows:

$P(T_1 \cup (T_2 \cap T_1^c) \cup (T_3 \cap T_2^c \cap T_1^c) ...) \gt 0.5$

We know the union of these events can be expressed as additions because they're disjoint, so:

$P(T_1) + ... P(T_n \cap T_{n-1}^c ... T_1^c) \gt 0.5$

We also know that in each, the event of $T_x$ and $T_y$ such that $x \ne y$ is independent. Then we can say for $P(T_n \cap T_{n-1}^c ... T_1^c)$ for example it becomes:

$P(T_n) P(T_{n-1}^c) ... P(T_1^c)$

So I just kept calculating the probabilities, which are 1/10, 1/9, 1/8, 1/7... 1 for the $T_n$'s and 9/10, 8/9, 7/8, 6/7... for the $T_n^c$'s. Iteratively doing this I found an answer.

However, this is pretty tedious. I was wondering if there was a closed form solution, or if this is even correct?

This is correct: if he makes $5$ attempts, the chance that he gets the correct number is the chance that he gets it on the first try $+$ the chance he gets it wrong on the first try and right on the second try, and so on up to the chance that he gets it wrong the first $4$ tries and right on the fifth try.
Here's an intuitive understanding: he'll go through all $10$ digits possible if he needs to. The correct digit is just as likely to be in the first $5$ digits he tries as it is to be in the last $5$ digits he tries, so it's a $\frac{1}{2}$ chance with $5$ tries. Thus, $6$ tries will make it more likely, which will push the probability over $.5$.