If I want the average number of PIN code attempts for a PIN of length 4 using the numbers 0-9, would I have to half my answer? So $10^4 = 10 000$ possible combinations, but the average number of attempts needed to guess correctly would be $5 000$?
Average Number of PIN Code Attempts
combinatorics
Best Answer
If I am correct in interpreting the problem as how many different attempts are required in order that the probability is $1/2$ that you will have guess the correct PIN, then your answer of $5000$ attempts is correct, as you would multiply the probability of guessing the PIN correctly on one attempt, which is $\frac{1}{10000}$ by that number $x$ whose product is $.5$; that is, we solve
$$ \frac{1}{10,000} x = .5 $$
which gives
$$ x = 5000$$