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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 $5 000$?

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    $\begingroup$ Are the successive failures leading to an ultimate success coordinated? Would a failed attempt be repeated? $\endgroup$ – DJohnM Jul 10 at 22:05
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    $\begingroup$ Are you essentially asking how many different guesses are required in order for the probability to be .5 that you have guessed the correct PIN? $\endgroup$ – mlchristians Jul 10 at 23:17
  • $\begingroup$ None of the answers have appropriate justification. Consider this post math.stackexchange.com/questions/206798/… $\endgroup$ – Dayton Jul 10 at 23:50
  • $\begingroup$ It should probably be noted that in practice you can expect to only have to guess about $1800$ times per PIN in the long run, with a typical number of guesses around $400$ and the most likely number of guesses being $1$. datagenetics.com/blog/september32012 $\endgroup$ – K B Dave Jul 11 at 0:02
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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

$$ x = 5000$$

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This question can be solved by first trying it for small numbers. What if there were not 10000 possible combinations, but 1, 2, 3 or 4?

$$ \begin{array}{ll} \text{Combinations} & \text{Average number of attempts} \\ 1 & 1 \\ 2 & \frac32 \\ 3 & 2 \\ 4 & \frac52 \\ \dots & \dots \end{array} $$

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