TITLE: Expected number of consecutive guesses to get a given sequence of numbers
QUESTION [5 upvotes]: I have a lock on my dorm door that's really stupid. Basically, it just checks whether or not the sequence of numbers I've put in is the combo, whether or not the lock was reset between guesses. So let's say my combo is 5556. Then I can input 555555556 into my lock and it'll unlock, without having to reset after inputting the first four numbers. 
I tried to calculate the expected number of random number guesses to eventually input the right combo by assuming each "guess" was independent. For example, the input of 123454321 has 6 "guesses": 1234, 2345, 3454, 4543, 5432, 4321. Assuming this, the expected length of input required would be 10,000, for 10^4 permutations of a 4 digit combo.
However, to check my work, I created a simulation with a queue object and random number generators and ran 100 trials per experiment over 100 experiments in Python. In every experiment, the average was always above 10,000 by a significant margin, ranging from 500-2000. 
I'm wondering, are the guesses really independent? What is the actual expected value?

REPLY [2 votes]: We can prove the following general result:

Given a code $C$ of $n$ digits, for each $1\le i\le n-1$, let $b_i$ be a number which is $1$ if the first $i$ digits of $C$ equal the last $i$ digits of $C$, and $0$ otherwise. The expected wait time for $C$ is 
  $$10^n+\sum_{i=1}^{n-1}b_i10^i.$$

For example, when $n=4$:

The expected wait time for codes like $aaaa$ is $11,110$.
The expected wait time for codes like $abab$ is $10,100$.
The expected wait time for codes like $abca$ is $10,010$.
The expected wait time for everything else is $10,000$.

To prove this, let us first assume that $b_i=0$ for all $i$, meaning no prefix of $C$ is also a suffix. 
Imagine a casino with a ten digit roulette wheel. It spins this wheel once per minute, except that the casino shuts down once the code $C$ appears over the course of $n$ consecutive spins. Players may place an $\$x$ bet on the outcome of the spin; if they are wrong, they lost $\$x$, and if they are right, they win $\$9x$, so the bet is fair.
Imagine that every minute, a new person enters the casino. They first place a $\$1$ bet on the first digit of $C$. If they win, they place a $\$10$ bet on the second digit of $C$, and in general people who have won $k$ times place a $10^k$ bet on the $(k+1)^{st}$ digit of $C$. Note that anyone who does not make it to the end of $C$ will lose exactly $\$1$; for example, if they make it to digit two then lose, their net winnings are $+9+90-100=-1$. Only a person who makes it all the way through to the end of $C$ will win big, a total of $10^n-1$. This can only happen to one person, because we stipulated the casino shuts down once $C$ appears in order.
Since all of these bets are fair, the total expected winnings of all the players is $0$. On the other hand, letting $T$ be the total number of spins, the actual winnings are $10^n-T$, since the first $T-1$ people lose $1$ and the last person wins $10^n-1$. Equating these two, we get that the expected number of spins is $10^n$. 
The full result comes from noting that when some of the $b_i$ are nonzero, then there are actually a couple more winners at the end of the game. Namely, the $i^{th}$ player from the end wins $10^i-1$ as long as the first $i$ digits of $C$ are equal to the last $i$ digits.