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A check-out counter at a supermarket will service one customer per unit time if there is anyone in line. Customers arrive at the line: in each unit of time, the probability that a single new customer arrives is $\frac13$, the probability that two arrive is $\frac13$, and the probability that no new customer arrives is also $\frac13$. There are initially three customers in line. Find the probability that the line empties before it has ten persons in it.

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  • $\begingroup$ This looks like another mathbook-type problem to me... $\endgroup$ – Mithical Sep 4 '16 at 19:22
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    $\begingroup$ Is there anything special or interesting about this question that makes it a puzzle and not just a simple maths textbook problem? Please see this meta post for criteria on the difference between maths puzzles (on-topic) and maths problems (off-topic). $\endgroup$ – Rand al'Thor Sep 4 '16 at 19:23
  • $\begingroup$ @Human Could you please tell about the book, its author, publisher, anything which would be helpful to cite the actual source of the puzzle(in case it is one). $\endgroup$ – ABcDexter Sep 4 '16 at 20:44
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There are initially three people in line. On average, the number of people doesn't change (since it either increases, decreases or remains the same with equal chances). Therefore, at any point in time, the average number of people in line is three.

Now, let $T$ be the first time the queue either empties or reaches 10 people, and let $p$ be the probability of the former. On the one hand, the expected number of people in line at time $T$ is three (as previously argued). On the other hand, the number is either 0 or 10 with probabilities $p$ and $1-p$ respectively, so the expected value can be calculated as $0\cdot(p)+10\cdot (1-p)=10(1-p)$. Therefore, $10(1-p)=3$, so $p=0.7$, for a 70% chance of an empty queue first.

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It's

7 in 10

Because

the distribution is unbiased, meaning that the probability that the queue gains or loses $k$ people every $3$ time units is equal.

Breakdown

$k=3: 000 | 222$
$k=2: 001, 010, 100 | 122,212,221$
$k=1: 011, 101, 110, 002, 020, 200 | 022, 202, 220, 112, 121, 211$
$k=0: 111, 012, 021, 102, 120, 201, 210$

Reasoning

After every $3$ time units, the queue has increased or decreased in size with equal probabilities, so with a starting position of $3$ we need the probability that a random walk reaches $0$ before it reaches $10$. This is equivalent to a random walk that moves towards $-1$ with probability $\frac7{10}$, and towards $1$ with probability $\frac3{10}$, so the answer is $\frac7{10}$.

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