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The Lost Boarding Pass is a famous puzzle as follows:

On a sold out flight, $100$ people line up to board the plane. The first passenger in the line has lost his boarding pass, but was allowed in, regardless. He takes a random seat. Each subsequent passenger takes his or her assigned seat if available, or a random unoccupied seat, otherwise.

What is the probability that the last passenger to board the plane finds his seat unoccupied?

It is not difficult to prove that the answer is $\frac 12$.

After seeing this I wanted to find the chance that the $k$ person to enter would end up at his spot. this resulted in the next function:

If $k\neq 1$ then:

$P(n,k) = \dfrac{n-k+1}{n-k+2}$, $n$ being the number of people.

and for $k=1$:

$P(n,1) = \frac 1n$

Now I'm interested in the chance that a number of people would sit at their place.

(If it helps I know the chance of the k person to enter to sit at the l one's spot if $l=1$ or $l>k$ is $\dfrac{1}{(n-k)(n-k+2)}$)

Thank you in advance.

gurtonn
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  • So $n$ people take seats according to this procedure, and you want to know the probability that at the end, exactly $m$ of those people (no more, no less) will be in their originally assigned seats; is that the question? – David K Apr 28 '15 at 12:47
  • i would like to know that as well but my original question is what is the probability that the {k_1, k_2.....k_m} would be in their place, for example: the chance that the first and any-other would be in their place is 1/n, and the chance that the second and third is (n-2)/n. – gurtonn Apr 28 '15 at 12:50

1 Answers1

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Since the person who lost their pass plays a special role, I find it easier to think about the problem in terms of $1$ loser and $n-1$ keepers.

If the loser happens to take their own seat, everyone else also gets to sit in their own seat, so the probability for any set of people including the loser to end up in their own seats is just $1/n$.

The probabilities of various keepers sitting in their own seats are independent, so the probability for any set of keepers to end up in their own seats is just the product of the individual probabilities.

To see this, let's look at the derivation of the individual probabilities. To simplify things, I'll number the keepers in reverse order of boarding, so keeper $1$ boards last, and keeper $j$ corresponds to your $k=n-j+1$.

The reason keeper $j$ has probability $j\,/\,(j+1)$ of sitting in their own seat is that at exactly one point before keeper $j$ boards, exactly one passenger will choose either the loser's seat or one of the seats of keepers $1$ through $j$, and keeper $j$ will lose their seat iff this choice falls on their seat out of these $j+1$ seats.

Now the probability of $j'\gt j$ taking $j$'s seat certainly depends on whether $j'$ gets their own seat – but that has no bearing on the derivation above. It makes no difference to $j$ whether it's $j'$ who makes that choice and potentially takes $j$'s seat, or someone else before or after $j'$. Independent of whether $j'$ gets their own seat, exactly one person will make that choice, and hence the probabilities for $j$ and $j'$ to get their own seats are independent.

This argument also carries through for more than two keepers, and thus, as claimed, the probability for any set of keepers to end up in their own seats is the product of the individual probabilities. For instance, the probability of all keepers to end up in their own seats is the product of all individual probabilities, which telescopes and yields $1/n$, the probability of the loser choosing their own seat (as it must).

joriki
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