When the doctor's surgery opens, on Monday morning, there are three people queueing outside. Each person is waiting to see a doctor. They each speak
to the receptionist and then take a seat in the waiting area.

The waiting area has five seats arranged in a single row, with seats side- by-side. Initially all seats are unoccupied. Each patients selects a seat
at random, with the proviso that they prefer not to sit next to a seat
that is already occupied by another patient.

What is the probability that all three patients get a seating position
they are happy with i.e. not seated next to another person?

--
David Entwistle

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On 15/06/2025 20:22, David Entwistle wrote:

When the doctor's surgery opens, on Monday morning, there are three people queueing outside. Each person is waiting to see a doctor. They each speak
to the receptionist and then take a seat in the waiting area.

The waiting area has five seats arranged in a single row, with seats side- by-side. Initially all seats are unoccupied. Each patients selects a seat
at random, with the proviso that they prefer not to sit next to a seat
that is already occupied by another patient.

What is the probability that all three patients get a seating position
they are happy with i.e. not seated next to another person?

.......assuming that they are perfect logicians with the faculty
of sight.
.......assuming that they are perfect logicians with the faculty
of sight
.......assuming that they are perfect logicians with the faculty
of sigh
.......assuming that they are perfect logicians with the faculty
of sig
.......assuming that they are perfect logicians with the faculty
of si
.......assuming that they are perfect logicians with the faculty of s .......assuming that they are perfect logicians with the faculty of .......assuming that they are perfect logicians with the faculty of .......assuming that they are perfect logicians with the faculty o .......assuming that they are perfect logicians with the faculty .......assuming that they are perfect logicians with the faculty .......assuming that they are perfect logicians with the facult
.......assuming that they are perfect logicians with the facul
.......assuming that they are perfect logicians with the facu
.......assuming that they are perfect logicians with the fac
.......assuming that they are perfect logicians with the fa
.......assuming that they are perfect logicians with the f
.......assuming that they are perfect logicians with the
.......assuming that they are perfect logicians with the
.......assuming that they are perfect logicians with th
.......assuming that they are perfect logicians with t
.......assuming that they are perfect logicians with
.......assuming that they are perfect logicians with
.......assuming that they are perfect logicians wit
.......assuming that they are perfect logicians wi
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.......assuming that they are perfect logicians
.......assuming that they are perfect logicians
.......assuming that they are perfect logician
.......assuming that they are perfect logicia
.......assuming that they are perfect logici
.......assuming that they are perfect logic
.......assuming that they are perfect logi
.......assuming that they are perfect log
.......assuming that they are perfect lo
.......assuming that they are perfect l
.......assuming that they are perfect
.......assuming that they are perfect
.......assuming that they are perfec
.......assuming that they are perfe
.......assuming that they are perf
.......assuming that they are per
.......assuming that they are pe
.......assuming that they are p
.......assuming that they are
.......assuming that they are
.......assuming that they ar
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.

100%.

Having seen the other two patients in the queue outside, they
will indulge their preference by avoiding seats B and D, both of
which increase the risk of breathing in neighbour-lurgies.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On Sun, 15 Jun 2025 21:23:10 +0100, Richard Heathfield wrote:

.......assuming that they are perfect logicians with the faculty of
sight.

Excellent.

Unfortunately the first two in the queue were work-shy individuals,
incapable of logical thought, oblivious to there surroundings and given to
acts of quite random seat occupancy (and completely unnecessary leg spreading)...

The third person was a distinguished elderly gentleman who just happened
to be having a spot of trouble with his water works. He had a keen and
active mind, fully capable of logical thought. Unfortunately for him, his
only role in this question is to endure the lack of consideration made by
the first two.

All completely hypothetical, of course.

--
David Entwistle

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On 17/06/2025 16:12, Charlie Roberts wrote:

On Mon, 16 Jun 2025 10:31:25 -0000 (UTC), David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:

David,

On Sun, 15 Jun 2025 21:23:10 +0100, Richard Heathfield wrote:

.......assuming that they are perfect logicians with the faculty of
sight.

Excellent.

Unfortunately the first two in the queue were work-shy individuals,
incapable of logical thought, oblivious to there surroundings and given to >> acts of quite random seat occupancy (and completely unnecessary leg
spreading)...

The third person was a distinguished elderly gentleman who just happened
to be having a spot of trouble with his water works. He had a keen and
active mind, fully capable of logical thought. Unfortunately for him, his
only role in this question is to endure the lack of consideration made by
the first two.

All completely hypothetical, of course.

Since I can't see anyone has answerd this yet...

P = P(3|1,2,3,4,5) * 1 +
P(1|1,2,3,4,5) * P(3,5|3,4,5) +
P(5|1,2,3,4,5) * P(1,3|1,2,3)

= 1/5 * 1 +
1/5 * 2/3 +
1/5 * 2/3

= 7/15

Is there another way on viewing the problem? What if the seats are
truly chosen at random, *without any patient knowing what the
others have chosen*. Translated, it is really asking that if there
are 5 cards numbered 1, 2, ... 5, what are chances that 2 and 4
are *not* picked in a random draw?

If they do not know what the others have chosen, this is sampling with replacement - two patients
could choose the same seat and one would have to sit on the others lap. Or all 3 might choose the
same seat!!

P = P(1,3,5|1,2,3,4,5)^2
= (3/5)^3
= 9/125

Again, to go further down the rabbit hole, is the draw done
with replacement or not?

As you worded it above, with replacement.

If no replacement, that implies patients know what the previous choices were (but not which patient
chose which). There is just one way of choosing 3 seats from 5 withoug replacement that works:
choosing 1,3,5.

P = 1 / C(3,5)
= 1/10


Mike.

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On Tue, 17 Jun 2025 17:27:47 +0100, Mike Terry wrote:

Since I can't see anyone has answerd this yet...

Thanks Mike. That was the answer I got and the one I was looking for.

Chat GPT looked at it differently. It found the valid solution of happy sitters: PEPEP and then worked out in how many ways it could arrange those three occupied seats in a row of five...

I'm happy that did not provide the correct answer.

--
David Entwistle

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