Re: The doctor's patients

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.