## Teaser 2951: Imprismed

**From The Sunday Times, 14th April 2019** [link]

A right regular prism has two ends with identical faces, joined by oblong rectangular faces. I have eight of them, with regular convex polygonal end-faces of 3, 4, 5, 6, 7, 8, 9 and 10 sides (triangle, square and so on). They sit on my flat desk (on oblong faces), and each prism has the same height.

I chose three prisms at random, and was able to slide them into contact, broadside, in such a way that the middle one overhung both others (and could be lifted without disturbing them). Also, I was able to slide one outer prism to the other side, and the new “middle” prism was overhung by both others (and so vertically “imprisoned” by them).

I was able to do all this again with three randomly chosen remaining prisms.

Give the prior chance of this double selection (as a fraction in lowest terms).

[teaser2951]

## Jim Randell 11:30 pm

on11 April 2019 Permalink |I’m not a great fan of this kind of puzzle. Partly because I don’t like probability much, but also because it is not easy to check that the answer you get works.

I also think the wording for this particular puzzle could have been clearer. I started out with an incorrect interpretation, and got a different answer. If we consider the selection of an appropriate “special” set of three prisms (that can be arranged in the manner described in the puzzle text), then I think the puzzle is asking for the probability of selecting two consecutive “special” sets from the 8 prisms.

Here is a Python program which works out the solution in 75ms.

Run:[ @repl.it ]Solution:The probability of choosing two “special” sets is 3 / 140.Here is my analytical solution:

There are 12 “special” arrangements

(centre; outside)that allow a further “special” arrangement to be made from the remaining prisms. They group into 6 sets of mutually distinct pairs:There are also 4 “special” arrangements that do not allow a further “special” arrangement to be made:

For the first choice of prisms there are

C(8, 3)= 56 possible arrangements, and we must choose one of the arrangements that is part of a “special” pair.So, there is a probability of 12 / 56 = 3 / 14 of choosing a suitable first set of prisms.

For the second choice there are

C(8 – 3, 3)= 10 possible arrangements, but only one of these arrangements will be “special” – the arrangement that goes with the first choice to make one of the six “special” pairs given above.So overall probability of choosing two suitable sets is (3 / 14) × (1 / 10) = 3 / 140.

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## Robert Brown 7:44 am

on20 April 2019 Permalink |Have you seen John Crabtree’s simple method? I’m also unhappy with probabilities, but wrote a simple program to count how many ways you can select 6 prisms from 8, and how many of these meet his algorithm. The ratio of these 2 numbers confirms the probability.

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## Jim Randell 2:39 pm

on20 April 2019 Permalink |@Robert: Thanks for your comment.

I try to provide a constructive solution where possible, so constructing the “special” sets, and then counting the number of desired outcomes from all possible outcomes seemed like the best way to understand the solution (and would be the approach least likely to contain a mistake). But it is good that other approaches are able to confirm the answer.

For me the more difficult part was firstly interpreting what the puzzle was actually asking, and then constructing the “over” relation (to determine when one prism goes over another). Once that was done, counting the possibilities to come up with an answer was the easier part.

I have got a manual approach to calculating the answer using probabilities, which I intend to post tomorrow along with the answer.

I did also write code to run a million random trials in selecting six of the prisms to confirm the number I found (see the [[

`teaser2951t.py`

]] file on @repl.it).LikeLike

## Robert Brown 11:42 am

on22 April 2019 Permalink |Jim

I wasn’t being critical of your method, which corresponds to that used by the various people who solved it manually. Just drawing your attention to John’s clever method, which doesn’t use any lists – just a couple of rules. Rule one – 6 random selections don’t include 4 or 8 (probability 1/28) Rule 2 – dividing the 6 into 1st 3, then 2nd 3, prisms 6 & 10 must not be in the same group (probability 3/5).

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## Jim Randell 2:00 pm

on22 April 2019 Permalink |It’s certainly neat. I suppose the issue I have is the need to demonstrate that the two rules generate exactly all possible pairs of “special” sets.

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## Robert Brown 7:42 am

on23 April 2019 Permalink |After Rule 1, abc & def must use all of the other 6. Then after Rule 2 every pair within abc or def has one member overhanging the other.

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