home *** CD-ROM | disk | FTP | other *** search
- Xref: sparky sci.math:14835 sci.physics:18757
- Newsgroups: sci.math,sci.physics
- Path: sparky!uunet!inmos!titan.inmos.co.uk!europa!stephen
- From: stephen@europa.inmos.co.uk (Stephen Collyer)
- Subject: Re: Three-sided coin
- Message-ID: <1992Nov12.144604.16373@titan.inmos.co.uk>
- Sender: stephen@europa (Stephen Collyer)
- Organization: INMOS Limited, Bristol, UK
- References: <5413@daily-planet.concordia.ca> <1992Nov10.032643.10467@galois.mit.edu>
- Date: Thu, 12 Nov 1992 14:46:04 GMT
- Lines: 44
-
- In article <1992Nov10.032643.10467@galois.mit.edu>, jbaez@riesz.mit.edu (John C. Baez) writes:
- |> In article <5413@daily-planet.concordia.ca> mckay@alcor.concordia.ca (John McKay) writes:
- |> >
- |> > von Neumann was asked, when about to enter a taxi,
- |> >
- |> >"How thick should a coin be to have equal (=1/3) probability
- |> > of landing on its head,tail,or edge?"
- |> >
- |> >He turned, told the taxi where to go, then answered and was
- |> >away.
- |> >
- |> >Surprisingly, making "reasonable" assumptions, this can be
- |> >done in your head.
- |>
- |> Well, this question has been seriously batted about before on sci.math
- |> and sci.physics, and the consensus was that there is simply not enough
- |> information to answer it. So I eagerly await what von Neumann's answer
- |> was supposed to have been - and what "reasonable" (note quotes!)
- |> assumptions go into it.
- |>
- |> My best shot at answer invoked an idea from stat mech, namely,
- |> equipartition. This says that the solution occurs when the *energy* of
- |> standing upright equals the energy of lying on edge. Guess this means
- |> the radius should equal half the thickness of the coin, or if you
- |> prefer, the diameter should equal the thickness.
-
- This seems wrong to me. Consider the cases where the coin is balanced on the
- edge between the head and the cylindrical section. It can fall either way,
- ending up on its on its head or its cylinder. Similarly for the other edge,
- it can end up on its tail or its cylinder. This would seem to weight the
- chances towards ending up on its cylinder, rather than its head or tail,
- if the diameter = thickness.
-
- Also if diameter = thickness, then A (tail) = A (edge) = 1/4 A (cylinder)
- (A = surface area) and surface area here is surely the determining factor.
-
- I guess we want to have A (tail) = A (edge) = A (cylinder) which occurs
- when thickness = 1/2 radius.
-
- |>
- |> However, I became convinced that in an actual coin toss conditions are
- |> not such that the equipartition principle is applicable.
-
- Steve Collyer.
-