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- Xref: sparky sci.math:15111 sci.physics:19055
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- From: rryan@panix.com (Rob Ryan)
- Subject: Re: 3-sided coin
- Message-ID: <1992Nov17.135016.3532@panix.com>
- Date: Tue, 17 Nov 1992 13:50:16 GMT
- References: <1dsj4gINNeuh@agate.berkeley.edu> <1e98b8INN6s@manuel.anu.edu.au>
- Organization: Panix, NYC
- Lines: 35
-
- In <1e98b8INN6s@manuel.anu.edu.au> butler@rschp2.anu.edu.au (Brent Butler)
- writes:
-
- >The solution to this problem is simple. When a coin is flipped it
- >will *come to rest* in a position that represents a minimum in its
- >(gravitational) potential energy. For a real coin this means one
- >of its two faces with equal probability - it will never *come to
- >rest* on its edge even though it has a finite area because this is
- >a metastable state.
-
- Yes, it will come to rest at a minimum, but it doesn't have to be an
- absolute minimum, but rather a local minimum would do as well. Resting on
- edge is a relative minimum and is therefore possible. You can think of the
- bouncing coin as a form of gradient descent. With a coin of standard
- dimensions, so little energy is necessary get it out of that local minimum
- that it is extremely unlike to settle there, but when you make the coin
- wider, it becomes more likely.
-
- >Our three sided coin, therefor, should be 1 coin diameter thick. This
- >is where the center of mass is at equal heights in each of the three
- >states and so each configuration represents an equally likely
- >stable state for the coin to *come to rest* after a toss.
-
- Incidentally, what if the coin is 0.9999 coin diameter thick. By your
- argument above, since resting on the side would become a "metastable"
- state, it would never come to rest on it's side, would it?
-
- Are you applying some non-standard interpretation of "come to rest"?
- Can you elaborate upon your argument here?
-
- --
- Rob Ryan
- Internet: rryan@panix.com or Robert_Ryan@brown.edu
- Bitnet: ROBR@BROWNVM
- Compu$erve: 70324.227@compuserve.com
-