Two comments:
(1) If the coins are numbered (on the non-zero side) 1, 2, 4, 8, ...
then each possible total occurs exactly once. If Derek's coins are
labelled 0 and X (X = 1, 2, 3, ...), these new coins are labelled 0 and
2^(X-1). I don't know if this observation is helpful, since I don't know
how Derek's "coins" are "really" generated, as it were.
(2) If the coins are unlimited in number, the left tail of the
distribution looks like this:
Total Number Coins
0 1 way 0000...
1 1 way 1000...
2 1 way 02000...
3 2 ways 12000..., 00300... [or, 12, 3.]
4 2 ways 13, 4.
5 3 ways 14, 23, 5.
6 4 ways 15, 24, 123, 6.
7 5 ways 16, 25, 34, 124, 7.
8 6 ways 17, 26, 35, 125, 134, 8.
9 8 ways 18, 27, 36, 45, 126, 135, 234, 9.
10 10 19, 28, 37, 46, 127, 136, 145, 235, 1234, 10.
11 12 1.10, 29, 38, 47, 56, 128, 137, 146, 236, 245, 1235, 11.
12 15 1.11,2.10,39,48,57,129,138,147,156,237,246,345,1236,1245,12.
13 18 1.12,2.11,3.10,49,58,67,1.2.10,139,148,157,238,247,256,346,
1237,1246,1345,13.
14 22 1.13,2.12,3.11,4.10,59,68,1.2.11,1.3.10,149,158,167,239,
248,257,347,356,1238,1247,1256,1346,2345,14.
etc.
and if the number of coins n is unlimited, the frequencies increase
without limit, so far as I can see. If the number of coins is limited,
the limitation imposes a right tail on the distribution, the maximum
total being n(n+1)/2, obtainable in only 1 way (all coins with non-zero
sides up); the maximum minus 1, and maximum minus 2, are also obtainable
in only one way (a single zero in place of 1, and a single zero in place
of 2, respectively). This suggests that perhaps the finite distribution
is symmetrical, which would imply a maximum frequency at
(n(n+1)/2)/2 if n(n+1)/2 is even, and equal maximum frequencies at
(n(n+1)/2)/2 +/- 1/2 if n(n+1)/2 is odd. I think.
On Sun, 24 Dec 2000, Derek Ross wrote:
> This is an odd distribution... I think it may somehow be related to the
> binomial distribution, but I'm not certain.
>
> The idea is difficult to explain, so here's a "real-life" example of
> generating the distribution:
>
> I have four coins in my hand. Each coin has a number on both sides. All
> coins have a zero on one side, and the other side has a number from 1 to 4.
>
> I then hurl the handful of coins at the ground, then add up the values of
> all the coins.
>
> Here's a list of every possible combination of ways the coins may land, with
> the corresponding total.
>
> Coins Total
> 0 0 0 0 0
> 1 0 0 0 1
> 0 2 0 0 2
> 1 2 0 0 3
> 0 0 3 0 3
> 1 0 3 0 4
> 0 2 3 0 5
> 1 2 3 0 6
> 0 0 0 4 4
> 1 0 0 4 5
> 0 2 0 4 6
> 1 2 0 4 7
> 0 0 3 4 7
> 1 0 3 4 8
> 0 2 3 4 9
> 1 2 3 4 10
>
> If you plot the histogram, you can see the faint shadow of a bell curve
> there.
>
> Histograms of totals:
> Total Quantity
> 0 1
> 1 1
> 2 1
> 3 2
> 4 2
> 5 2
> 6 2
> 7 2
> 8 1
> 9 1
> 10 1
>
>
> Well then, with all that being said, my question: what kind of distribution
> is this anyway? I know I can calculate everything by hand for a small number
> of coins, but say I had 100 coins, then I would need an equation to generate
> the distribution.
>
> Any help would be greatly appreciated...
>
> Derek Ross.
>
>
>
>
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Donald F. Burrill [EMAIL PROTECTED]
348 Hyde Hall, Plymouth State College, [EMAIL PROTECTED]
MSC #29, Plymouth, NH 03264 (603) 535-2597
Department of Mathematics, Boston University [EMAIL PROTECTED]
111 Cummington Street, room 261, Boston, MA 02215 (617) 353-5288
184 Nashua Road, Bedford, NH 03110 (603) 471-7128
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