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<DIV><FONT face=3DArial color=3D#0000ff size=3D2><SPAN class=3D214055615-28=
092004>Glad=20
we have cleared up the "quarter of a quarter" problem. Assuming that =
bell=20
weights randomly end in any number of pounds from 0 to 27 then there is a 4=
/28 =3D=20
1/7 probability that the weight will end in 0,7,14 or 28. This is a=20
reasonable assumption given teh range of weights we are considering but wou=
ld=20
not be true for very small bells. </SPAN></FONT></DIV>
<DIV><FONT face=3DArial color=3D#0000ff size=3D2><SPAN=20
class=3D214055615-28092004></SPAN></FONT> </DIV>
<DIV><FONT face=3DArial color=3D#0000ff size=3D2><SPAN class=3D214055615-28=
092004>For a=20
ring of six to all end in such a number the probability would be 1/7^6 whic=
h is=20
0.000008499 in other words less than a 1 in 100 000 chance - be VERY=20
suspisious. Even for a ring of four this is less than a 1 in 2000=20
chance.</SPAN></FONT></DIV>
<DIV><FONT face=3DArial color=3D#0000ff size=3D2><SPAN=20
class=3D214055615-28092004></SPAN></FONT> </DIV>
<DIV><FONT face=3DArial color=3D#0000ff size=3D2><SPAN class=3D214055615-28=
092004>For=20
all six bells to end in 0 (Shaftsbury) the probability would be 1/28^6 whic=
h is=20
0.000000002 or around a 1 in 500 000 000 chance. Statistically I do n=
ot=20
believe that Shaftsbury all ended in 0.</SPAN></FONT></DIV>
<DIV><FONT face=3DArial color=3D#0000ff size=3D2><SPAN=20
class=3D214055615-28092004></SPAN></FONT> </DIV>
<DIV><FONT face=3DArial color=3D#0000ff size=3D2><SPAN class=3D214055615-28=
092004>For=20
any two bells out of eight to end in 0 is a less than 4% chance - worth=20
checking!</SPAN></FONT></DIV>
<DIV><FONT face=3DArial color=3D#0000ff size=3D2><SPAN=20
class=3D214055615-28092004></SPAN></FONT> </DIV>
<DIV><FONT face=3DArial color=3D#0000ff size=3D2><SPAN=20
class=3D214055615-28092004>Andrew</SPAN></FONT></DIV><!-- |**|end egp html =
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