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Old 07-12-2009   #7 (permalink)
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CraigD
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Algebraic notation's traditional implied multiplication

Quote:
Originally Posted by Turtle View Post
.. anyway, i keep looking at your 'clue' post & either i'm missing something even more obvious or you have made an error. to whit:
Quote:
Originally Posted by Craig
For example, the expected value of a fair 6-sided die roll is
1\frac16+2\frac16+3\frac16+4\frac16+5\frac16+6\frac16 = \frac{21}6 = 3.5
which can be confirmed by noting that, after many die roles, the average roll approaches 3.5.
since 1\frac{1}6 = \frac{7}6 and 2\frac{1}6 = \frac{13}6, aren't we already up to \frac{20}6 and with the 4 other terms aren't we at \frac{95}6 ?
I believe you’re missing the traditional algebraic implied multiplication operator, Turtle.

That is
a\frac{b}{c}
means
a \cdot \frac{b}{c}
, not
a +\frac{b}{c}
, as you’re reading it.


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