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Difference between revisions of "User:DVO"

 
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*  Created [[Chain Rule]] article
 
*  Created [[Chain Rule]] article
 
*  Created [[Fundamental Theorem of Calculus]] article
 
*  Created [[Fundamental Theorem of Calculus]] article
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== Random Math Problem ==
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Suppose you have a fair six-sided die and you're going to roll the die again and again indefinitely.  What's the expected number of rolls until a <math>1</math> comes up on the die?
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The probability that it will take one roll is <math>\frac{1}{6} </math>.
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The probability that it will take two rolls is <math>\left(\frac56 \right)\left(\frac16 \right) </math>.
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The probability that it will take three rolls is <math>\left(\frac56 \right)^2 \left(\frac16 \right) </math>.
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The probability that it will take four rolls is <math>\left(\frac56 \right)^3 \left(\frac16 \right) </math>.
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And so on.
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So, the expected number of rolls that it will take to get a <math>1</math> is:
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<math>1\cdot \frac{1}{6} + 2\cdot \left(\frac56 \right)\left(\frac16 \right) + 3\cdot \left(\frac56 \right)^2 \left(\frac16 \right) + 4 \cdot \left(\frac56 \right)^3 \left(\frac16 \right) + \cdots</math>.
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What's the sum of this infinite series?  It looks kind of like an infinite geometric series, but not exactly.  Factoring out a <math>\frac16</math> makes it look a bit more like an infinite geometric series:
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<math>\left(\frac16 \right) \left(1 + 2\cdot \left(\frac56 \right) + 3\cdot \left(\frac56 \right)^2 + 4 \cdot \left(\frac56 \right)^3 + \cdots \right)</math>
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This is similar to a geometric series, which we know how to sum.  But we have those annoying factors of <math>2</math>, <math>3</math>, <math>4</math>, etc. to worry about.  Maybe we could play around with the formula for a geometric series to get a formula for this series.  The formula for a geometric series is:
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<math>1 + r + r^2 + r^3 + r^4 + \cdots = \frac{1}{1-r}</math>.
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Differentiating both sides of this with respect to <math>r</math>, we find:
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<math>1 + 2r + 3r^2 + 4r^3 + \cdots = -(1-r)^{-2}(-1) = \frac{1}{(1-r)^2}</math>.
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So, we find that
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<math>\left(\frac16 \right) \left(1 + 2\cdot \left(\frac56 \right) + 3\cdot \left(\frac56 \right)^2 + 4 \cdot \left(\frac56 \right)^3 + \cdots \right) = \frac16 \frac{1}{(1-\frac56)^2} = \frac16 (36) = 6</math>.
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Which seems like a very reasonable answer, given that the die has six sides.

Revision as of 01:11, 28 June 2006

Personal info

Name: Daniel O'Connor

(full name: Daniel Verity O'Connor)

Location: Los Angeles


Contributions


Random Math Problem

Suppose you have a fair six-sided die and you're going to roll the die again and again indefinitely. What's the expected number of rolls until a $1$ comes up on the die?

The probability that it will take one roll is $\frac{1}{6}$.

The probability that it will take two rolls is $\left(\frac56 \right)\left(\frac16 \right)$.

The probability that it will take three rolls is $\left(\frac56 \right)^2 \left(\frac16 \right)$.

The probability that it will take four rolls is $\left(\frac56 \right)^3 \left(\frac16 \right)$.

And so on.

So, the expected number of rolls that it will take to get a $1$ is:

$1\cdot \frac{1}{6} + 2\cdot \left(\frac56 \right)\left(\frac16 \right) + 3\cdot \left(\frac56 \right)^2 \left(\frac16 \right) + 4 \cdot \left(\frac56 \right)^3 \left(\frac16 \right) + \cdots$.

What's the sum of this infinite series? It looks kind of like an infinite geometric series, but not exactly. Factoring out a $\frac16$ makes it look a bit more like an infinite geometric series:

$\left(\frac16 \right) \left(1 + 2\cdot \left(\frac56 \right) + 3\cdot \left(\frac56 \right)^2 + 4 \cdot \left(\frac56 \right)^3 + \cdots \right)$

This is similar to a geometric series, which we know how to sum. But we have those annoying factors of $2$, $3$, $4$, etc. to worry about. Maybe we could play around with the formula for a geometric series to get a formula for this series. The formula for a geometric series is:

$1 + r + r^2 + r^3 + r^4 + \cdots = \frac{1}{1-r}$.

Differentiating both sides of this with respect to $r$, we find:

$1 + 2r + 3r^2 + 4r^3 + \cdots = -(1-r)^{-2}(-1) = \frac{1}{(1-r)^2}$.

So, we find that

$\left(\frac16 \right) \left(1 + 2\cdot \left(\frac56 \right) + 3\cdot \left(\frac56 \right)^2 + 4 \cdot \left(\frac56 \right)^3 + \cdots \right) = \frac16 \frac{1}{(1-\frac56)^2} = \frac16 (36) = 6$.

Which seems like a very reasonable answer, given that the die has six sides.

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