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 == | ||
+ | |||
+ | 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? | ||
+ | |||
+ | The probability that it will take one roll is <math>\frac{1}{6} </math>. | ||
+ | |||
+ | The probability that it will take two rolls is <math>\left(\frac56 \right)\left(\frac16 \right) </math>. | ||
+ | |||
+ | The probability that it will take three rolls is <math>\left(\frac56 \right)^2 \left(\frac16 \right) </math>. | ||
+ | |||
+ | The probability that it will take four rolls is <math>\left(\frac56 \right)^3 \left(\frac16 \right) </math>. | ||
+ | |||
+ | And so on. | ||
+ | |||
+ | So, the expected number of rolls that it will take to get a <math>1</math> is: | ||
+ | |||
+ | <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: | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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: | ||
+ | |||
+ | <math>1 + r + r^2 + r^3 + r^4 + \cdots = \frac{1}{1-r}</math>. | ||
+ | |||
+ | Differentiating both sides of this with respect to <math>r</math>, we find: | ||
+ | |||
+ | <math>1 + 2r + 3r^2 + 4r^3 + \cdots = -(1-r)^{-2}(-1) = \frac{1}{(1-r)^2}</math>. | ||
+ | |||
+ | So, we find that | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | Which seems like a very reasonable answer, given that the die has six sides. |
Revision as of 00:11, 28 June 2006
Personal info
Name: Daniel O'Connor
(full name: Daniel Verity O'Connor)
Location: Los Angeles
Contributions
- Created Chain Rule article
- Created Fundamental Theorem of Calculus article
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 comes up on the die?
The probability that it will take one roll is .
The probability that it will take two rolls is .
The probability that it will take three rolls is .
The probability that it will take four rolls is .
And so on.
So, the expected number of rolls that it will take to get a is:
.
What's the sum of this infinite series? It looks kind of like an infinite geometric series, but not exactly. Factoring out a makes it look a bit more like an infinite geometric series:
This is similar to a geometric series, which we know how to sum. But we have those annoying factors of , , , 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:
.
Differentiating both sides of this with respect to , we find:
.
So, we find that
.
Which seems like a very reasonable answer, given that the die has six sides.