Difference between revisions of "User:DVO"
(→Questions about Multiple View Geometry) |
(→For Prof. Welch) |
||
Line 1: | Line 1: | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
== Personal info == | == Personal info == | ||
Revision as of 16:58, 26 March 2011
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 Problems
The following problems are part of a discussion I'm having with someone I know. Please don't comment about them, and most importantly please don't correct any errors or give any hints about better solutions.
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.
What's the expected number of rolls it will take in order to get all six numbers at least once?
As a first step, let's find out the expected number of rolls it will take in order to get both and
at least once.
Let be an integer. What's the probability that it will take
rolls to get a
and a
?
Consider two different events: Event is the event that the
roll is a
, and the first
rolls give at least one
and also no
's. Event
is the event that the
roll is a
, and the first
rolls give at least one
and also no
's. The probability that it will take
rolls to get a
and a
is
.
What is ?
Let be the event that the
roll is a
. Let
be the event that the first
rolls give at least one
. Let
be the event that the first
rolls give no
's.
.
Event is independent of the event
, so:
.
is
, but what is
?
Events and
aren't independent. But we can say this:
. (Here
is the complement of the event
. I have used the fact that
.)
.
.
Thus, $P(B \cap C)= \left(\frac56 \right)^{n-1} \left(1- \frac{ (\frac46)^{n-1} }{ (\frac56)^{n-1} } \right)
= \left(\frac56 \right)^{n-1} - \left(\frac46 \right)^{n-1}$ (Error compiling LaTeX. Unknown error_msg).
So .
We have found . We're close to finding the probability that it will take n rolls to get both a
and a
.
. Thus the probability that it will take
rolls to get both a
and a
is
.
Okay.....
So what's the expected number of rolls it will take to get both a and a
?
It is:
.
The expected number of rolls is .
Using a similar (but a little more complicated) method, I get that the expected number of rolls until all six numbers appear is .
I also found that the expected number of rolls until ,
, and
appear is
.