10at10 Club
Main Discussion Area => KFOG's 10@10 => Topic started by: RGMike on December 20, 2013, 10:02:04 AM
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Xmas or not?
Not. Jeebus -- THREE '90s sets in one week. I'm outta here.
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Lisa Loeb leads off with "I Do". Don't groan too loud everybody.
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Not. Jeebus -- THREE '90s sets in one week. I'm outta here.
Is this the first time this has happened for the 90s?
I do remember one week when AL did 3 70s sets (and 2 themes).
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Not. Jeebus -- THREE '90s sets in one week. I'm outta here.
Is this the first time this has happened for the 90s?
I do remember one week when AL did 3 70s sets (and 2 themes).
Well, it does lend creedence to the idea that the wheel is random.
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And the GSers are going to go apeshit over Sugar Ray. It's "Fly".
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Not recognizing #3. Apparently it's Toad The Wet Sprocket, "Throw It All Away".
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Everclear's "I Will Buy You A New Life" is #4.
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Sister Hazel: F- on the Necessary List.
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And the GSers are going to go apeshit over Sugar Ray. It's "Fly".
Apeshit mad or apeshit glad?
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And the GSers are going to go apeshit over Sugar Ray. It's "Fly".
Apeshit mad or apeshit glad?
One guess. Sarah McLachlan's "Building A Mystery" is a tremendous step up, of course pretty much any song would be.
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And apparently it's the unedited version. ;D
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And the GSers are going to go apeshit over Sugar Ray. It's "Fly".
Apeshit mad or apeshit glad?
One guess. Sarah McLachlan's "Building A Mystery" is a tremendous step up, of course pretty much any song would be.
Mad?
I'll BOS Building a Mystery.
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And apparently it's the unedited version. ;D
LOL! Go Renee! :D
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#7: Jonny Lang, "Lie To Me".
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#7: Jonny Lang, "Lie To Me".
BOS, because it's the first track that caught my attention. Nice guitar wankery.
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Not recognizing #8. Big Head Todd & the Monsters - Resignation Superman... thanks Rad
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BOS, because it's the first track that caught my attention. Nice guitar wankery.
[/quote]#7: Jonny Lang, "Lie To Me".
15 years old when he recorded it.
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So far this is closest to my BOS, although nothing has really jumped out at me yet.
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#9: INXS' last hit with Michael Hutchence, "Elegantly Wasted".
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Radiohead's "Karma Police" closes.
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Good/Necessary
1. Loeb B+//B+
2. Sugar Ray C/F
3. Toad C/B+
4. Everclear B/D
5. ShittyHazel F/F-
6. Sarah A/B
7. Jonny B/B
8. BHTATM C/B
9. INXS B+/B
10. Radiohead B+/B+
Apart from two egregious LNs, not as bad as it could've been.
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Per Renee: BOS: Tie between Toad and Radiohead.
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5. ShittyHazel F/F-
Bwahahahaha! Damn, I hated them. I remember writing to the then-PD of KFOG and asking why they had to play every Hootie-clone that came down the pike. I got a really snotty reply.
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Xmas or not?
Not. Jeebus -- THREE '90s sets in one week. I'm outta here.
Ugh. I was really hoping for a '60s set. I'm not upset that I'll be missing tonight's replay.
A weak week, what with three '90s sets, and generally high LNQs overall. Tuesday's 1972 was
the one standout, and that's a strong shoulder year to the incredible '69-'71 musical peak.
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Not. Jeebus -- THREE '90s sets in one week. I'm outta here.
Is this the first time this has happened for the 90s?
I do remember one week when AL did 3 70s sets (and 2 themes).
Well, it does lend creedence to the idea that the wheel is random.
It may be "more random" than before, but it's demonstrably not completely random.
Here's an illustrative little puzzle: assume there are forty years (1963 through 2002 inclusive)
on the wheel. If it was truly random, how often would there be a week with at least one
(or more) duplicated years?
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Not. Jeebus -- THREE '90s sets in one week. I'm outta here.
Is this the first time this has happened for the 90s?
I do remember one week when AL did 3 70s sets (and 2 themes).
Well, it does lend creedence to the idea that the wheel is random.
It may be "more random" than before, but it's demonstrably not completely random.
Here's an illustrative little puzzle: assume there are forty years (1963 through 2002 inclusive)
on the wheel. If it was truly random, how often would there be a week with at least one
(or more) duplicated years?
While this wasn't exactly in the same calendar week, we got 1993 on both July 5 & 11 of this year.
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Not. Jeebus -- THREE '90s sets in one week. I'm outta here.
Is this the first time this has happened for the 90s?
I do remember one week when AL did 3 70s sets (and 2 themes).
Well, it does lend creedence to the idea that the wheel is random.
It may be "more random" than before, but it's demonstrably not completely random.
Here's an illustrative little puzzle: assume there are forty years (1963 through 2002 inclusive)
on the wheel. If it was truly random, how often would there be a week with at least one
(or more) duplicated years?
Assuming a five day week, and assuming no theme sets or repeats, something like:
(1/40)+(2/40)(39/40)+(3/40)(38/40)+(4/40)(37/40)
I believe the third and fourth days require extra terms, but my head's a bit sore right now.
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Just listened on SoundCloud...
2013-12-20 - Friday! It's 1997 today on 10@10, the year Ellen DeGeneres came out, (YAY ELLEN!) on her sitcom.
1. Lisa Loeb - I Do
(TV: Seinfeld - George sings for his answering machine)
2. Sugar Ray - Fly
3. Toad The Wet Sprocket - Throw It All Away (BOS)
(Sports: Mike Tyson bites Evander Holyfield's ear, refereed by Mills Lane)
4. Everclear - I Will Buy You A New Life
5. Sister Hazel - All For You
6. Sarah McLachlan - Building A Mystery
7. Jonny Lang - Lie To Me
8. Big Head Todd & The Monsters - Resignation Superman
9. INXS - Elegantly Wasted
(Movie: Good Will Hunting - why should I work for the NSA?)
10. Radiohead - Karma Police (BOS)
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Well, it does lend creedence to the idea that the wheel is random.
It may be "more random" than before, but it's demonstrably not completely random.
Here's an illustrative little puzzle: assume there are forty years (1963 through 2002 inclusive)
on the wheel. If it was truly random, how often would there be a week with at least one
(or more) duplicated years?
No math nerds or puzzle enthusiasts?
Spoiler alert: The solution is presented below, after a couple of hints. Stop reading now if you
still want to try to solve it.
Since we are assuming a random distribution, with each possible arrangement of five 10@10
shows drawn from a pool of forty different years being equally likely, then the probability of
a week with one or more duplicated years is simply the number of arrangements with duplicates
divided by the total number of different arrangements. For example, if there were only ten
different arrangements, and three of them contained duplicates, then the probability would
be 3/10 = .30, or a 30% chance that a randomly chosen week would contain duplicate years.
The actual numbers are different of course. What eases their calculation is first knowing that
the probabilty of a week with duplicates is equal to one minus the probability of a week
without duplicates, i.e., if we know the probability of one we can compute the other, and
vice versa. The second piece is the realization that it is easier to determine the number of
arrangements without duplicates than the number of arrangements with duplicates.
The total number arrangements is simply 40 to the fifth power (40^5), since we have a choice of
forty different years for each of five days. The number of arrangements without duplicates is
calculated as follows: on the first day, we a choice of any of 40 different years. On the second
day, there are only 39 different years to choose from, as one was used on Monday. Similarly,
on the third day we have only 38 years for our choice, as one was used on Monday and a second
on Tuesday. We have 37 choices on the fourth day and finally 36 choices on the fifth day. The
total number of arrangements without duplicates is 40 x 39 x 38 x 37 x 36.
So the probability we will have week without duplicates is the above number divided by 40^5,
which is .7711... or just over 77%. Thus the probability that there would be a week with
duplicates is 1 - .7711 or .2289 -- almost 23%. If the wheel was truly random, (and ignoring theme
sets), we could expect a week with one or more duplicated years twice every nine weeks, or just
a little less than once a month.
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Well, it does lend creedence to the idea that the wheel is random.
It may be "more random" than before, but it's demonstrably not completely random.
Here's an illustrative little puzzle: assume there are forty years (1963 through 2002 inclusive)
on the wheel. If it was truly random, how often would there be a week with at least one
(or more) duplicated years?
No math nerds or puzzle enthusiasts?
Spoiler alert: The solution is presented below, after a couple of hints. Stop reading now if you
still want to try to solve it.
Since we are assuming a random distribution, with each possible arrangement of five 10@10
shows drawn from a pool of forty different years being equally likely, then the probability of
a week with one or more duplicated years is simply the number of arrangements with duplicates
divided by the total number of different arrangements. For example, if there were only ten
different arrangements, and three of them contained duplicates, then the probability would
be 3/10 = .30, or a 30% chance that a randomly chosen week would contain duplicate years.
The actual numbers are different of course. What eases their calculation is first knowing that
the probabilty of a week with duplicates is equal to one minus the probability of a week
without duplicates, i.e., if we know the probability of one we can compute the other, and
vice versa. The second piece is the realization that it is easier to determine the number of
arrangements without duplicates than the number of arrangements with duplicates.
The total number arrangements is simply 40 to the fifth power (40^5), since we have a choice of
forty different years for each of five days. The number of arrangements without duplicates is
calculated as follows: on the first day, we a choice of any of 40 different years. On the second
day, there are only 39 different years to choose from, as one was used on Monday. Similarly,
on the third day we have only 38 years for our choice, as one was used on Monday and a second
on Tuesday. We have 37 choices on the fourth day and finally 36 choices on the fifth day. The
total number of arrangements without duplicates is 40 x 39 x 38 x 37 x 36.
So the probability we will have week without duplicates is the above number divided by 40^5,
which is .7711... or just over 77%. Thus the probability that there would be a week with
duplicates is 1 - .7711 or .2289 -- almost 23%. If the wheel was truly random, (and ignoring theme
sets), we could expect a week with one or more duplicated years twice every nine weeks, or just
a little less than once a month.
Did you not see my response?
The chance of Days 1 and 2 being the same are 1 out of 40 (same set repeated).
The chance of having at least one repeat in Days 1-3 would be that 1/40 plus 2/40 times the complement of 1/40, or 39/40, or (1/40) + (2/40)(39/40), and so forth, which would eventually result in 1- (40)(39)(38)(37)(36)/40^5.
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Did you not see my response?
The chance of Days 1 and 2 being the same are 1 out of 40 (same set repeated).
The chance of having at least one repeat in Days 1-3 would be that 1/40 plus 2/40 times the complement of 1/40, or 39/40, or (1/40) + (2/40)(39/40), and so forth, which would eventually result in 1- (40)(39)(38)(37)(36)/40^5.
I'm sorry, but I can only give you partial credit. But you should be able to make it up on the final
exam.
Your expression would not reduce to mine. You were on the right track when you mentioned
extra terms were needed. The probability is an additional fortieth for each additional day, but
only for the percentage of the time when the existing days didn't already have a repeat. In
other words, the sum of:
Day1 = 0
Day2 = 1/40 * (1 - Day1) = 1/40 * (1 - 0) = 1/40
Day3 = 2/40 * (1 - Day2) = 2/40 * (1 - 1/40) = 2/40 * 39/40 = 78/1600
Day4 = 3/40 * (1 - Day3) = 3/40 * (1 - 78/1600) = 3/40 * 1522/1600 = 4566/64000
Day5 = 4/40 * (1 - Day4) = 4/40 * (1 - 4566/64000) = 4/40 * 59434/64000 = 237736/2560000
Unfortunately, this sums to 0.2379..., which doesn't agree with my result. Either I've made a
mistake in my calculations, or else summing the probabilities of successive days is not equivalent
to counting all the five-day combinations with repeated years. I will have to ponder this further.
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I asked in another forum and got the answer, which seems obvious in retrospect. I wasn't
subtracting the probability for *all* the previous days in my expressions. They are:
Day1 = 0
Day2 = 1/40 * (1 - Day1) = 1/40
Day3 = 2/40 * (1 - Day2 - Day1) = 78/1600
Day4 = 3/40 * (1 - Day3 - Day2 - Day1) = 4446/64000
Day5 = 4/40 * (1 - Day4 - Day3 - Day2 - Day 1) = 219336/2560000
And these do sum to 0.2288... which is the result I got with the other method, which is a
more elegant solution, as they say in the math biz.