http://mostlikelytosuck.blogspot.com/
seriously, SOME of you really need to come up with some less defeatist blog names!!! kids these days...
Official Poll Results:
Most of us are not injured and running. Let us rejoice!
We think that Bolt and El Guerrouj are dirty and that Big Bekele is clean. We're not sure about Seb or Geb.
Kara Goucher wears the pants. Adam carries the baby sling.
If our coach had a lesser opinion of our athletic potential than ourselves, a slim plurality of us would want him or her to let us know.
When given the choice between being the world's best and dominating runner in one distance event or ten billion dollars, we would pick the former...barely.
If we could be any Caltech runner besides ourselves, we would most likely be Ian. Garrett, Katherine, Mark, and Rosen come in second.
Four out of four voters agree: everybody should update his or her log more often. I mean, seriously.
If we could be a world-class runner in one distance event, we'd pick either the 800m, the 1500/Mile, or the Marathon. The 10000m, steeplechase, and ultra-marathon come in second.
The cross-country course at Bush's Pasture Park in Salem, Oregon is more enjoyable than those in Chino, La Mirada, and even Riverside. It's also more enjoyable than "Your Mom," but that's not a legitimate cross-country course.
Markkimarkkonnen and Billy 'Reclaimed Water' Zdon are even money for being the craziest Caltech runners.
"Ribwich" is the most under-utilized nickname for a really skinny runner. "Pork Chop" and "Bones" are pretty good, too.
Alan Webb was more disappointing than Adam Goucher at the 2008 US Olympic Trials.
The 1500/Mile is the most fun distance event to race.
Out of the entire North Field Track Club, Markimarkkonnen is the most likely to fail a gender test.
Katherine should run the 1500 - 5000 double at conferences.
Katherine should focus on the 5000m this track season, based on the Goldilocks Principle (it's juuuuust right.)
Sara Hall, or Both Ryan and Sara At The Same Time
In the past four weeks:
77% of us have consumed alcohol, 77% have used Gatorade, 55% have used Endurox R4, 44% have shaved our legs, have smoked pot and 11% have used EPO.
Tyson Gay is the best 100m runner in the world (right now).
Matt Tegenkamp is the runner most likely to break Bob Kennedy's American 5000m record.
This is a pipe. Also it is Ian's Bones (see right). I
wonder if Ian's hollow bones means he can fly like a bird.
Highway 39 (or some variation) is a totally sweet ride and is the favorite of many.
Alan Webb will get First Place (GOLD, BABY!) at the 2007 Osaka World Athletics Championship!!! (If we are wrong we suck).
This Nerd-Jock Blog is Amazingly Awesome, the best idea ever, but it borders dangerously as a huge waste of time, which is why we're on it, voting.
The Last Stage of the Tour de France is a fine tradition honoring the great acomplishments of the previous weeks' racing.
Most anticipated log: Ian's Exercise Log
We think that Bolt and El Guerrouj are dirty and that Big Bekele is clean. We're not sure about Seb or Geb.
Kara Goucher wears the pants. Adam carries the baby sling.
If our coach had a lesser opinion of our athletic potential than ourselves, a slim plurality of us would want him or her to let us know.
When given the choice between being the world's best and dominating runner in one distance event or ten billion dollars, we would pick the former...barely.
If we could be any Caltech runner besides ourselves, we would most likely be Ian. Garrett, Katherine, Mark, and Rosen come in second.
Four out of four voters agree: everybody should update his or her log more often. I mean, seriously.
If we could be a world-class runner in one distance event, we'd pick either the 800m, the 1500/Mile, or the Marathon. The 10000m, steeplechase, and ultra-marathon come in second.
The cross-country course at Bush's Pasture Park in Salem, Oregon is more enjoyable than those in Chino, La Mirada, and even Riverside. It's also more enjoyable than "Your Mom," but that's not a legitimate cross-country course.
Markkimarkkonnen and Billy 'Reclaimed Water' Zdon are even money for being the craziest Caltech runners.
"Ribwich" is the most under-utilized nickname for a really skinny runner. "Pork Chop" and "Bones" are pretty good, too.
Alan Webb was more disappointing than Adam Goucher at the 2008 US Olympic Trials.
The 1500/Mile is the most fun distance event to race.
Out of the entire North Field Track Club, Markimarkkonnen is the most likely to fail a gender test.
Katherine should run the 1500 - 5000 double at conferences.
Katherine should focus on the 5000m this track season, based on the Goldilocks Principle (it's juuuuust right.)
Sara Hall, or Both Ryan and Sara At The Same Time
In the past four weeks:
77% of us have consumed alcohol, 77% have used Gatorade, 55% have used Endurox R4, 44% have shaved our legs, have smoked pot and 11% have used EPO.
Tyson Gay is the best 100m runner in the world (right now).
Matt Tegenkamp is the runner most likely to break Bob Kennedy's American 5000m record.
This is a pipe. Also it is Ian's Bones (see right). I
wonder if Ian's hollow bones means he can fly like a bird.Highway 39 (or some variation) is a totally sweet ride and is the favorite of many.
Alan Webb will get First Place (GOLD, BABY!) at the 2007 Osaka World Athletics Championship!!! (If we are wrong we suck).
This Nerd-Jock Blog is Amazingly Awesome, the best idea ever, but it borders dangerously as a huge waste of time, which is why we're on it, voting.
The Last Stage of the Tour de France is a fine tradition honoring the great acomplishments of the previous weeks' racing.
Most anticipated log: Ian's Exercise Log
13 comments:
IS THAT DAVID CUPCAKE ROSEN?!!??!
But you're right. I mean, look at the trend, and my trend I mean the two examples provided: Garrett and Dave. I guess they're roommates this summer, but "runhardloseanyways" and "mostlikelytosuck" are pretty pessimistic.
thank goodness cupcake is here. rosen figure this out for me:
with replacement, makes x draws from a bag of y balls. if x>y, what is the probability that each ball was drawn at least once?
there are y^x possible sequences of drawing the balls, but I don't know how to count how many of those have all the balls in them.
PS - hi
That sounds like a goddamn AIME question. The strategy is always something about pigeons and holes...
Yo Mark,
I've got a solution to this problem, but I don't have the time to fully explain it at the moment (I've got jury duty tomorrow, so I gotta hit the sack). Also, it requires some notation that can't really be expressed in the comment box here. I'll find some way of hosting the solution as a .pdf on my server tomorrow, and I'll put up a link to it here so that the entire Caltech XC community can partake of the joys of combinatorics :-).
sweet, thanks dude. the original problem actually came up during dinner conversation, and rephrased to make sense out of context was,
"everyone in the world has a telephone and the telephone number of everyone else. a single person is seeded with a rumor. he calls two other people at random and tells them the rumor. they then call two people each at random, etc. what is the expectation value of the number of cycles necessary for everyone in the world to hear the rumor?"
Hey Dave and I share our own little quirks as runners that unique allow us to have such names. For instance, Dave is fast, but spends his summer losing at robotics competitions instead of putting his running talent to use. I, on the other hand, have little talent, but train as though I have some hope of success. Hence our light-hearted blog names have a surprising amount of accuracy.
Ok, for your viewing pleasure, the solution to Mark's original question is available at http://bag.caltech.edu/documents/MarkProblem.pdf
If you're unfamiliar with multinomial coefficients or the Multinomial Distribution, Wikipedia has excellent articles on both.
damn, that shit is HOT. i took an entire 100-level class on probability and stats and i didn't even know of such a distribution. obviously, i was aware that such a distribution must exist as a solution to the problem, but had never seen it before.
although, now that i think about it a little more, i'm sorry to say that it didn't really help.
in fact, I had already figured out the answer given. what i don't know how to do is take "the summation evaluated all sequences" which satisfy the condition k1+k2+...kb = d
for example, how would i numerically evaluate the expression for k=6*10^9 and d=2^35?
"for example, how would i numerically evaluate the expression for k=6*10^9 and d=2^35?"
Hmm... Well, first let start by copping out. Try to keep in mind that I'm a mathematician, so my responsibilities only extend as far as saying "a solution exists", and sometimes (if I'm not too lazy) coming up with a complicated formula for the answer that looks good enough for someone to give me money for it. Actually doing USEFUL things (like calculating numbers) isn't really my field...
One way that you can try to simplify the problem is by considering a sequence to be a set equipped with an additional ordering. We have the advantage in this case that all of our sequences will have the same length. Instead of considering all sequences of positive integers of length b whose sum is d, we can simply consider NONDECREASING sequences of positive integers of length b whose sum is d; once we have found such a sequence, we look at the multinomial probability for that number, and multiply by b! (the number of ways of permuting b distinct elements) to capture all of the other sequences that utilize that same set of numbers (thus reducing the computational complexity of the problem by many, many orders of magnitude).
However, the basic difficulty that you're running up against here is that you're looking at a combinatorial problem involving huge numbers. I know that this frequently comes up in quantum physics and thermodynamics when considering large ensembles of particles, so I think the physicists may have a leg up when it comes to developing techniques to approximate these kinds of combinatorial problems (for example, Stirling's Approximation comes to mind). In any case, we've reduced to the case where we need only find distinct SETS of b positive integers whose sum is d.
I haven't looked in to this yet, but the first method of attack that comes to mind for trying to find such sets would be some kind of recurrence relation. I say this because the base case is trivial; if b = d, then clearly, every number in our set of b positive integers must be 1. If d = b + 1, then exactly one of the numbers must be 2, etc. You can see based on these examples how although calculating the possible sets directly might be difficult, it might not be so bad if you consider a recurrence relation (a recursive mathematical relation, like the Fibonacci numbers). Sometimes, if you're lucky, it turns out that the equation that specifies the recursion can actually be solved explicitly for the nth term, which then gives a closed-form solution (this approach can actually be done in the case of the Fibonacci numbers).
More generally, if I were going to do this problem myself (and the numbers were smaller) I would simply write a computer program that would just exhaustively calculate every such possible set of positive integers and then evaluate the multinomial probability. But again, you run into the whole computational complexity problem. I don't know if there is some clever way you'll be able to get around this. Katherine' a computer scientist, and they spend a lot of time worrying about "trivial" shit like this. You might consider asking her for some ideas....
Sorry it wasn't quite as helpful as you would have liked...
it's cool. i'm pretty sure the answer to the telephone problem is 42
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