
Harvard Physics Problem of the Week (2002-2004) - rfreytag
https://www.physics.harvard.edu/academics/undergrad/problems
======
liamcardenas
Wow I love this!

I have a very limited understanding of physics, but I do plan on tackling the
problems in the book linked at the top of that page, by the same author. It
just requires basic algebra and sometimes a bit of calculus:

The Green-Eyed Dragons and Other Mathematical Monsters

> This book is a collection of 57 very challenging math problems with detailed
> solutions. It is written for anyone who enjoys pondering difficult problems
> for great lengths of time. The problems are mostly classics that have been
> around for ages. They are divided into four categories: General, Geometry,
> Probability, and Foundational, with the Probability section constituting
> roughly half the book. Many of the solutions contain extensions/variations
> of the given problems. In addition to the full solution, each problem comes
> with a hint. For the most part, algebra is the only formal prerequisite,
> although a few problems require calculus.Are you eager to tackle the
> Birthday Problem, Simpson’s Paradox, the Game-Show Problem, the Boy/Girl
> Problem, the Hotel Problem, and of course the Green-Eyed Dragons? If so,
> this book is for you! You are encouraged to peruse the problems via either
> the Look Inside feature on Amazon, or the author's Harvard webpage (where
> all of the problems are posted), to gauge whether the level of difficulty is
> right for you.

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pseudoramble
I had a hard time understanding part of the explanation to Problem 6 [1]. The
statement:

> Consider the following game. You flip a coin until you get a tails. The
> number of dollars you win equals the number of coins you end up flipping.
> (So if you immediately get a tails, you win one dollar; if you get one heads
> before a tails, you win two dollars, etc.) What is the expectation value of
> your winnings?

And then the beginning of the explanation [2]:

> There is a 1/2 chance that you win one dollar, a 1/4 chance that you win two
> dollars, a 1/8 chance that you win three dollars, etc. Therefore...

It took me a few times of reading it to understand this. If you choose to play
this game, you have a 100% chance of making at least $1. Why state it that
there's a 1/2 chance you win $1? It doesn't seem to help answer the problem
because assuming you're playing you will always always win $1. The reason they
say this (I think) is that there's really a 50% chance you win $1 and a 50%
chance you win $2. But again, this doesn't help you with expected winnings -
you know you've got something on the first pass. Maybe it help explains Part
2?

Anyway, cool concept!

[1]:
[http://web.physics.harvard.edu/uploads/files/undergrad/probw...](http://web.physics.harvard.edu/uploads/files/undergrad/probweek/prob6.pdf)
[2]:
[https://www.physics.harvard.edu/uploads/files/undergrad/prob...](https://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol6.pdf)

~~~
jackcarter
> Why state it that there's a 1/2 chance you win $1? It doesn't seem to help
> answer the problem because assuming you're playing you will always always
> win $1. The reason they say this (I think) is that there's really a 50%
> chance you win $1 and a 50% chance you win $2.

There is a 50% chance of winning _exactly_ $1. There's a 25% chance of winning
_exactly_ $2. It sounds like you might be interpreting it to mean "at least
$1" instead of "exactly $1".

~~~
pseudoramble
Thanks - This is exactly what I was doing thinking back on it! I found it
surprising because I landed on the same answer they did even with my
misunderstanding of the statement. Ultimately I ended up figuring out this
sequence:

> 1 + 1/2 + 1/4 + 1/8 + ... = 2

But this now seems like I got kind of lucky, and it might not always hold up
generally.

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melling
He’s got several self-published books:

[http://www.people.fas.harvard.edu/~djmorin/book.html](http://www.people.fas.harvard.edu/~djmorin/book.html)

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agumonkey
I read "of the weak"..

