Harvard Physics Problem of the Week (2002-2004) | Hacker News Reader

Harvard Physics Problem of the Week (2002-2004)

physics.harvard.edu

101 pointsrfreytag8 comments

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.

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... [2]: https://www.physics.harvard.edu/uploads/files/undergrad/prob...

what's the sum of n/2^n?

Edit: generatingfunctionology to the rescue

> 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".

Understanding this really helped me understand probability in college. The __exactly__ is the difference

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.

He’s got several self-published books:

http://www.people.fas.harvard.edu/~djmorin/book.html

I read "of the weak"..

Reply on news.ycombinator.com