> 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...