TITLE: Expected value of the number of bills
QUESTION [7 upvotes]: Every day I put 1 or 2 dollars in the piggy bank with probability $1/2$. 
What's is expected value number of 2-dollar bills when in the piggy bank will be for the first time at least 100 dollars ?
I know what is going on. Let $X$- number of 2-dollar bills.
1) $X=1$ :
1 dollar- 98 bills and 2 dollars- 1 bill or 1 dollar - 99 bills and 2-dollars- 1 bill
2) $X=2$
1 dollar- 96 bills and 2 dollars- 2 bills or 1 dollar- 97 bills and 2 dollars- 2 bills
etc.
Unfortunately, I can't think of a quick way.

REPLY [5 votes]: You either reach $100$, or you reach $99$ and then draw a $2$-dollar bill. The probability to reach $100$ with $k$ $2$-dollar bills and $100-2k$ $1$-dollar bills  is $\binom{100-k}k2^{k-100}$. The probability to reach $99$ with $k$ $2$-dollar bills and $99-2k$ $1$-dollar bills is $\binom{99-k}k2^{k-99}$. Thus the expected number of $2$-dollar bills is
\begin{eqnarray*}
&&\sum_{k=0}^{50}k\binom{100-k}k2^{k-100}+\frac12\sum_{k=0}^{49}(k+1)\binom{99-k}k2^{k-99}\\
&=&\frac2{27}\left(299+301\cdot2^{-100}\right)+\frac1{27}\left(305-605\cdot2^{-100}\right)\\
&=&
\frac{301-3\cdot2^{-100}}9\\
&\approx&\frac{301}9\\
&=&33\frac49\;.
\end{eqnarray*}