Expected value of number comprising randomly selected digits and a decimal point You are given a bag with eleven tiles, one each marked with a different digit $0$ to $9$ or a decimal point.  The tiles will be pulled from the bag one at a time and placed in sequence on the table to produce a single decimal number.  This resulting number will be your prize.  Compute the expected value of your prize.  

Here is my solution to the problem:

The average value considering only the digits is $4999999999.5$ (i.e., if the decimal point is the last chosen).  Each candidate position for the chosen decimal point tile has an equal $\dfrac{1}{11}$ chance of being the case.  Therefore, the answer is $\dfrac{1}{11} \big(4999999999.5(10^{0}+10^{-1}+\cdots+10^{-11}\big)\big) = 505050505$.
The last equation is incorrect. 10 digits have 11 positions to put the decimal point. So it should be sum of series (10^0, 10^-1, 10^-2....10^-10). 

10^-11 should not appear in the equation
The last equation is incorrect. 10 digits have 11 positions to put the decimal point. So it should be sum of series (10^0, 10^-1, 10^-2....10^-10). 

10^-11 should not appear in the equation