Expected value in coin flipping process

Question

You flip a coin, and if the result is tails, you lose. If the result is heads, you get to play again. What is the expected value of throws before you lose?

My Approach

The expected value is the sum of all the outcomes multiplied by their respective probabilities:$$\sum_{i=1}^{n}V_iP_i$$So for this problem:$$\sum_{i=1}^{\infty}i(\frac{1}{2^i})=0.5+0.5+0.375+…$$I can’t figure out how to find the sum, even though I know it converges.

Answer 1

In general, starting with a geometric series ($\sum_{j=0}^{\infty} x^j = 1/(1-x)$), you can evaluate many related series by taking derivatives or integrals.

• When you need to shift the exponent, multiply or divide by a power of $x$;
• When you need to multiply the coefficient by the exponent, take a derivative;
• When you need to divide the coefficient by the exponent, take an integral.

In this case, $$ \sum_{j=0}^{\infty} j x^j=x\frac{d}{dx}\sum_{j=0}^{\infty} x^{j}=x\frac{d}{dx}\frac{1}{1-x}=\frac{x}{(1-x)^2}. $$ Substituting $x=1/2$ gives $x/(1-x)^2=(1/2)/(1/2)^2=2.$