Coin Flipping

To find the expected value what we do is we multiply each case of winning some dollars with their respective probability of happening and add all, finally divide it by total probability that is 1.

If we get tails on 1st attempt that is we win 1 dollars the probability is = $\frac{1}{2}$

If we get tails on 2nd attempt , we win 2 dollars the probability is equivalent to getting one head first and on the next attempt getting tail = $\frac{1}{2}.\frac{1}{2}=\frac{1}{{2}^{2}}$

If we get tails on 3 rd attempt, we win 3 dollars the probability is equivalent to getting 2 heads first and on the next attempt getting a tail = $\frac{1}{2}.\frac{1}{2}.\frac{1}{2}=\frac{1}{{2}^{3}}$

Generalising we get :

Probability of winning $n$ dollars is $= \frac{1}{{2}^{n}}$

Our expected value is $= 1.\frac{1}{2}+2.\frac{1}{{2}^{2}}+3.\frac{1}{{2}^{3}}+..........=\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { n }{ { 2 }^{ n } } } }$

Now we know that :

$\frac { 1 }{ 1-x } =\lim _{ n\rightarrow \infty }{ \sum _{ k=0 }^{ n }{ { x }^{ n } } }$

Differentiating both sides with respect to x we get :

$\frac { 1 }{ { (1-x) }^{ 2 } } =\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ n{ x }^{ n-1 } } }$

$\Rightarrow \frac { x }{ { (1-x) }^{ 2 } } =\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ n{ x }^{ n } } }$

Put $x=\frac{1}{2}$ to get :

$2=\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { n }{ { 2 }^{ n } } } }$ =Expected Value