# Expected Value with Fair Coin Toss Game

You start with 100 dollars. When you flip a coin and receive a head, you gain 100 dollars. If you flip the coin and received a tail, you lose half of your current money. What is the expected amount of money you will have after 4 flips?

I know that if you gained 100 dollars on heads and lost 50 dollars on tails you could say: E(final money) = (100 * 0.5) - (50 * 0.5).

I also know that if you doubled your money on heads and halved your money on tails then E(final money) = 100 * 2^(# of heads) * (1/2)^(#of tails)

However, this problem seems to be mixing both addition and multiplication, so I am unsure of how to combine them into a single equation to receive a solution.

• There are so few scenarios that you can just do the computation by hand. – lulu Sep 14 '18 at 0:16
• There are 16 possibilities in total so just calculate each case. – abc... Sep 14 '18 at 0:19
• If $\mu_n$ is the expected money after $n$ flips, then $\mu_{n+1} = \frac{1}{2} (\mu_n + 100) + \frac{1}{2} (\frac{1}{2} \mu_n)$. – Daniel Schepler Sep 14 '18 at 0:22

Let $E_n$ be the expected money after n flips. Using recursion,

$$E_{n+1}=\frac12(\frac12E_{n}+(E_n+100))=\frac34E_n+50$$

$E_1=125$

$E_2=\frac{375}4+50=\frac{575}4$

$E_3=\frac{3\cdot575}{16}+50=\frac{2525}{16}$

$E_4=\frac{3\cdot2525}{64}+50=\frac{10775}{64}=168\frac{23}{64}$

Therefore your money increase by $68\%$. $:)$

Let's call $X_n$ a random variable that represents the number of money after $n$ coin flips. We have $$\mathbb{E}[X_{n+1} | X_n] = \frac{X_n+100}{2} + \frac{X_n}{4} = \frac{3X_n}{4} + 50$$ Moreover $$\mathbb{E}[X_{n+1}] = \mathbb{E}[\mathbb{E}[X_{n+1}|X_n]] = \frac{3}{4}\mathbb{E}[X_n] + 50$$ (I used standard properties of Conditional Expectation)