Coin flipping game - expected value

Question

Given a weighted coin that lands on heads 55% of the time, you flip the coin until you get you get your first tails. For each heads, you make \$1. When you flip tails you lose \$0.90. What is the expected value of this game?

So far I have come up with the following: Let $X$ be a random variable that takes value 1 with probability 0.55 and -0.90 with probability 0.45. Then $E[X] = 1 \times 0.55 - (0.45 \times 0.90) = 0.145$. I have no idea if this is correct or not though? Would appreciate any help!

Answer 1

Consider getting a tails as a success and getting heads a failure, then the probability distribution of $r$ failures followed by one success is a geometric distribution with probability of success $p=P(\text{tails})=0.45$.

Let $X$ denote this random variable giving the number of failures before the first success. Then the value of the game is given by $X-0.90$, since each failure awards $\$1$ and the final success takes away $\$0.90$. Then the expected value is given by$$E[X-0.90]=E[X]-0.90=\frac{0.55}{0.45}-0.90\approx0.32$$

---

Alternatively, note that the value earned on getting $x$ heads followed by one tails is $x-0.90$. The probability of getting $x$ heads consecutively and then one tails is $0.55^x(0.45)$. The expected value is$$\sum_{x=0}^\infty(x-0.90)\cdot0.45\cdot(0.55)^x$$

Answer 2

An alternative approach to Shubham Johri's, giving the same answer

• If you get a heads, you receive $\$1$ and get to flip again

• If you get a tails, you pay $\$0.90$ and stop

• So $E[X]=0.55 \times (1+E[X]) +0.45\times (-0.90)$

• Giving $E[X] = \frac{0.55 - 0.45\times 0.90}{1-0.55} = \frac{29}{90} \approx \$0.32$