# Coin flipping game - expected value

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!

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$$\sum_{x=0}^\infty(x-0.90)\cdot0.45\cdot(0.55)^x$$

• The expression at the end of your post is flawed. Try computing it and you will see that you do not get the correct result. – Math1000 Dec 4 '19 at 4:34
• @Math1000 It matches my first answer. – Shubham Johri Dec 4 '19 at 4:40
• Oh, my bad. I converted the decimals into fractions incorrectly. Never mind then. – Math1000 Dec 4 '19 at 4:42

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