I am a little embarrassed to ask this question, but for the life of me I cannot figure it out.
A man pays \$1 a throw to win a \$3 doll. His probability of landing a throw is 0.1. What is the probability that two throws will be required to win the doll? What is the probability that $x$ throws is needed?
The book said the probability that $x$ throws is required is $0.1(0.9)^{x-1}$. How did they get this?
I tried using a binomial distribution, but I don't know the number of trials
EDIT:
I have another similar problem, but i am having problems with determining the success rate
Three men toss coins to see who pays for coffee. If all three match, they toss again. Otherwise the "odd man" pays for the coffee. What is the probability that they will need to do this more than once? And at most twice?
I basically thought that outcomes like HHH is equal to TTT and likewise any combination of HHT = TTH.
So essentially we either stop after the first toss or we toss again and that makes the success probability equal to 1/2 and the failure is equal to 1/2.