# What will be the sample space of this problem?

5 tosses of a coin are made We receive 1 dollar for every coin toss, up to and including the first time a head comes up. Then, we receive 2 dollars for every coin toss, up to the second time, a head comes up. More generally, the dollar amount per toss is doubled each time a head comes up.

What will the sample space be and how should I calculate it if I want to calculate the probability that I will make 10 dollars?

I think that in this game there will be only one unique way to win a certain dollar amount (I am unable to prove this concretely) For example, the only way to win 7 dollars is T T H T H , the only way to make 6 dollars is T T T H H .

I was thinking that the sample space would be A = P(5,1) + P(5,2) ... + P(5,5) . This will account for all the ways in which heads can occur in 5 tosses. Since I think there is only one unique way to get a certain dollar amount, the probability would be 1 /A.

Is my way of thinking correct?

I would just take the sample space to be $$\{TTTTT, TTTTH, TTTHT, TTTHH, \ldots \}$$. In other words, the set of all possible heads / tails sequences of length 5.
The sample space could be $$S:=\{H,T\}^{10}=\{H,T\}\times \{H,T\} \times ... \times \{H,T\}$$ and every outcome in $$S$$ has the same probability of $$\frac{1}{2^{10}}$$ (assuming independence of the coins). The amount of money won can be described by stochastic variable $$X:S\rightarrow \mathbb{R}$$.
Now to calculate the probability that $$X=10$$, we see that there is only two ways that this can happen, and it is if we rolled only tails or only tails and the last one a head, therefore $$P(X=10)=P(\{(T,T,...,T,T),(T,T,...,T,H)\})=\frac{2}{2^{10}}=\frac{1}{2^9}$$.
• I wrote the answer when the question said 10 coin flips, but the concept is the same. For any k $P(X=k)=\frac{ |\{s\in S\:|\: X(s)=k\}|}{2^n}$ and if there is no ways that $X=k$ then the probability is simply 0. Jul 31 '19 at 10:14