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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?

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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.

One advantage of this choice is that all the outcomes are equally likely.

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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 $P(X=10)=P(\{(T,T,...,T,T),(T,T,...,T,H)\})=\frac{2}{2^{10}}=\frac{1}{2^9}$.

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  • $\begingroup$ I am assuming that you have answered for the case when there are 10 tosses. It was easy to intuitively figure out the possibilities where we could win 10 dollars, but it might not be that easy all the time, for example, if there are 100 tosses and we are asked to calculate P(X = 132 ) . There will be some dollar amounts that can't be won . For example , there is no way to win 11 dollars if we toss the coin 10 times . How can we mathematically take these cases into account ? $\endgroup$
    – CuriousMan
    Jul 31 '19 at 10:07
  • $\begingroup$ 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. $\endgroup$ Jul 31 '19 at 10:14
  • $\begingroup$ How do we know whether the dollar amount is possible or not without using intuition? $\endgroup$
    – CuriousMan
    Jul 31 '19 at 10:30

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