1. Gambling

(The text of the problem is presented in its original form)
Ann and Bob are gambling at a casino. In each game the probability of winning a dollar is 48 percent, and the probability of losing a dollar is 52 percent.
Ann decided to play 20 games, but will stop after 2 games if she wins them both. Bob decide to play 20 games, but will stop after 10 games if he wins a least 9 out of the first 10.
What is larger: the amount of money Ann is expected to loose, or the amount of money Bob is expected to loose?

My difficulty is that the problem seems quite ambiguous as presented: does it mean that Ann will stop after the first two games if she wins, and if she loses she will play the whole 20 games even if a sequence of two wins appear again?
Same for Bob: is it only the first 10 games or the first "good sequence" appearance we are talking about here?
(The latter seems more realistic for the construction of a stopping time.)
Finally, is the game considered fair? e.g. if one bets a dollar, does he receives $\displaystyle \frac{1}{0.48}$ if he wins, or do we need to take into account the non-zero (negative) expectation when constructing the martingale?
But maybe there are some common rules for this kind of problem which I'm not familiar with.

2. Well, you're probably worrying too much about this innocent problem The most meaningful interpretation is that Ann stops either after the second game or after the 20th one, and similarly for Bob. And we know that 1 dollar is won or lost, thus we may deduce that $1 is bet and one receives$2 after a winning game (and $0 otherwise), but this doesn't matter to solve the question. 3. In that case, it boils down to using the total probability formula. One could formalize it using the stopping times:$\displaystyle \tau_A=\{2 $if two first wins,$\displaystyle 20$otherwise$\displaystyle \}$.$\displaystyle \tau_B=\{10 $if nine or ten first wins,$\displaystyle 20$otherwise$\displaystyle \}$. Then the expectations are:$\displaystyle \mathbb{E}(A)=\mathbb{E}(A|\tau=2).\mathbb{P}(\tau =2)+\mathbb{E}(A|\tau=20).\mathbb{P}(\tau=20)\displaystyle =\frac{1}{4}.2+\frac{3}{4}(-0.04\times20)=0.5-0.6=-0.1\displaystyle \mathbb{E}(B)=\mathbb{E}(B|\tau=10, 9W).\mathbb{P}(\tau=10, 9W)+\mathbb{E}(B|\tau=10, 10W).\mathbb{P}(\tau=10, 10W)$$\displaystyle +\mathbb{E}(B|\tau=20).\mathbb{P}(\tau=20) \displaystyle =(\frac{10}{2^{10}}.8+\frac{1}{2^{10}}.10)+(1-\frac{11}{2^{10}})(-0.04\times20)=-0.7 B is expected to loose more money... as expected. ...Could this be enough? 4. Originally Posted by akbar ...Could this be enough? Not quite: you forgot to take the conditioning into account in 2 terms: \displaystyle E[A|\tau_A=20]\neq 20\times (-0.04), it is less since at least one of the two first games is lost. And same with Bob. 5. There is also the fact the "coin" is biased. It is probably easier to subdivide the game sequences according to the number of wins. Which make the stopping times even more useless and a calculator even more needed. Here is my second attempt: \displaystyle \mathbb{E}(A)=\mathbb{E}(A|\tau=2).\mathbb{P}(\tau =2)+\mathbb{E}(A|\tau=20,1W1L).\mathbb{P}(\tau=20, 1W1L)$$\displaystyle +\mathbb{E}(A|\tau=20,2L).\mathbb{P}(\tau=20,2L)\displaystyle =(0.48)^2.2+2(0.48)(0.52)(0+18\times(-0.04))+(0.52)^2(-2+18\times(-0.04))\approx-0.634\displaystyle \mathbb{E}(B)=\mathbb{E}(B|\tau=10, 9W).\mathbb{P}(\tau=10, 9W)+\mathbb{E}(B|\tau=10, 10W).\mathbb{P}(\tau=10, 10W)$$\displaystyle +\sum_{k=1}^8\mathbb{E}(B|\tau=20,k W).\mathbb{P}(\tau=20,k W) \displaystyle =(0.48)^{10}.10+10(0.48)^9(0.52).8+\sum_{k=1}^8\le ft( \begin{array}{c} 10 \\ k \end{array} \right)(0.48)^k(0.52)^{10-k}$$\displaystyle [(k-(10-k))+10\times(-0.04)]\approx -0.782\$