Lets suppose we are in a casino, having two slot machines. Our strategy is to flip a coin to decide which machine to start on. (Without loss of generality, we say that we play the slot one first with probability $c_1$ and play the slot two first with probability $1-c_1$)
Then, we play on that machine until losing one game, then we switch to another slot machine to continue playing, we play a total n games (n is finite here).
Suppose the probability to win in slot one and slot two is i.i.d $Bernoulli(p_1)$ and $Bernoulli(p_2)$, respectively, and each win just gain 1 dollar, and each loss just get nothing.
Then, what should be the expected earning using this strategy???
I approached it by iteration (drawing a tree), hoping that there will be some pattern and if we find the expected winning in each turn, then we can sum the n rounds up because the trials are independent.
The expected winning in round 1 is: $$c_1\cdot p_1 + (1-c_1) \cdot p_2$$
The expected winning in round 2 is: $$c_1 \cdot[p_1^2+(1-p_1)\cdot p_2]+ (1-c_1)\cdot[p_2^2+(1-p_2)\cdot p_1]$$
The expected winning in round 3 is: $$c_1 \cdot \Big[p_1^3 + p_1(1-p_1) p_2 +(1-p_1)p_2^2 + (1-p_1)(1-p_2)p_1\Big] + (1-c_1) \Big[p_2^3 + p_2(1-p_2) \cdot p_1 + (1-p_2)p_1^2+(1-p_2)(1-p_1)p_2\Big]$$
I counted the expected winning up till round 4, which is: $$c_1 \cdot \Big[p_1^4 + p_1^2(1-p_1)p_2 + p_1(1-p_1) p_2^2 + p_1(1-p_1)(1-p_2)p_1 +(1-p_1)p_2^3 + (1-p_1)p_2(1-p_2)p_1 + (1-p_1)(1-p_2)p_1^2 + (1-p_1)^2(1-p_2)p_2\Big]$$ $$+ (1-c_1) \cdot \Big[p_2^4 + p_2^2(1-p_2)p_1 + p_2(1-p_2) p_1^2 + p_2(1-p_2)(1-p_1)p_2 +(1-p_2)p_1^3 + (1-p_2)p_1(1-p_1)p_2 + (1-p_2)(1-p_1)p_2^2 + (1-p_2)^2(1-p_1)p_1\Big]$$
This calculation seems endless so it does not work well. How could I find the expected winning in this case?