Casino turns 50% of your losses into "free play", are odds in your favor? As a limited-time promotion, if you gamble during your first week at this casino, and you suffer a net loss of money, the casino will give you half of your losses (up to a certain amount) as "free play", or cash only usable for more gambling.
It seems to me that for a simple game of slot machines, where the payback is mandated to be $95\%$ (avg) of what you bet, this tips the odds in your favor.
What is an optimal/very good strategy (highest % average return) using slot machines (or anything) for taking advantage of this scenario? Are the odds in your favor with this strategy?
Consider a machine with $95\%$ payout and a $50/50$ binary outcome (lose it all or not):
It seems to me the simplest strategy, finding a machine that takes $1.00$ and has a $50\%$ chance of paying $1.90$ and $50\%$ chance of paying $0.00$, is in one's favor, since the half the time you would lose, you can get half of it back and bet $0.50$ for a $50/50$ payout of $0.95$ or $0$ (on a bet of $0.50$), for a total average outcome of $1.90\times0.5 + 0.95\times0.25 + 0\times0.25 = 1.1875$ on a $1.00$ bet, or a profit of $0.1875$.
I know one can solve the problem as a generalized binary (money or no money) two-action case (just like above) given a mandated payout of $95\%$ and some probability of winning (like $50\%$ above), and maximize the average outcome.
Losses incurred with "free play" are not $50\%$ redeemable like with the original funds.
 A: Take (American) roulette as an example, with outcomes $\color{green}{0, 00}$ along with $1$-$36$, half red, half black.
If you put $\$100$ on black, you have an $\dfrac{18}{38} \approx 47.37\%$ chance of walking away with $\$200$, and a $\dfrac{20}{38} \approx 52.63\%$ chance of walking away having $50$ "casino credits", with which you can bet again.
The same applies when you bet again, only at this point, your $47.36\%$ chance is only to walk away even, in the other case you've lost your money for good.
Now your expected value of such a scenario is (approximately)
$$\frac{18}{38}(\$100) + \frac{18}{38} \cdot \frac{20}{38}(\$0) + \frac{20}{38}\cdot\frac{20}{38}(-\$100) \approx \$19.67.$$
If an extremely crude tree of outcomes will help, then you're in luck!

Note though that your probability of walking away doubling your money is exactly the same - you just now have a non-zero chance of not losing everything that's distinct from doubling your money. This is what's responsible for any sequence of bets having a positive expected value.
A: Your analysis is good.  You can do even better if you find a machine that pays an infrequent large jackpot.  Say you find one that pays 100 for a bet of 1, but only pays it $0.0095$ of the time.  If you play once your expectation is $0.95$.  You will almost surely lose.  Your second play has an expectation of $0.475$ with a chance that you do it of $0.9905$, so your overall expectation is $0.95+0.475\cdot 0.9905$, a little greater than $1.42$.  The limit for infinite odds is $1.425$, so we are almost there.  
For the second play, it doesn't matter what machine you use as long as the expectation is $0.95$
A: The way you've defined the game, it might not be possible to ever get above your initial amount of money. (So you probably need to stipulate more in your problem statement). Suppose with probability $1$ you lose $5\%$ of what you bet whenever you play (so 0.95 ratio of expected value for playing, like you said), but then you get half of your losses back whenever you want due to the rules you stipulated. Then if you play at all, you will always wind up with less money than what you started with, regardless of how much you play or what you wager.
A: Given that a casino covers 50% of losses and you must re-gamble it and
Given that a machine has only two outcomes: 


*

*You lose all your money with probability $p$. 

*You win otherwise

*The machine must pay out on average 95% of what you bet


One can show that if you stop playing when you win, the expected (risky!) payout is
$$
0.95 (1.5 - 0.5p) = E\space [MoneyOut / MoneyIn]
$$
which can be greater than $1$.
For the case where you always win ( p=1 ), the problem is restricted since you cannot lose and thus cannot take advantage of the promotion, meanwhile even "winning" you still lose 5% due to the 95% payout, and you can see that plugging p=1 in yields 0.95. 
For p=0.5, where you lose half of the time, we get 1.1875 like in the question example; the benefit of the promotion begins to show.
If the bet is extremely skewed (p=0.01) where you almost certainly lose, the average payout reaches its maximum: 1.425 or 0.95*1.5, which is significantly more than 1.
This makes sense. The promotion only works if you lose the first bet. To maximize the average payout, counter-intuitively you must maximize your probability of losing, at least during the first bet.
