Math Answers
You wrote:
I can set $1000 aside every month as play money for the casino. So, if I go to the casino and flatbet $50 until I either lose it all or win $200, then what is the likelihood that I will win $200? I would switch tables once I got down $500. Further, would I be better off flatbetting a different amount than $50? And, how much am I losing per hour if I flatbet $50 and don�t count? Also, at what point should I switch tables?
Basically, Big player answered your question below, and gave the main points. I will simply expand on this.
If you were playing an break-even coin-tossing game, then the answer would be 1000/(1000+200) = 5/6, or about 83%. 80%. In 100 months, you would expect to win $200 in 83 of the months, for a total gain of 16,600; you would lose 1000 in 16 times for a loss of 16,000. Your overall profit would be zero.
But this is not break-even coin tossing game. First, there is the house advantage working on you. The more you play, the more it costs you. Another way to say this is hat you probability of successes less than 83%, because of the house edge.
A second problem concerns the boundaries. How do you handle splits and double downs. If you play BS, these will cause you to sometimes reach $250 or even more on your winning months. That is, your effective Stop-Win is grater than $200, so your probability of reaching it is less.
There is a similar problem on the other end. What do you do if you have lost $950, and you have a $50 out. Would you double it? Would you have the cash to double it?
There is one other consideration. My earlier calculations were based on always reaching the boundry each month. That is, you always get to either +200 or -1000. In reality, there will be month where will lose money, but not lose the entire $1000. You lose $700, win back $500, lose $300 more, win $100, etc., etc. If you play enough you will eventually hit boundary, you may not have enough time every month to do it. This would also complicate he calculation.