SOLVED: See comment by henry. Thanks!
- 511 Dollar is the budget and the man tries a very known strategy. Betting 1$ on red {with a probability of 18/37} and if he loses he doubles it. And if red comes on, stop playing. So he works against green and black {19/37}.
Say X is the profit at the end of the day, then what is the "entity" and the Probability mass function?
I am not sure what "entity" really means in english but this is my answer, it may become clearer:
E = {-511, -255, -127, -63, -31, -15, -7, -3, -1,1} The reason I have this is because he can never make more profit than 1$.
Because, for example, if he wins the first one, he leaves. If he loses the first one and wins the second, he doubled and won just a 1$ in the end. If he loses, he continues to double, til he reaches 511 Dollar because that is his budget.
And the pmf:
X-> -551 | -277 | -127 etc.
P(X) -> {19/37} ^9 | {19/37}^8 | {19/37}^7 etc.
etc. because you'd need to lose 9 times in a row to lose your whole budget. 8 times in a row to lose -277 etc.
My question here is how do you calculate the probability of winning 1$, because there are multiple ways of doing so: 1. Win the first game 2. Lose the first, win the second. etc. 9. Lose the first 8, win the ninth
I've tried this way: 18/37 + {19/37} * {18/37} + {19/37}² *{18/37} .. but I really don't think this is the right way. Ive tried addition and multiplication but I think there is some wrong thinking here.
For the Expected value I would do it this way: -511*{19\over 37}^9 - 255*{19\over 37} etc. But my problem would be the P(X) of 1$ so to speak.
It would be nice if you can tell me the wrong doing here and maybe give me a tip on how to continue. I know the question is long but I think its not too difficult.