I just take a probability course this month and I find an interesting question that I do not know to solve, as following:
Suppose a casino machine costs \$10 to play game. for the game, you will get nothing with probability 80%, get \$10 with chance 15%, and get \$20 with chance 5%.
| X | $-10 | $10 | $20 |
|------------|-------------|--------------|-------------|
| P. | 0.8 | 0.15 | 0.05 |
Question, you have $40 and keep using them until you have no money, what is the chance you will play at least 7 games?
P(play at least 7) = 1 - P(Lose in less than 7)
You can't lose in less than 4 plays
The only way to lose in 4 plays is to have four \$10 losses in a row $$P(4) =(0.8)^4$$
You can't lose in 5 plays
The only way to lose in 6 plays is to have one \$10 win and five \$10 losses. your last play must be a loss but the single win could occur in any of the first 5 plays
$$ P(6) = 5(0.8)^5(0.15)$$
Can you take it from there ?