Well, the binomial distribution does not really fit here, as it measures the number of successes without placing any weight on losses.
This is wrong. It fits perfectly as there is only two possible outcomes win or loss and when n is large a reasonable approximation to it is given by the normal distribution.
Well, it fits perfectly in measuring the number of wins, but it does not take into account the losses incurred.
If you wanted to calculate the probability this way, you would need to look at the discrepancy between the number of wins, and not the overall profit:
Let's look at it from the perspective of the gambler this time (it doesn't matter)
Consider the binomial distribution of his wins with the same conditions. The expected value of this distribution is 8910 (which would mean house won 9090 times, and therefore house profit is 36000), and the variance around 4500. However, in order to have house profit be lower or equal to -3000, gambler must have won at least 9007.5 times. As you said, with an n this big the distribution is essentially normal (as with every normalized sample average of independent and identically distributed finite-variance random variables - this is what the central limit theorem states). Therefore, plugging it into the normal distribution with mean 8910 and variance 4500, we get that the probability of this occurring is around 7.3%, which is what I got previously
