Tony estimates his chance of winning a particular hand of blackjack at a casino is 0.45, his probability of losing is 0.5, and his probability of breaking even on a hand is 0.05. He is playing at a \$10 table, i.e. on each play, he either wins \$10, loses \$10, or breaks even, with the stated probabilities. Tony plays 100 times. What is the approximate probability that he has won money on the 100 plays of the game in total?

Let X be his net gain in a particular play of the game
E(X)= -0.5, Var(X)=94.75
Let W be his net gain in 100 games
W=X1+X2+...+X100
So E(W)= -50, Var(W)=9475

I think we're supposed to find P(W>0) by normal approximation? I believe W is a DISCRETE r.v. with possible values -1000,-990,-980,...-10,0,10,...,980,990,1000, so the continuity correction should be applied.

If we don't apply the correction, there is just absolutely no reason why not to say the probability in question is P(W≥10) since it's exactly the same as P(W>0), but if you do normal approximation with the value of 10, the answer would be way off. Without continuity correction, there is just no consistent way to answer.

So now there are at least THREE possible ways to answer this question: two ways without continuity correction P(W>0), P(W≥10), and another way with the correction P(W>5), all of which will give really different answers. Not sure what to do in an exam situtation.

So which way should we answer this problem? Any help is much appreciated!