I would like to find the answer to the following question. A casino offers games with one of the following 21 p’s [0; 0.05; 0.10 .... 0.90; 0.95; 1] and one of the following 21 v’s (values) [0; 5; 10 ... 90, 95, 100]. The person taking part in the game either wins the v or he gets nothing.
In sequence A, 98 games are offered. The total expected value (ev) of the 98 games is: 1785.75 euro. I assumed that the variance (var) could be computed by summing: ev * (v-ev). Thus, a game with a p of 0.5 and a v of 80 has an ev of 40 and a var of 1600. The total var is 31862.94.
In sequence B, 98 other games with different p’s and v’s are offered. The total ev is 1276 euro and the total var is 48585.75.
I assume that both binomial probability distributions can be approximated with a normal distribution with means of 1785.75 and 1276 and sd’s of 178.5019 and 220.4218 (square root of the variance).
I would like to know the probability based on the normal approximations that sequence B leads to more v than sequence A. The difference between the means is 509,75 euro which is between 2 and 3 sd’s. I am not looking for very exact answers. I would just like to know how the probability that V(B) > V (A) can be approximated.