# Overall Probability for Multiple Independent Events on Single Trial

Am working on a probability problem where a game is described using the hypergeometric distribution. The sample successes are $3...7$, the number of samples is $7$, population successes are $10$, and the number in population is $55$. Summing the probabilities gives $\sim0.1040683327$ or "overall odds" of about $1$ in $9.61$.

Here's the tricky part. For each play, there are $3$ sets of $7$ numbers, and only one set may be chosen by the player (the other $2$ are chosen randomly). Basically, $3$ independent trials on a single slip. So how do you calculate the overall odds of winning per play for each level ($3...7$)? According to the operator it is $1$ in $3.6$. Using the Binomial distribution, taking the probability of at least one success and averaging it with the probability of exactly one success gives overall odds of exactly $1$ in $3.6$. However, this is not the closest approximation compared to using the Rule of Complements on the sum of probability from the hypergeometric distribution mentioned earlier. That result comes out to be $\sim 0.2808414272$ or about $1$ in $3.56$. The closest approximation seems to come from using a formula derived from the Poisson distribution...this gives overall odds of $1$ in $3.58$. Keep in mind I am trying to calculate the closest probability of each success level so that the overall odds come out to be nearest $1$ in $3.6$. Now, rounded to a couple decimal places both methods I tried come to the same results, but I just want to make sure I am doing this correctly.

If this isn't clear, basically I want to calculate the actual probability of winning for each tier based on the fact that 3 trials occur in the same play. Obviously it isn't as simple as dividing each odds by 3 since there is the chance of winning more than once.

Thanks