Suppose I enter a daily lottery where my odds of winning are $p=1/625$. If I enter this lottery daily, my odds of winning on at least one of the days that I played become better than 50% on day 433, since $1 – [(1 – p)^{432}] = 0.4993$, but $1 – [(1 – p)^{433}] = 0.5001$.
Just to make things slightly more complicated, say there are actually six independent lotteries daily, and each has the same p odds of winning. Now if I enter all of them daily, my odds of winning any of the lotteries on any given day rise to $1 – (1 – p)^6$, and I should exceed even odds of winning any of lottery entries on any of the days I played on day 73. Please correct me if I've flubbed at this point.
Now, suppose further that you can get a ticket to each of these six daily lotteries in a bundle, and they are always free to enter, but you can only play each at most once per day (bundling all six tickets doesn't change that). The payout for each of the lotteries is $100.
Best Answer
So the expected value of a month's supply of 6-ticket bundles is the same as the expected value of 180 tickets (because the lotteries and days are independent).
How many times we win/lose the lottery is stochastic, and we notate $X$ for this variable. Since this question is about how many time we win or lose in 180(= n) trials with chance $p$ for each trial, we know that $X \sim$bin$(n,p)$. Now the expectation of the random variable $X$ is known (do you know what the expectation of a binomial distribution is?).
A more intuitive approach is weighing each possible outcome with is probability of occurrence. So if we win three times, the winning would be 300 and the probability of occurrence would be $\binom{180}{3}p^3(1-p)^{177}$, do you understand why?. Thus the weight of winning three times (X=3) is $300 \binom{180}{3}p^3(1-p)^{177}$. Now if you do this for each possible amount of winnings $(\{0,1,2,...,180\})$ and sum them all up, this will be you expectation of $X$.
So to answer your question — If you can determine the distribution and parameters of your random variable, then use this in order to 'know' the corresponding expected value.
An alternative is weighing each possible outcome with is probability of occurrence.
Edit on your question:
So we were talking about calculating the probability of some event, in this case 177 losses and 3 wins. But there are a lot of different combinations that would satisfy our event of 3 wins and 177 losses. For example, we could win the first 3 and lose the last 177. But we could also win, lose, win, lose, win, and lose the last 175 lotteries.
Each of these events can occur with probability $p^3(1-p)^{177}.$ So therefore we multiply the amount of combinations that satisfy our event, with $p^3(1-p)^{177}$.
So how many combinations are there for 3 wins out of 180 lotteries? Well the first win could be on one of the 180 lotteries, the second on one of the 179 (because it can't win on the same day as the previous ticket) and the third on one of the 178 left.
This gives us $180 \times 179 \times 178$ groups of 3 out of 180. But wait, we counted each combination more then once. We have counted each combination $3 \times 2 \times 1$ times, do you understand why?
We divide the $180 \times 179 \times 178$ by $3 \times 2 \times1$ to get the amount of unique combinations. We can write $\dfrac{180 \times 179 \times 178}{3 \times 2 \times 1}$ as $\dfrac{180!}{3!(180-177)!}=\dbinom{180}{3}$. The '!' is called a factorial.