I were playing Hearthstone the other day (a card game).
I got a card that gave me 50% chance to draw a extra card, when it where to be my turn.
Then i got another one of those. It made me wonder what my chances then would be to draw a extra card, if i had two of them on the battleground.
On
If you are curious and want to learn more how this works you can read about binomial distribution, which basically is this sum:
$$(p+q)^n = \sum_{k=0}^n{n \choose k} p^kq^{n-k}$$
for probabilities, which add up to 100%, we have: $p+q=1$ so $q=1-p$ so it becomes
$$(p+q)^n = \sum_{k=0}^n{n \choose k} p^k(1-p)^{n-k}$$
The probability you are looking for is 1 minus one of the end points in this sum: $$1-\left({1 \over 2}\right)^2=1-{1\over 4}= {3\over 4}$$
The probability that both $50-50$- chances fail is $$\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$$ because the events are indepenent.
So, the probability of at least one success is $$\ 1-\frac{1}{4}=\frac{3}{4}$$
Another way is to consider the possible outcomes
$$SS,SF,FS,FF$$
where $S$ is a success and $F$ is a failure. Each of the sequences has the same probability , namely $\frac{1}{4}\ $ , and in $3$ of the $4$ cases, you have at least one success.