Is there something wrong with this probability formula? In another question someone answered a probability question using a formula that seemed to work for that particular example. I applied that formula to a coin toss, and it worked out. So I know the probability of winning both of two coin tosses is:
$$0.5^2 = 0.25$$
and applying this formula of 1 minus the abovementioned probability gives you the probability of guessing the result of the coin toss at least once:
$$1 - 0.5^2 = 0.75$$
In the particular answer I'm referring to it was in response to winning at least one of 5 games, each of which has a 50% chance of a win, so the formula they gave was:
$$1 - 0.5^5 = 0.96875$$
I tried this with a lottery problem and something went really wrong. So I then tried with a die throw example, and got the following:
$$1 - (1/6)^2 = 0.972$$
And so my result tells me that rolling a 6-sided die twice gives a higher probability of selecting the correct result in at least one throw, compared to the 0.75 probability of predicting the correct result at least once in two coin tosses. And this is obviously wrong.
My guess is that the 1-minus formula only works for when the probability is 50%? Is that the problem?
 A: The probability to roll a number $i$ with a fair six-sided die in one toss is constant  $\frac16 \ \forall \ \ i\in \{1,2,3,4,5,6\}$ Suppose $X$ is the random variable for  rolling   $x$ times number $i=3$.
Now we can calculate  the probability that we roll the number $3$ at least two times in two roll. Here we can apply the converse probability $P(X\geq 1)=1-P(X=0)$
For $P(X=0)$ we roll two times not number three. The probability to roll not number three is $1-\frac16=\frac56$. Thus the probability to roll not number three twice is $\left(\frac56\right)^2$.
Therefore $P(X\geq 1)=1-P(X=0)=1-\left(\frac56\right)^2=\frac{11}{36}\approx 30.56\%$
The reason why your calculation works at coin toss is that the probability to win a toss and the probability not to win a toss are both $0.5$
A: The "1 minus" formula works for any probability, but you have to use the right probability.
For example, if you roll a six-sided die twice, the probability 
of guessing the number correctly at least once is
$$ 1 - (5/6)^2 \approx 0.306.$$
That's because $5/6$ is the probability of guessing wrongly on a given roll,
and we're looking for the probability that you do not guess wrongly twice.
So: guessing wrongly once, $5/6$; guessing wrongly twice, $(5/6)^2$; 
not guessing wrongly twice, $1 - (5/6)^2$.
You'll often see the formula written as
$$ 1 - (1 - p)^n,$$
which is the probability that you have a "success" at least once in a series of $n$ attempts, where $p$ is the probability of "success" on a single attempt.
If $p = 1/2$ then $1 - p = 1/2$ as well, so you won't notice the difference between using $1 - p$ and using $p.$
