Say you have a 1 in 6 chance of winning a card game. The more times you play, the higher the odds of you winning.
$$P(\text{win over 1 trial}) = 1/6 \\
P(\text{win over 2 trials}) = 1/6 + 1/6 \\
… \\
P(\text{win over n trials}) = n/6$$
From this, you can deduce you'd have to play six times in order to have won once. However, this isn't the case. Even though these events are in a series, the odd always remain $1/6$; there is indeed a chance you won't win even after 6 trials, 10 trials, 100 trials, $\infty$ trials, etc.
Provided the odds of winning one game is $1/6$, how many games would you have to play to ensure you've won at least once? I think what I'm asking is the trial where the confidence of winning has the widest range.
How does the addition of probabilities take into account true randomness?
Best Answer
I think there is a problem with your calculation of probability. One way of thinking about your question is: "After playing $n$ times what are the odds I have still not won anything?". Chaining events is done by multiplying the probabilities. So, $$ \begin{array}{lll} P(\text{loss over 1 trial}) = 5/6&\implies& P(\text{some wins over 1 trial}) = 1 -5/6 = 1/6\\ P(\text{loss over 2 trial}) = (5/6)^2&\implies& P(\text{some wins over 2 trials}) = 1 -(5/6)^2 = 1/4\\ \end{array} $$ As you can guess, this generalizes for any $n$ so $$ P(\text{some wins over $n$ trials}) = 1 -(5/6)^n $$ So, if you play forever, then the limit of the probability that you win something is $\lim_{n\to\infty} 1 -(5/6)^n = 1$. Unfortunately, there is never any way of guaranteeing a win (ever) for a finite number of trials. If you want 50% chance of winning something, you can solve for $n$: $$ 1 -(5/6)^n = \frac{1}{2} \implies n = \frac{\log 2}{\log 6 - \log 5} \approx 3.802 $$ So if you play 4 times odd are in your favor to win something. :)