Probability of $5\%$ Okay, I am playing a game. I have $5 \%$ chance of upgrading a weapon, so $95\%$ chance of it failing when I try to upgrade. How many times do I need to try to upgrade in order to guarantee a $100\%$ success rate? 
My idea was that if I do $1-0.05$ and get $0.95$ and multiple it over and over, I would get the answer.
My friends disagree. Please also show me how you got the answer as well
 A: Probability of failure: .95
Probability of n- failures in a row: $.95^n$.
Probability of success at least once in n tries: $1 - .95^n$.
Number of times to guarentee 100% chance of success: Solution to $1-.95^n = 1$.
Solve $1-.95^n = 1$
$.95^n = 0$
$n = \log_{.95} 0 = \frac {\ln 0}{\ln .95} = -19.495725746223689348556622494416*\ln 0$.
Enjoy....
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In seriousness (the above was meant as a tongue in cheek joke), the Have a P percent chance of success is.
$1 - .95^n = P/100$
$.95^n = (1 - P/100)$
$n = \ln (1 - P/100)/\ln .95 = -19.49\ln(1 - P/100)$.
So if you want a 95 percent chance of success, you must try
$n = -19.49\ln(.05) = -19.49*-2.99 \approx 59$ attempts$
For $P = 100$, however, you need to find $\ln 0$ which is infinitely undefined.
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100% is impossible.  But you know how people say "I'm 99.999 percent certain"?  Of course, they are just making shite up, but it does lead to the question: What degree of certainty do you need to say, "I am certain I'll have success".
Is less than 1 in a billion chance of failure good enough for you?
Then you need to attempt $n = \ln(10^{-9})/\ln{.95} = -9*\ln(10)/\ln .95 = 405$ tries.  Your probability of failure will be less than 1 in a billion.
If 1% failure rate is acceptable you need $n = -2*\ln(10)/\ln .95 = 90$ attempts to have a 99% success rate.
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Just for giggles.  To have a 50-50 percent success rate you need $n = \ln(.5)/\ln(.95) = 13.51$ tries.
A: You cannot guarantee success, no matter how many times you try.  You are correct that the chance of failure in $n$ tries is $0.95^n$.  If we put in $n=1000$, we get about $5.29 \cdot 10^{-23}$, which is a very small number but still greater than $0$.  You can still fail after $1000$ tries.
A: It is not possible to achieve $100\%$ success rate. However, the chance of failing $n$ times are $0.95^n$, which tends to $0$ as $n$ goes to infinity. So the chance of succeding after at most $n$ tries is the complementary of that, which is $1-0.95^n$, which of course tends to $1$ (or $100\%$) as $n$ goes to infinity.
A: This probability at the $n^{th}$ trial is $1-0.95^n$. You never reach exactly 100 %
