What is the minimum sample size required for this problem? Suppose I have a gambling strategy that gives me the odds of winning the casino (regretfully, this does not exist). I have a 51% chance of winning on each bet. How many bets should I place to give me a 99% chance of winning in the casino?
The bets are independent. Every bet size is the same at USD1. A winning bet wins USD1. A losing bet loses USD1. The win amount is the same as the loss amount.
Some formula would be helpful.
 A: The probability that a single bet doesn't win is $0.49$. Assuming the bets are independent, then the probability that $n$ bets all don't win is $0.49 \times 0.49 \times \ldots \times 0.49 = (0.49)^n$. You want to know what $n$ is when that probability is only $1 - 0.99 = 0.01$, i.e. you want to know when $(0.49)^n \leq 0.01$, which occurs when $n \leq \frac{\log 0.01}{\log 0.49} \approx 6.456$, so you need at least 7 bets.
Of course, that doesn't let you know how much you'll win, or have to pay, but you were asking just about the probabilities.
A: ConMan's answer concerns the probability that you win at least once out of n attempts. If winning as in 'winning more than losing' is what is of interest, you would need the binomial distribution, see https://math.stackexchange.com/questions/439281/probability-of-an-event-happening-n-or-more-times
For P(k>n/2) >= 0.99, n would have to be 1030 or higher. See this little script in R:
winProb     <- 0
n           <- 0
successRate <- 0.51
while(winProb <= 0.99){
  winProb <- 0
  n       <- n+1 
  for(k in c((floor(n/2)+1):n)){
    winProb <- winProb + choose(n, k)*successRate^k*(1-successRate)^(n-k)
  }
}
print(n)

