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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.

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2 Answers 2

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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)
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  • $\begingroup$ Thanks and upvoted. I think your answer is more logical. ConMan's answer of only 7 bets seems too small. $\endgroup$
    – curious
    Nov 23, 2016 at 15:47
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    $\begingroup$ May I ask why is binomial distribution used? Why not some other distribution like the more popular normal distribution? $\endgroup$
    – curious
    Nov 23, 2016 at 15:51
  • $\begingroup$ Just want to point out that the previous answer is correct and answers a perfectly valid way of reading your question, just a different one than you are interested in, apparently. :-) $\endgroup$
    – k-di-b
    Nov 23, 2016 at 16:08
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    $\begingroup$ Ah, that was meant to be a line break - sorry, I am new here. As for why the binomial distribution, well, that is what it does. It models the number of occurrences in binary situations given the number of trials and the probability of the occurrence. A normal distribution can be used in many situations, but in this case is a rather poor approximation, so it will give you a rather poorly approximated answer. $\endgroup$
    – k-di-b
    Nov 23, 2016 at 16:15
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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.

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  • $\begingroup$ Thanks and upvoted. I am surprised only 7 bets is required. This is good news. $\endgroup$
    – curious
    Nov 23, 2016 at 5:51

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