“At least one” is the phrase that separates students who compute from students who think. At least one success in $n$ attempts can happen in a dizzying number of ways — one success, two, three, all of them — but it fails in exactly ONE way: everything misses. So count the cheap side and subtract: $P = 1 - (1-p)^{n}$, read P equals one minus, one minus p, to the n. Here $p$ is the chance one attempt succeeds and $n$ is how many attempts are made; $(1-p)^{n}$ is the chance every single one misses. Never add the chances — $n \times p$ sails past 1 and takes your credibility with it.

A near cousin asks WHEN the first success lands: $P(X = k) = (1-p)^{\,k-1} p$, the geometric distribution, where $k$ is the trial number carrying that first success. Read the exponent carefully — $k-1$ misses come first, and then the hit. Off-by-one there is the whole trap.

Finally, the question probability exists to answer: is it worth it? $E = pW - (1-p)L$, the expected value of a wager, read E equals p W minus, one minus p, L. $W$ is what you GAIN when it wins, $L$ is what you lose when it does not, and $p$ is the chance of the win; $E$ is the average result of one play, in dollars, and it is the only number that settles the argument. A single number on American roulette pays 35 to 1 at a chance of 1 in 38, which values every dollar staked at about −5.3 cents. The wheel is not cheating. It is just charging admission, and the arithmetic prints the price.