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This might be a stupid question.

If I roll a 1d6, it has 1/6 changes of rolling a 1. What are the chances of rolling a 1 if I roll the 1d6 two or three times in a row? On either roll.

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    $\begingroup$ Can you be more precise about what you mean? The probability of rolling 3d6 and showing one 1 is different than the probability of rolling 3d6 and showing two 1s is different than the probability of rolling 3d6 and showing three 1s. Likewise, it's different from rolling a d6 until a certain number of 1s are shown. Are you asking about one of these events or a different event? $\endgroup$ – Sycorax Sep 4 at 13:39
  • $\begingroup$ This is a simplified version of a question with an important history, de Méré's Problem. $\endgroup$ – whuber Sep 4 at 15:01
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The chance of at least one one is one minus the chance of no ones.

If you roll one die, you have a chance of $\frac{5}{6}$ to not roll a one.

Die rolls are independent. Therefore, chances multiply: the chance of rolling no one on two rolls is $\left(\frac{5}{6}\right)^2$, on three rolls $\left(\frac{5}{6}\right)^3$.

Therefore, the chance you are looking for is

$$ 1-\left(\frac{5}{6}\right)^k$$

for $k\in\{2,3\}$, the number of rolls.

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