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  "body": "\n## Statement\n\nLet $p \\in (0, 1]$ and let $X$ have the [[geometric distribution|0XCVV5Q500]] with parameter $p$, so $P(X = k) = (1 - p)^k p$ for $k = 0, 1, 2, \\dots$. Then\n$$E[X] = \\frac{1 - p}{p}.$$\nFor the number $T = X + 1$ of trials up to and including the first success, $E[T] = 1/p$.\n\n## Remarks\n\nA proof uses the [[tail sum formula|0Y6QNF1J49]]: $P(X \\ge n) = (1 - p)^n$, so $E[X] = \\sum_{n \\ge 1} (1 - p)^n = (1 - p)/p$.\nThe variance of $X$ and of $T$ is $(1 - p)/p^2$.\n\n## Explanations\n\n### L1 human\n\nIf each trial succeeds with probability $p$, you need on average $1/p$ trials to get the first success.\nFor a fair die, you wait on average $6$ rolls for the first six.\n",
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