| English | 意味 |
|---|---|
| geometric/ˌdʒiːəʊˈmetrɪk/ | 幾何分布 |
| geometric random variable/ˌdʒiːəʊˈmetrɪk ˈrændəm ˈveərɪəbl/ | 幾何確率変数 |
Waiting for the first success
- The geometric 几何 distribution counts trials until the first success.
- A geometric random variable 几何随机变量 $X$ = the number of trials to get the first success.
- Example: how many rolls until you get your first six?
- Unlike binomial, $n$ isn't fixed — you stop the moment you succeed.
The geometric setting
- Same trial conditions as binomial, but a different stopping rule:
- Repeated independent trials, each with the same probability $p$ of success.
- You keep going until the first success, then stop.
- "Until first success" is the signal word for geometric.
The geometric formula
-
$$P(X = k) = (1-p)^{k-1}\,p$$
- To succeed first on trial $k$: fail the first $k-1$ times, then succeed.
- $(1-p)^{k-1}$ = those failures; $p$ = the success on trial $k$.
- The probabilities shrink as $k$ grows — early success is most likely.
The geometric mean
- The mean (expected number of trials) is:
-
$$\mu_X = \frac{1}{p}$$
- If $p = 0.2$, expect $1/0.2 = 5$ trials on average until the first success.
- Rarer successes ($small$p$) mean you wait longer — makes intuitive sense.
Geometric vs. binomial comes down to the stopping rule. Binomial: fixed $n$, count successes. Geometric: run until the first success, count trials. If a problem fixes the number of tries, it's binomial; if it says "until the first…", it's geometric. And the geometric mean is $1/p$ — not $np$.
Roll a die until the first six; $p = 1/6$.
- $P(X = 3) = (5/6)^2 (1/6) = \tfrac{25}{36}\cdot\tfrac{1}{6} \approx 0.116$ — first six on the $3$rd roll.
- Mean: $\mu_X = 1/(1/6) = 6$ rolls expected until the first six.
- Makes sense — a six comes up on average once every six rolls.
A geometric random variable counts trials until the first success, with independent trials at constant $p$. Then $P(X=k)=(1-p)^{k-1}p$ and the mean is $\mu_X = \frac{1}{p}$. The key contrast with binomial: geometric runs until the first success (variable trials), binomial has a fixed $n$.
最初の成功が出るまでサイコロを振る
ターゲットが出るまで繰り返し振る — これが幾何的なカウントである。
幾何変数の p = 0.2 のとき、試行の平均回数 1/p を求めよ。
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
μ = 1/p = 1/0.2 = 5.
どの状況が幾何分布（二項分布ではない）か？
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
'最初の成功まで' → 幾何分布；固定されたn → 二項分布。
サイコロを振り、最初の6が出るまで続ける（p = 1/6）。投擲の期待値を求めよ。
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
μ = 1/p = 1/(1/6) = 6.
幾何分布の式 P(X=k) = (1−p)^(k−1)·p において、(1−p)^(k−1) の部分は最初の k−1 回の失敗を表す。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
k−1 回失敗して、k 回目の試行で成功する。
幾何確率変数の平均は 1 ÷ ___ （成功率を表す文字）である。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
μ = 1/p.