試手氣遊戲期望值

Evaluating expected outcomes

I'm analyzing expected costs based on how soon 3 heads show up in 5 tosses: 240, 320, 400, or 480. Now I need to calculate E[X].

Clarifying event probabilities

I'm piecing together that events A_3, A_4, and A_5 denote getting 3 heads for the first time on the 3rd, 4th, and 5th tosses.

Examining the ambiguity

The phrase "前3次連續擲得3個正面者" could mean "the first 3 tosses are heads in a row." This makes me think the probability for event E1 is 1/8.

Assessing coin toss odds

I'm working through calculating the probability of getting exactly 2 heads in the first 3 tosses, followed by a head on the 4th toss, which equals 3/16.

Calculating probabilities

I'm piecing together the probability of getting exactly 2 heads in the first 4 tosses and then a head on the 5th toss, which is 3/16. Noticing the exclusion of earlier heads.

Ensuring accurate probabilities

I determined that the probability for event E2 is 3/16, confirming 3 heads in the first 3 tosses and a head on the 4th.

Assessing probabilities

I found the probability for event E3 is 3/16: precisely 2 heads in the first 4 tosses, then a head on the 5th, without earlier accumulations.

Breaking down probabilities

I determined that the probability of not achieving 3 heads in 5 tosses is 1/2. This entails considering 0 to 2 heads in 5 tosses.

Summing probabilities

I verified that the probability of achieving 3 heads in 5 tosses is 1/2, combining the outcomes of exactly 3, 4, or 5 heads.

Thought about投擲硬幣遊戲期望值 for 22 seconds

先明確規則與事件的定義：

• 若「前 3 次擲出的硬幣皆為正面」，則只需 240 元。
• 否則繼續擲第 4 次，若「到第 4 次為止正面數累積滿 3 個」，則只需 320 元。
• 否則繼續擲第 5 次，若「到第 5 次為止正面數累積滿 3 個」，則只需 400 元。
• 若 5 次擲完仍無法累積 3 個正面，則需 480 元。

由於硬幣均勻，單次擲出正面或反面的機率均為 $\frac{1}{2}$。定義下列事件：

1. $E_1$：前 3 次的結果即出現 3 個正面
2. $E_2$：前 3 次未出現 3 個正面，但到第 4 次時累積正面總數達 3
3. $E_3$：前 4 次未出現 3 個正面，但到第 5 次時累積正面總數達 3
4. $E_4$：5 次擲完仍無法累積 3 個正面

1. 計算各事件的機率

• $E_1$：第 1、2、3 次皆為正面
  前 3 擲全正面的機率

  $$P(E_1) = \left(\tfrac12\right)^3 = \tfrac{1}{8}.$$

• $E_2$：第 4 次才累積到 3 個正面
  先決條件是「前 3 擲中只有 2 個正面（才能到第 4 擲才滿 3 個）」，再加上「第 4 擲為正面」：

  $$P(E_2) = P(\text{前 3 擲中有 2 個正面}) \times P(\text{第 4 擲為正面}) = \binom{3}{2}\left(\tfrac12\right)^3 \times \tfrac12 = 3 \cdot \tfrac{1}{8} \times \tfrac12 = \tfrac{3}{16}.$$

• $E_3$：第 5 次才累積到 3 個正面
  先決條件是「前 4 擲中只有 2 個正面」，再加上「第 5 擲為正面」：

  $$P(E_3) = P(\text{前 4 擲中有 2 個正面}) \times P(\text{第 5 擲為正面}) = \binom{4}{2}\left(\tfrac12\right)^4 \times \tfrac12 = 6 \cdot \tfrac{1}{16} \times \tfrac12 = \tfrac{3}{16}.$$

• $E_4$：5 擲完仍無法累積 3 個正面
  其機率即是「以上三事件都沒發生」的補事件機率：

  $$P(E_4) = 1 - \left(P(E_1) + P(E_2) + P(E_3)\right) = 1 - \left(\tfrac{1}{8} + \tfrac{3}{16} + \tfrac{3}{16}\right) = 1 - \tfrac{1}{2} = \tfrac{1}{2}.$$

2. 各事件對應的金額

• $E_1$ 對應金額 $240$ 元
• $E_2$ 對應金額 $320$ 元
• $E_3$ 對應金額 $400$ 元
• $E_4$ 對應金額 $480$ 元

3. 計算期望值

令隨機變數 $X$ 表示顧客最終支付的金額，則

逐項計算：

1. $240 \times \tfrac{1}{8} = 30$
2. $320 \times \tfrac{3}{16} = 320 \times \tfrac{3}{16} = 20 \times 3 = 60$
3. $400 \times \tfrac{3}{16} = 25 \times 3 = 75$
4. $480 \times \tfrac{1}{2} = 240$

加總後：

因此，顧客購得一個玩偶所花金額的期望值為 405 元。