# 擲硬幣的期望值問題?

(1)首次連續擲出四次同面

(2)首次連續擲出四次正面

### 1 個解答

• 4 月前
最愛解答

(2) M = 首次連續擲出四次正面所需投擲次數的期望值

M = P{HHHH}×4 + P{HHHT}×(4+M) + P{HHT)×(3+M)

+ P{HT}×(2+M) + P{T}×(1+M)

= (1/16)×4 + (1/16)×(4+M) + (1/8)×(3+M)

+ (1/4)×(2+M) + (1/2)×(1+M)

∴ (1-1/16-1/8-1/4-1/2)M = 4/16+4/16+3/8+2/4+1/2

∴ (1/16)M = 15/8

∴ M = 30

(1) N = 首次連續擲出四次同面所需投擲次數

E[N] = (1/2)E[N|H] + (1/2)E[N|T]

式中 E[N|H] 表開頭擲出 H(正面) 條件下, 擲出連續4個同面

所需總投擲數之期望值.

在 P{H} = P{T} = 1/2 假設下,

E[N|H] = E[N|T]

= (1/2)E[N|HH] + (1/2)E[N|HT]

= (1/2)E[N|HH] + (1/2)(1+E[N|T])

= (1/2)E[N|HH] + (1/2)(1+E[N|H])

E[N|HH] = E[N|TT]

= (1/2)E[N|HHH] + (1/2)E[N|HHT]

= (1/2)E[N|HHH] + (1/2)(2+E[N|T])

= (1/2)E[N|HHH] + (1/2)(2+E[N|H])

E[N|HHH] = E[N|TTT]

= (1/2)E[N|HHHH] + (1/2)E[N|HHHT]

= 4/2 + (1/2)(3+E[N|T])

= 2 + (1/2)(3+E[N|H])

∴ E[N|H] = (1/2)E[N|HH] + (1/2)(1+E[N|H])

= (1/4)E[N|HHH] + (1/4)(2+E[N|H]) + (1/2)(1+E[N|H])

= (1/4)E[N|HHH] + (3/4)E[N|H] + 1

= (1/8)E[N|HHHH] + (1/8)(3+E[N|H]) +(3/4)E[N|H]+1

= 1/2 + (7/8)E[N|H] + 11/8

= 15/8 + (7/8)E[N|H]

∴ E[N|H] = 15 = E[N|T]

∴ E[N] = 15