統計擲銅板的問題

設p=0.4, q=1-p=0.6
若連擲n次均無連續H出現,共有二種case
(1)H(頭)結尾: 設機率為a(n), 本題答案= p*a(n-1)
(2)T(尾)結尾: 設機率為b(n), 則
a(1)=p, b(1)=q
a(n+1)= p b(n)
b(n+1)=q a(n)+ q b(n),
a(n+2)= pq a(n)+ q a(n+1), a(1)=p, a(2)=pq
so, a(n+2)= 0.6 a(n+1)+ 0.24 a(n), a(1)=0.4, a(2)=0.24 ---(A)
solve the difference equation, then
a(n)=Aα^n+Bβ^n, A=p(q-β)/[α(α-β)], B=p(α-q)/[β(α-β)]
where α,β= 0.3+/- √0.33 are roots of x^2=0.6x+0.24
so, A=0.2/√0.33= -B
Ans: pa(n-1)=0.08/√0.33 *[α^(n-1)- β^(n-1)], n=2,3,4,...