數學問題(銀幣)(機會率)

Let a = expected number of throws to first head.

We must make 1 throw at least and we have probability 1/2 of a head
and probability 1/2 of returning to a, so

a = (1/2)1 + (1/2)(1 + a)

(1/2)a = 1

a = 2.

Let E = expected number of throws to 2 consecutive heads.

Consider that we have just thrown a head and what happens on the next throw. We are dealing with the (a + 1)th throw, with probability 1/2 this is not a head and we return to E.

So E = (1/2)(a + 1) + (1/2)(a + 1 + E)

(1/2)E = a + 1

E = 2(a + 1)

and now putting in the value a = 2 we get E = 2(3) = 6

Expected throws to 2 consecutive heads is 6.

1/2*1/2=1/4
因為擲一次系公的機會率系1/2,那麼要連續2次都要系公的便是1/2*1/2=1/4的機會率。
如果講到平均要幾次,則是實驗概率的問題,要通過做實驗來確認,所以,一般不會這樣問的。
希望可以答到你。