The first situation.
It is equal to the lottery question, and the probability is fair every time. The probability of drawing a in the k th time is1/n.
The cost of drawing A for the k time is 2k- 1 (only give him 1 yuan for drawing A, don't give him another dollar to go to the recreation room).
The mathematical expectation of total expenditure is1/n (1+3+5+7+... 2n-1) = n.
The second situation.
The probability of drawing A for the kth time is the probability of not drawing A for the previous k- 1 time: [(n-1)/n] (k-1) * (1/n).
The cost of drawing a for the k th time is K.
The mathematical expectation of the total expenditure is: K * [(n-1)/n] (k-1) * (1/n) (k from1to infinity).
The above formula is the product of arithmetic progression k and geometric progression [(n-1)/n] (k-1) * (1/n). Sum formula a1b1(1-q)-(a1+nd-d) b1qn/(1-q)+db2 [/kloc-q]