抽蓝球红球，蓝结束红放回继续，平均结束游戏抽取次

假设蓝x个，红y个，那么$p_1=\frac{x}{x+y},p_2=\frac{y}{x+y}$
次数为$1\ast p_1+2\ast p_2\ast p_1+...+n\ast p_2^{n-1}{p}_1$n趋向于无穷大
$$E=p_1[1+2\ast p2+...n\ast p_2^{n-1}]$$
假设$[1+2\ast p2+...n\ast p_2^{n-1}]=S$
那么$$S-p_{2}S=1+p_2+p_2^2+...+p_2^{n-1}-n\ast p_2^n=\frac{1-p_2^n}{1-p_2}-n\ast p_2^n$$
$$E=(1-p_2)S=\frac{1-p_2^n}{1-p_2}-n\ast p_2^n$$
当n趋向于无穷大时次数等于$$\frac{1-0}{1-p_2}=\frac{1}{p_1}$$
下面用python代码进行验证

p1=0.4
p2=1-p1
E=0
cnt=10000
for i in range(1,cnt+1):
    E+=i*p1*p2**(i-1)
print(E)

E=2.5000000000000004 结果相符