y*****d
发帖数: 415 | 1 You already lost 12 times on a slot machine that has a payout rate of 96
cents on the dollar and the probability to win on the first run is 1%. What
is the probability that you will win the 13th time? | y*****d
发帖数: 415 | 2 payout rate means, in a long run, if you pay $100, you can get $96 back. | w********r
发帖数: 290 | 3 My two cents:
Define a variable X_n, at nth time X_n=0 if lose, X_n=1 if win, and X_i is
independent of X_j, for any i≠j.
From the question, we have
P(X_1=1)=0.01, lim_(n->∞)P(X_n=1)=0.96.
So, intuitively, we can set
P(X_n=1)=0.96*(1-1/f(n))
where f(n)>0 is increasing and unbounded in n, furthermore 0.96*(1-1/f(1))=0
.01, i.e. f(1)=96/95.
Then f(n) could be defined naively as
f(n)=96*n/95.
Given above assumption,
P(X_13=1|X_1=0,..X_12=0)=P(X_13)=0.96*(1-95/(96*13))=1153/1300=0.88692 | M******e
发帖数: 8 | 4 那我定义别的fn的话，比如96n^2/95，最后的概率就不一样。
谁有更好的解法呢？
=0
【在 w********r 的大作中提到】
: My two cents:
: Define a variable X_n, at nth time X_n=0 if lose, X_n=1 if win, and X_i is
: independent of X_j, for any i≠j.
: From the question, we have
: P(X_1=1)=0.01, lim_(n->∞)P(X_n=1)=0.96.
: So, intuitively, we can set
: P(X_n=1)=0.96*(1-1/f(n))
: where f(n)>0 is increasing and unbounded in n, furthermore 0.96*(1-1/f(1))=0
: .01, i.e. f(1)=96/95.
: Then f(n) could be defined naively as
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