The problem is like this:↵
You are on the origin of a number axis, each time, you can move forward, backward or stay with probability $P_f, P_b, P_s=1-P_f+P_s$ respectively, i.e., if you are on $x$ now, the next step you can move to $x+1$ with probability of $P_f$ and etc.↵
And the question is: after n steps, what's the mathematical expectation of the maximal number you have reached.↵
↵
Here's an example when $n = 2, P_f = 0.25, P_b = 0.25, P_s = 0.5$:↵
~~↵
~~~↵
maximal number: 1↵
fs: 0.25*0.5 = 0.125↵
sf: 0.5*0.25 = 0.125↵
fb: 0.25*0.25 = 0.0625 (since you have arrived at number 1, even if you go back, you will have maximal number 1)↵
~~~~~↵
~~↵
↵
~~~↵
maximal number: 2↵
ff: 0.25*0.25 = 0.0625↵
~~~~~↵
↵
since you are on the origin, all other path will have maximal number equal to 0↵
↵
Thus the expectation is: 1*(0.125+0.125+0.0625)+2*0.0625=0.4375↵
↵
↵
↵
There is a dp solution on the problem, but I just can't quite figure out why, perhaps you can just think it over first before I give the answer.
You are on the origin of a number axis, each time, you can move forward, backward or stay with probability $P_f, P_b, P_s=1-P_f+P_s$ respectively, i.e., if you are on $x$ now, the next step you can move to $x+1$ with probability of $P_f$ and etc.↵
And the question is: after n steps, what's the mathematical expectation of the maximal number you have reached.↵
↵
Here's an example when $n = 2, P_f = 0.25, P_b = 0.25, P_s = 0.5$:↵
~~~↵
maximal number: 1↵
fs: 0.25*0.5 = 0.125↵
sf: 0.5*0.25 = 0.125↵
fb: 0.25*0.25 = 0.0625 (since you have arrived at number 1, even if you go back, you will have maximal number 1)↵
~~~
~~
↵
~~~↵
maximal number: 2↵
ff: 0.25*0.25 = 0.0625↵
~~~
↵
since you are on the origin, all other path will have maximal number equal to 0↵
↵
Thus the expectation is: 1*(0.125+0.125+0.0625)+2*0.0625=0.4375↵
↵
↵
↵
There is a dp solution on the problem, but I just can't quite figure out why, perhaps you can just think it over first before I give the answer.
