Proving the Average Winning Amount in a Game with n Moves and Probability p

[trenekas]

Hello. I have a problem with one task. The task is:

Suppose that you playing the game. You have n moves. on each move you win the game with probability p. Your winning amount is equal to move number. For example if you win in first move your winning amount is 1, if you win in n move, your winning amount is equal to n, and if you not win, your winning amount is 0. Need to prove that average winning amount is: (1+n(1-p)^n+1-(n+1)(1-p)^n)/p

My try:

1/n \sum k*p(1-p)^k-1, for k=1 to n.
And tryed to do something but nothing goes on. For example:
1/n(n*p*(1-p)^n-1+(n-1)*p*(1-p)^n-2+...+(n-n+1)*p*(1-p)n-n
but can't get the right answer.
Thanks for helping.

 

[tiny-tim]

hello trenekas!

(try using the X² button just above the Reply box )

what is ∑ ka^k-1 ?

(and why are you dividing by n ?)

 

[trenekas]

ok i don't need divide by n :) :D but later ill try to solve that because now I'm in the university lol. when i'll back home :D

 

[trenekas]

from here the answer is very difficult: http://www.wolframalpha.com/input/?i=sum+k*a^(k-1)+from+k+=+1+to+n
but how got this answer without wolfram?

 

[tiny-tim]

what is ∑ ka^k-1 ?

what does ka^k-1 remind you of?

 

[trenekas]

derivative of a^k:D ok thanks. now ill try myself.

 

[trenekas]

ok. its time to ask the help :D
after few rearrangements:
p * Ʃ k(1-p)^k-1
So k(1-p)^k-1=((1-p)^k)'
First of all i calculated the sum of ((1-p)^k)
Ʃ((1-p)^k)= (1-(1-p)ⁿ⁺¹)/p
When calculated derivative of that:
-(n+1)(1-p)ⁿ*p-1(1-(1-p)ⁿ⁺¹)/p²
And all this multiply by p.
And i got:
-(n+1)(1-p)ⁿ*p-1(1-(1-p)ⁿ⁺¹)/p=-(n+1)(1-p)ⁿ*p-1+(1-p)ⁿ⁺¹)/p
but it is different than i need to get. Where is mistake? Or no mistake? Thanks

 

[tiny-tim]

hi trenekas!

it's a bit difficult to tell, because you're using 1-p instead of a,

but i think if you replace the p in …

-(n+1)(1-p)ⁿ*p

by p-1, and then adjust the rest to make up for it, you'll get the same as wolfram

 

[trenekas]

hi trenekas!

it's a bit difficult to tell, because you're using 1-p instead of a,

but i think if you replace the p in …


by p-1, and then adjust the rest to make up for it, you'll get the same as wolfram

but i can't replace because of rules of composition functions. maybe somwhere else is mistake.

 

[tiny-tim]

yes you can, just subtract (n+1)(1-p)ⁿ from the first part, and add it to the second part (total zero)

 

[trenekas]

oh :D thanks!