Question 1200991
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A local club sells 100 tickets every week from which a winning ticket is randomly selected. 
Assume that every week all 100 tickets are sold and we start afresh the next week with a new lottery. 
If each ticket costs €3 and you buy 2 tickets every week, how much would you expect to pay overall 
on tickets by the time you win the lottery for the first time?
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<pre>
From week to week, at each lottery, we have a binomial experiment 
with the probability of the success (winning at each new week/lottery) of 


    p = P(1st ticket wins) + P(1st ticket does not win, but the 2nd ticket wins) = 

      = {{{(1/100)}}} + {{{(99/100)*(1/99)}}} = {{{1/100 + 1/100}}} = {{{2/100}}} = {{{1/50}}} = 0.02.


Repeating this experiment every week (formally, infinitely many times) and paying 2*3 = 6 euros 
every week for 2 tickets, we get the mathematical expectation of the amount of the money 
we pay overall on tickets by the time you win the lottery for the first time


    E = 6p + 12(1-p)*p + 18(1-p)^2*p + 24(1-p)^3*p + 30(1-p)^4)*p + . . . =

      = 6p*(1 + 2(1-p) + 3(1-p)^2 + 4(1-p)^3 + 5(1-p)^4 + . . . ).


The formula of the infinite sum in the brackets is classic

        1 + 2*x + 3x^2 + 4x^3 + 5x^4 + . . . = {{{1/(1-x)^2}}}.   (*)


You may find its proof everywhere (see the relevant info at the end of my post).


In our case, x = 1-p, and we get

    E = {{{(6p)*(1/(1-(1-p))^2)}}} = {{{(6p)*(1/p)^2}}} = {{{6/p}}} = {{{6/((1/50))}}} = 6*50 = 300 euros.


<U>ANSWER</U>.  You expect to pay 300 euros overall on tickets by the time you win the lottery for the first time.
</pre>

Solved.


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Below is the link to the formula (*), first one, which I found in the Internet.

https://www.toppr.com/ask/en-us/question/find-the-sum-of-the-series-1-2x-3x2-4x3-dotsgiven-that/


You may find many sources for it, and, without any doubts, you may find it
in any serious textbook on combinatorics - it is like a PREREQUISITE for solving this problem.