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Can you help solve this problem?

Place contains $N$ coupons $(N>3)$. $3$ tickets have prizes $ \$10000, \$1000, \$500$ respectively. $X$ is the number of coupons drawn before a prize-containing coupon is chosen.

How do I find $E[X]$? If $Y$ is the number of coupons left in the place when all $3$ prize-containing coupons were drawn. What's $E[Y]$? What's $\mathrm{Var}(X-Y)$?


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up vote 3 down vote accepted

Label the non-winning tickets $1$, $2$, and so on up to $N-3$. let $X_i=1$ non-winning ticket $i$ is drawn before any winning ticket is drawn. Let $X_i=0$ otherwise.

Let $$X=X_1+X_2+\cdots+X_{N-3}.$$ The sum of the $X_i$ is the number of tickets drawn before a winner is drawn. We really mean before. So if you want to include the draw in which the first winning ticket is obtained, the random variable you are interested in is $X+1$. But of course if we find $E(X)$, we just add $1$ to find $E(X+1)$.

The $X_i$ are not independent. However, expectation always has the following linearity property: $$E(X)=E(X_1+X_2+\cdots +X_n)=E(X_1)+E(X_2)+\cdots +E(X_n).$$

Now we find $\Pr(X_i=1)=E(X_i)$. This is the probability that non-winning ticket $i$ is drawn before any winner. All orders of the tickets are equally likely. The probability ticket $i$ comes before any winner is therefore $\dfrac{1}{4}$

Now we are finished! We have $$E(X)=\frac{N-3}{4}.$$

If you want to count also the draw on which we got the first winner, add $1$.

A nearly identical argument will take care of $E(Y)$. We can define random variable $Y_i$ to be $1$ if all the winners are before non-winner $i$, and $0$ otherwise. Or more simply, film the drawing and play the movie in reverse!

I have not thought about the variance problem.

Added The variance is not too bad. Note that $E(X-Y)=0$, so we want $E((X-Y)^2)$. Express $X$ and $Y$ in terms of the $X_i$ and $Y_i$, and expand. We have $X_i^2=X_i$ and $Y_i^2=Y_i$. But there is a mess of cross terms to deal with. We will need the probability that $X_i=1$ and $X_j=1$ ($i\ne j$), and also the probability that $X_i=1$ and $Y_j=1$ (all pairs $(i,j)$). These are all accessible.

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Hint: what is the probability that the first coupon is not a winner? what is the probability that the first ten coupons are not winners? Can you use this to find $E[X]$?

Can you find an easy relation between $N, E[X], E[Y]$ that will save the work of calculating $E[Y]$?

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Hum... E[X] = (N-i)-3/N-10? I don't get the easy relation you talked about... – user1172558 Dec 12 '12 at 5:33
@user1172558: $E[X]$ can't depend on $i$, only on $N$ and constants. There is no $i$ in the problem. For the easy relation, think about drawing the tickets in the reverse order. – Ross Millikan Dec 12 '12 at 5:36
Do you recommend any readings to help me? Maybe a book? I need to solve this problem, but I can't find a way to think about it. – user1172558 Dec 12 '12 at 5:55

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