TITLE: Bingo Probability Problem
QUESTION [10 upvotes]: A Bingo card has 25 squares with numbers on 24 of them, the center being a free
square. The integers that are placed on the Bingo card are selected randomly and
without replacement from 1 to 75, inclusive. When a game is called “cover-up" is
played, balls numbered 1 to 75, inclusive, are selected randomly and without
replacement until a player covers each of the numbers on a card. Let X equal the
number of balls that must be drawn to cover all the numbers on a single card.
(a) Find p.m.f. of X
I got (24/X)((51 choose (x-24))/(75 choose x)) and I'm pretty certain it's right.
(b) Find the mean and variance of X
mean = 72.96
variance = 5.725 
(c) If there are 183 people playing the game together, assume indepedence, and let Y
be number of balls that must be drawn to cover all the numbers on a single card of
the winner (the first person(s) to have his card coverd), find p.m.f. of Y. 
I have no idea how to find the pmf of Y
(d) Find the Mean and Variance of Y.
I'm pretty sure I can get it after I find the pmf of Y.

REPLY [2 votes]: (a)
If $m\ge 24$ balls have been drawn, the probability that exactly $k$ of your numbers are covered is:
${24 \choose k}\cdot {51 \choose m-k}$ out of a possible ${75 \choose m}$ ways to draw
Thus $P(X\le m)=\frac{ {24 \choose 24} \cdot {51 \choose m-24}}{{75\choose m}}=\frac{{51 \choose m-24}}{{75\choose m}}$
Now:
$P(X=m)=P(X\le m)-P(X\le m-1)=\frac{{51 \choose m-24}}{{75 \choose m}}-\frac{{51 \choose m-25}}{{75 \choose m-1}}=\frac{{m-1 \choose 23}}{{75 \choose 24}}$
(b) $E[X]=\sum_{m=24}^{75} mP(X=m)=\frac{24}{{75\choose 24}}\sum_{m=24}^{75} {m\choose 24}=\frac{24}{{75\choose 24}}\cdot {76 \choose 25}=\frac{24\cdot 76}{25}=72.96$
$E[X^2]=\sum_{m=24}^{75} (m^2+m-m)P(X=m)=\frac{24\cdot 25}{{75 \choose 24}}\sum_{m=24}^{75} {m+1 \choose 25}-E[X]$
$=\frac{24\cdot 25}{{75 \choose 24}}\cdot {77 \choose 26}-E[X]=\frac{24 \cdot 77 \cdot 76}{26}-\frac{24\cdot 76}{25}=\frac{24\cdot 76\cdot (76\cdot 25-1)}{25\cdot 26}$
So $Var[X]=E[X^2]-(E[X])^2=\frac{24\cdot 76 \cdot 51}{25^2\cdot 26}\approx 5.725$
(c) Let $X_1,X_2,\dots, X_{183}$ be the independent copies of $X$ for each individual player
$Y=\min (X_1,\dots , X_{183})$ so $P(Y\ge m) = P(X_1,\dots X_{183}\ge m)=\left[ P(X\ge m)\right]^{183}$
$=\left[1-P(X\le m-1)\right]^{183}=\left[1-\frac{{51\choose m-25}}{{75\choose m-1}}\right]^{183}$
Thus we  can get $P(Y=m)=P(Y\ge m)-P(Y\ge m+1)=\left[1-\frac{{51\choose m-25}}{{75\choose m-1}}\right]^{183}-\left[1-\frac{{51\choose m-24}}{{75\choose m}}\right]^{183}$
This is signficantly less pretty to manipulate, perhaps you could use the approximation $(1-\alpha)^n \approx 1-n\alpha$ to get the sums to go through the same motions?
for (d) $E[Y]=\sum_{m=24}^{75} m\left(\left[1-\frac{{51\choose m-25}}{{75\choose m-1}}\right]^{183}-\left[1-\frac{{51\choose m-24}}{{75\choose m}}\right]^{183}\right)\approx 62.1453$
and $E[Y^2]=\sum_{m=24}^{75} m^2 \left(\left[1-\frac{{51\choose m-25}}{{75\choose m-1}}\right]^{183}-\left[1-\frac{{51\choose m-24}}{{75\choose m}}\right]^{183}\right)\approx 3868.619796$
so $Var[Y]=E[Y^2]-(E[Y])^2\approx 3868.619796 - 62.1453^2=6.5814$