Can you keep this raffle fair? So there is a car draw in my area. There are 3221 participants in the draw. The winner is decided by a trustee drawing each digit from a separate drum. So from the first drum there is 0-3, the second 0-9 (this is the same for the third and the fourth drum). So if I draw 0-4-0-7 the winner is the ticket holder of 407.
My first suspicion was those that have a ticket in the 3 thousands have a better chance of winning, correct me if I'm wrong. If it is the case that the probability is askew then how can one keep this draw fair? While maintaining the concept of drawing digits from a drum.
If this is not possible could anyone give suggestions on other draw methods that could work? Just putting all tickets in one bucket isn't an option though.
Thank you!
 A: Whether or not the raffle is fair, depends on how you draw the ticket. Let's say I draw, from the four different bins, the numbers $3, 5, 1, 0$. This is an invalid number, so the price cannot be assigned. If I simply restart the whole procedure, each time drawing four numbers until a valid number shows up, then each participant has the same probability of winning.
However, we can also start drawing from left to right, and only redraw the number which resulted in an invalid sequence. Note that the first number is always valid, but the validity of the second number depends on the first number: if I first drew a $3$, I cannot draw the number $3$ to $9$. Indeed, in this case, the probability that a random person whose ticket number starts with a $3$ has a probability of: 
$$\frac{1}{4} \cdot \frac{1}{222} = \frac{1}{888} > \frac{1}{3221}$$
of winning the raffle. Note that, using this procedure, the two people holding the tickets $3220$ and $3221$ have the highest probability of winning the raffle:
$$\frac{1}{4} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{72}$$
A: Unfortunately, 3221 is prime, so there's no way to get a perfectly fair draw by drawing a fixed finite number of times successively from drums with any fewer than that many items in each.
If you're willing to keep drawing randomly from a drum containing the digits 0-9 until one of three stopping rules is satisfied, however, you can get each of the numbers 0-3220 with equal probability (of $\ \frac{1}{3221}\ $).  Typically you will only need to draw 4 or 5 times, but with exponentially decreasing probability you may need to keep drawing more and more digits.
After drawing $\ r\ $ digits $\ d_1, d_2, \dots, d_r $, calculate the decimal fraction $\ D_r = \sum_{i=1}^r \frac{d_i}{10^i}\ $.  The stopping rules are:


*

*If $\ D_r + \frac{9}{10^{r+1}} < \frac{1}{3221}\ $, call the number drawn to be $\ 0\ $.

*If $\ D_r > \frac{3220}{3221}\ $, call the number drawn to be $\ 3220\ $.

*If $\ D_r > \frac{i}{3221}\ $ and $\ D_r + \frac{9}{10^{r+1}} < \frac{i+1}{3221}\ $, for some $\ i\in \left\{1,2,\dots,3219\right\}\ $, call the number drawn to be $\ i\ $.


You could, of course, carry out a similar procedure using a different base for the the fraction, with the larger the base chosen the smaller the expected number of draws needed.
The reason why this works is that if you keep drawing forever, the number $\ D = \sum_{i=1}^\infty \frac{d_i}{10^i}\ $ will be uniformly distributed over the unit interval.  Therefore, for any $\ i\in\left\{0,2,\dots,3220\right\}\ $, $\ \mathrm{Prob}\left(\frac{i}{3221} < D < \frac{i+1}{3221}\right) = \frac{1}{3221}\ $.  But once one of the stopping rule criteria has been satisfied, you will know with certainty which of these 3221 events has occurred.
The probabilities that $4, 5, 6, \dots$ draws will suffice are $\ 0.678, 0.9678, 0.99678, \dots$, and so on, and the expected number of draws needed is $4+\frac{3220}{9000} \approx4.36\ $.
