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This question is (I think) simpler than most posed here, but it's beyond my ability to solve.

I'm trying to calculate the probability of various outcomes for a charter school lottery. There are two classes of entrants: students otherwise zoned for under-performing schools, and all other students.

Let x=the number of students zoned for under-performing schools

Let y=the number of students not zoned for under-performing schools

Let z=the number of spaces available in the school

It's easy enough to determine the expected number of students admitted from each class, but how can I determine the probability of various other outcomes. For instance, how likely is it that the number of students admitted from a class is half the expected number?

For what it's worth, I'm a math teacher who has never had a stat, probablity, or discrete math course, although I've taught myself the basics. If you have any suggestions for further self-guided study in probability I'd appreciate them, but right now I just need to solve this practical problem.

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  • $\begingroup$ This question needs to provide the specifics of the lottery in order to be answerable. $\endgroup$
    – whuber
    Sep 21, 2012 at 18:17
  • $\begingroup$ What specifics? Is it now possible to answer in terms of x, y, and z? $\endgroup$ Sep 21, 2012 at 19:00
  • $\begingroup$ Which specifics? Suppose that the lottery has x+y entrants, and z identical prizes, and that each entrant gets exactly one entry/chance/bean in the hat. Is it not possible to answer in terms of x, y, and z? One case I'm interested in is x=80, y=120, and z=200, but I'd like a general method or rule so that I can play around with other ways to run the lottery (giving greater weight to students zoned for underperforming schools, for instance, would effectively increase the value of x, though it wouldn't be quite that simple, since you can't admit the same student twice). $\endgroup$ Sep 21, 2012 at 19:09
  • $\begingroup$ When $x=80$, $y=120$, and $z=200$, then all entrants will get a prize! To describe a lottery you need, at a minimum, to specify how many tickets are entered, the chances with which each is drawn, whether tickets are replaced after drawing, what kind of prize each ticket can win, and rules for terminating the drawing (e.g., whether a fixed number of draws will be made or if drawing continues until some event occurs). To compute chances in the lottery you need to know how many tickets of each type are entered. This requires more information than just $x$, $y$, and $z$. $\endgroup$
    – whuber
    Sep 21, 2012 at 19:21
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    $\begingroup$ I thought that the other things you called out as lacking were implicit in the description I gave. each ticket has the same chance of being drawn. Tickets would not be replaced after drawing, because no one is admitted twice. Every prize is the same. Tickets are drawn until every prize is awarded. $\endgroup$ Sep 21, 2012 at 19:29

2 Answers 2

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In a simple lottery, all $z$ element subsets of the box of tickets have equal chances of being drawn. When there are $x$ tickets of type "X" and $y$ tickets of type "Y", the number of these subsets is written $\binom{x+y}{z}$ (read "$x$ plus $y$ choose $z$"). This binomial coefficient has a simple, well-known formula,

$$\binom{x+y}{z} = \frac{(x+y)(x+y-1)\cdots(x+y-z+1)}{z(z-1)\cdots(2)(1)}.$$

This is the $k^\text{th}$ entry (counting from $0$) in the $x+y^\text{th}$ row of Pascal's Triangle.

Each $k$ from $0$ through the smaller of $x$ and $z$ is a possible number of the "X" tickets drawn. The number of ways to draw $k$ tickets from category "X" and the remaining $z-k$ tickets from category "Y" is the product $\binom{x}{k}\binom{y}{z-k}$. Whence the chances of drawing exactly $k$ tickets from category "X" are the ratio of this number and the total number of $z$ element subsets:

$$\frac{\binom{x}{k}\binom{y}{z-k}}{\binom{x+y}{z}}.$$

This is known as "the" hypergeometric distribution. (For every combination of $x$, $y$, and $z$ describing such a lottery, there is a hypergeometric distribution giving the chances for all possible numbers of tickets $k$ that can be drawn from category "X".)

For example, with $x=80$, $y=120$, and $z=20$, the chances for $k=0, 1, \ldots, 20$ are

0.000018, 0.000289, 0.0021, 0.0097, 0.0304, 0.0705, 0.1247, 0.1724, 0.1894, 0.1668, 0.1184, 0.0679, 0.0314, 0.0116, 0.0034, 0.00078, 0.000137, 0.000018, 1.57e-6, 8.62e-8, 2.2e-9

Here is a plot of these chances:

Plots

The exact chances are shown by the heights of the black dots. For reference, the red curve plots the density function for the normal distribution with the same expectation ($8$) and same variance ($864/199$). It can be used (with a so-called "continuity correction") to approximate the chances. Those approximate values are indicated by the heights of the thin red lines. A close look shows that the red lines are a tiny bit short to the left of $k=8$ and a tiny bit tall to the right--but the approximation looks quite good. Using this normal approximation sometimes is easier than doing the calculations, but its chief virtue is in allowing us to use our experience with and intuition about normal distributions to understand this one. (The normal approximation will be reasonable provided both $x$ and $y$ are $5$ or greater and neither is large compared to the other.)

With these chances in hand, the probability of any stipulated event can be computed by adding the chances of its components. For instance, the chance that five or fewer of the tickets are drawn from the "X" pool is the chance that $k$ is one of $0, 1, 2, 3, 4,$ or $5$. The sum of those component chances is

$$0.000018+ 0.000289+ 0.0021+ 0.0097+ 0.0304+ 0.0705 = 0.1130,$$

or about $11$%.

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Are you selecting separately from the zoned and non-zoned pools? It seems you are randomly selecting from the total pool of zoned and non-zoned students. If so, you can calculate the precise probability for any outcome using the binomial theorem. If the numbers get unwieldy, you can use the normal approximation to the binomial to calculate the probability of an outcome as extreme or more extreme than a given value. Both these approaches are in most introductory statistics textbooks. See, for example, http://jwilson.coe.uga.edu/EMAT6680Fa06/Crumley/Normal/Normal1.html

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  • $\begingroup$ The binomial serves as a pretty good approximation. The absolute error of any particular outcome is always less than 0.1%; the relative error is as high as 40% in the right tail. $\endgroup$
    – cardinal
    Sep 21, 2012 at 20:13
  • $\begingroup$ Did you mean the normal is a good approximation? The binomial yields exact probabilities, yes? $\endgroup$
    – Joel W.
    Sep 21, 2012 at 21:45
  • $\begingroup$ Hi, Joel. To answer your questions in the comment: No I didn't, and, no, they don't. :-) (But, in this case the normal is also a reasonably good approximation.) $\endgroup$
    – cardinal
    Sep 22, 2012 at 14:43

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