Recently I was called in for jury duty, my first time in a courtroom as a potential juror, and the first time in a court in many years. It was an enlightening if somewhat disheartening experience.

Jury duty in Los Angeles starts with a mailing demanding your attendance at a courthouse for some particular week. Each day of that week you have to call or check a website to see if your presence is required (otherwise you are to go to work as usual like a good worker-bee). If summoned you end up in a large room with about 100 other people also not wanting to be there. Jury service pays $12/day (but nothing on the first day!), soon to be less than the minimum wage in Los Angeles for one hour of work, and employers are not required by law to offer any compensation, so it is of little wonder why most people want to avoid jury duty.


I had been assigned jury duty previously and was not actually called in but this time was different. I was called in on Monday.

Like most of the other jurors, I hoped to be dismissed after the first day. If you are not called in by the end of the day, you are dismissed for the rest of the week (and not called upon again for at least the next year). Not that I am opposed to jury duty, rather the timing was simply poor (isn't it always?), and from what I had read mathematicians and scientists are typically dismissed because our standard of beyond a reasonable doubt is much more stringent than other people's. Apparently prosecutors don't like that, or maybe we're just a curmudgeonly lot. If I was destined to be dismissed, sooner was preferred to later. Nevertheless I made an honest attempt at service and was not actively trying to get dismissed.

Along with the other potential jurors, I sat in the jury room waiting for 5pm to come around, reading what turned out to be a interesting book. At 2pm all the jurors were called up (one by one) to sit on the tile floor of a long open hall outside a court room (seriously there were maybe 20 seats on benches for 50+ people) for about two hours until we could enter a court room to begin jury selection. The courtroom itself also did not have enough seats for all the jurors.

While waiting outside the courtroom, the judicial assistant (JA) began explaining the process of jury selection. Juror names being secret inside the courthouse, each juror was assigned a nine-digit number, presumably at random, and we would all be referred to by the last 4 digits. Once inside the court room, the judge would talk to us for a while, and was rather personable. He told us that when people find out he is a judge, invariably they ask "How do I get out of jury duty?"

While we were still seated on the floor outside the courtroom, the JA explained that if our four digit number was called, we were to step up into the juror box. She said if your number is called and someone else gets up, speak up! Previously more than one juror has had the same last four digits.

This is not prevented by the jury duty selection system for a simple reason -- it's possible to move your jury duty date multiple times, and the nine-digit number is assigned when you are first informed of jury duty by mail, so I assume that the system just assigns the numbers randomly and hopes for the best. This particular courthouse goes through several hundred jurors per week, so it only takes a few months to go through the 10000 possible last four digits.

Anyway, the JA goes on to tell us that in the previous incident in which two jurors had the same last four digits, one of the other jurors was "really OCD" and had said that the chance of two jurors with the same last first digits was "less than winning the lottery". (All people with some number sense have a mental issue, amirite?) I presume we're talking about winning a serious jackpot, the chance of which is typically 1 in a few hundred million. Presumably Mr. Allegedly OCD thought that there was a 1 in \(10000^2\) (1 in 100 million) chance of any two randomly chosen four digit numbers being equal.

Immediately this struck me as not quite right. I quickly realized that this scenario was a variant of the infamous birthday problem, and I knew the probability was much higher. But what was it exactly?

The Birthday Problem

What is the birthday problem? The idea is simple -- in a room of say \(N=30\) people, what is the probability that any two have the same birthday? Most people would say "very small" -- after all there are 365 days of the year and just 30 people, so a rough estimate is 1 in 32 (assuming each day has the same number of births). The true probability that at least one pair of 30 people share a birthday is about 73%. For just \(N=23\) people the probability exceeds 50%.

Come again?

While it is unlikely that any of the other people in the room has the same birthday as you, this game is played out for all 30 people in the room, and the probability that any two people have the same birthday is much greater than \(N \,/\, 365\). Here's what the probability looks like as a function of the number of people present:

The python code to produce this plot:
import matplotlib
from matplotlib import pyplot

font = {'size': 20}
matplotlib.rc('font', **font)

def birthday_probabilities(n, d):
    Computes the probability of a collision for k people with births drawn
    from d days, for 0 < k < n.

    results = []
    s = 1.
    for i in range(0, n):
        s *= (1. - float(i) / d)
        results.append(1. - s)
    return results

def birthday_plot(n=30, d=365):
    Make a plot of the birthday problem probabilities.

    # Compute the values
    domain = list(range(0, n))
    probs = birthday_probabilities(n, d)
    values = [round(100 * p, 3) for p in probs]

    # Make the plot
    figure, ax = pyplot.subplots()
    ax.plot(domain, values, linewidth=2)
    ax.set_title("Probability of Two People with the Same Birthday")
    ax.set_xlabel("Number of People")
    ax.set_ylabel("Probability (%)")

if __name__ == "__main__":
    birthday_plot(50, 365)

Anyway back to the tile floor. My phone had a very poor data connection inside the courthouse, so I was unable to look up the formula for the birthday problem probabilities and generalize. I had time on my hands, and recalled the derivation from several years ago when I taught finite math at the University of Illinois, so I spent a few minutes working out the formula. The basic idea is simple -- rather than calculate the probability that any two people have the same birthday it's easier to calculate the probability that this doesn't happen, and substract from one (since \(Pr(not \, A) = 1 - Pr(A)).\) This reasoning goes like this:

\[\begin{align} 1 &- \left(1 - \frac{1}{365}\right)\left(1 - \frac{2}{365}\right) \cdots \left(1 - \frac{N-1}{365}\right) \\ & = \frac{365!}{365^N (365-N)!} \end{align}\]

Let's call \(Pr(N, D)\) the chance that at least one pair of the \(N\) individuals collide out of \(D\) days; the formula above is \(Pr(N, 365)\). To calculate the analogous probability for the juror duty selection, just note that there are \(D=10000\) possible last-four-digit juror birthdays and approximately \(N=50\) to \(N=100\) jurors.

This is what the plot looks like:

As you can see for 100 jurors, the probability of two jurors out of 100 having the same last four digits is almost 40%! That's a lot more likely than winning the lottery. Even for 50 jurors the probability is not negligable, about 12%. Here's a widget you can play with varying \(N\) and \(D\).

Collision Probabilities \(Pr(N, D)\) (%) vs. \(N\)

\(N\) (People): \(D\) (Pigeonholes):

Made with ChartJS and this script.

FYI, there is a nice approximation that makes it a bit easier to see how the parameters \(N\) and \(D\) affect the probability: \[ Pr(N, D) \approx 1 - e^{\frac{-N^2}{2D}} \]

I made some plots for the JA after the first day, and also included one for the case of using the last five digits, which drops the probability to about 1% for 50 jurors. Note that this isn't the probability that any particular number that the judge calls will have two people associated to it, that's smaller of course. But jury selection tends to go through a lot of jurors, so the chance that it happens over the course of jury selection is fairly close to \(Pr(N, D).\)

On the second day I gave the plots to the JA, and she shared them with some of the judges over lunch. I was dismissed in the afternoon by the prosecutor for (I presume) unrelated reasons (read on if you are interested).

When we first sat on the floor, each of the jurors was given a questionnaire. Answering yes to any required giving an explanation in open court (or sidebar if sensitive) as it potentially disqualifies a juror from serving. Many if not most of the questions (IMO) could be reasonably be answered yes by most people, and everyone spoke up about at least one question, so the process was very time consuming. The questions included (going from memory):

There were also a number of questions about philosophical and religious beliefs, including whether you can follow the law as instructed by the judge even if you disagree with it, and if you can separate the question of guilt from the potential punishment that is applied. Any answer that suggested that one knew about, or would consider acting in accordance with jury nullification led to dismissal. Finally there was a meta-question: Do you disagree with any of the statements of law on this questionnaire? I don't even know how to begin answering this question, and it was number 2 out of about 20 (why not the last?).

I was also struck by various aspects of the jury process. Upon entering the court, everyone swears to uphold the law and such without being told beforehand what that entails. When answering questions, the judge would tell potential jurors that e.g. the law says that jurors cannot be biased in various ways. When pressed to say they would be unbiased, jurors would say "I'll try my best", and the judge eventually explained that that was not good enough. Given that the vast majority of people have numerous cognitive biases that they are unaware of, it seems a bit much to expect people to swear to do more than their best to remain unbiased.