Probability Jury Problem

Question

A bailiff who works in the court system asks 12-person juries when they are first sequestered, to break the ice, to play the following game. The jurors sit around a circular table that could be thought of as a clock, with the fore-man (fore-woman) sitting at position 12, and the other 11 jurors, in clockwise order, sitting in positions 1, 2, …, 11. Each juror is asked to write the month of his/her birthday with a marker on a sheet of paper and put it face-down on the table. The bailiff first turns over the sheet of paper with the fore-person's month on it. The jurors proceed around the room, in order, turning over their sheets of paper until a month's name is repeated. The juror who turns over the sheet where a repeat first occurs wins a prize. The only way the fore-person can win is if the jurors make it all the way around the table with no repeats (i.e. all 12 months show up exactly once). The bailiff asks the jurors, before they play, which position is most likely to win. Here are some of the answers which he has received over the years. Do you think any are correct? How would you explain?

(a) All 12 positions are equally likely to win.
(b) Position 1 is the best, and chances decrease monotonically from there.
(c) Position 6 is the best because it is in the middle of the first 11.
(d) Position 3 is best.
(e) Position 4 is best.

Answer 1

Let $N$ denote the winning position. Then for $n=0,1,\ldots, 12,$ $$\begin{align*}P(N > n) &= \text{probability that none of the first } n+1 \text{ birthday months match} \\ &= \frac{12}{12}\cdot\frac{11}{12} \cdots \frac{12-n}{12} \\ &= \frac{12!(12-n)}{12^{n+1}(12-n)!} \left(=\frac{12!}{12^{n+1}(11-n)!} \text{ if } n\neq 12\right)\end{align*}$$

Then for $n=1,2,\ldots,12,$ $$\begin{align*}P(N=n) &= P(N>n-1) - P(N>n) \\ &= \frac{12!}{12^n(12-n)!} - \frac{12!(12-n)}{12^{n+1}(12-n)!} \\ &= \frac{12!\cdot n}{12^{n+1}(12-n)!} \end{align*}$$

[From here on, I'll write $P(n)$ as a shorthand for $P(N=n)$]

Therefore, for $n=1,2,\ldots,11,$ $$P(n+1) = \frac{(n+1)(12-n)}{12n}\cdot P(n)$$

Therefore, for $1\leq n \leq 11,$ $$\begin{align*}P(n+1) \left\{\begin{array}{c}< \\ = \\ >\end{array}\right\} P(n) &\iff \frac{(n+1)(12-n)}{12n} \left\{\begin{array}{c}< \\ = \\ >\end{array}\right\} 1 \\ &\iff (4+n)(3-n) \left\{\begin{array}{c}< \\ = \\ >\end{array}\right\} 0 \end{align*}$$

This allows us to see that $$P(1) < P(2) < P(3) = P(4) > P(5) > P(6) > \ldots > P(12)$$ so that positions $3$ and $4$ are equally likely to be the winning positions, and they are the best two positions.

Answer 2

You asked about how to find this in R

You could do something like:

twelve <- 12
probwin <- numeric(twelve)
for (n in 1:twelve){
  if(n == 1){
    probwin[n] <- 1 / twelve
  }else{
    probwin[n] <- (1 - sum(probwin[1:(n-1)])) * n / twelve 
  }
}
probwin

which would give

 [1] 8.333333e-02 1.527778e-01 1.909722e-01 1.909722e-01 1.591435e-01
 [6] 1.114005e-01 6.498360e-02 3.094457e-02 1.160421e-02 3.223393e-03
[11] 5.909554e-04 5.372322e-05

though the one line corresponding to $\frac{12!\, n}{(12-n)!\, 12^{n+1}}$

(1:12) * factorial(12) / factorial(12-(1:12)) / 12^((1:12)+1)

should give the same result.

By inspection, the third and fourth positions are both best with probability $\frac{55}{288}$, though using the which.max function in R would probably only identify one of these due to floating point precision issues