In a comment here, Aleks points us to Gary C. Ramseyer’s Archives of Statistics Fun. In addition to various amusing items, he has the following fun-looking probability demo. For over a decade now, I’ve been collecting class-participation demonstrations for probability and statistics, and it’s not every day or even every week that I hear of a new one. Here it is:

From Gary Ramseyer:

This month we [Gary Ramseyer] will present another neat probability experiment that can easily be conducted by students as a short assignment or an in-class project. . . That experiment and the current one represent excellent opportunities for an instructor to highlight the value of the Monte Carlo method in estimating probabilities for unusual variations in experiments that don’t lend themselves to the usual formulas.

Let’s state the current question in simple language:

If a penny is flipped until a head first appears, what is the probability that this first head occurs on an odd-numbered trial (i.e.,first, third, fifth, etc.)?

At first blush, a typical student would reason that since the first head is just as likely to occur on an odd trial as it is on an even trial (second, fourth, sixth, etc.), the probability is obviously .5. But wait! Another student mentions that maybe the probaility should be somewhat greater than .5 since the first opportunity for a head to pop up is on the very first trial, and an odd trial continues to preceed an even trial after the first two. At this point the band-wagon effect sets in and students begin to incrementally up their estimates slightly from .5. But after many values are offered, a hush settles over the room and students begin to look at one another and shrug their shoulders. No one is really sure!

Enter Captain Sigma (the instructor)! With a flourish of his cape and a wink of his eye, he quietly suggests that this is a problem that just begs for empirical data. He urges each student to take about five minutes at home and repeat the experiment 10 times, tally how many times the first head appears on an odd trial, and bring the data to the next meeting. The students happily concent to this simple task ( Someone in the back of the room asks, “How many points is it worth?”) and they all eagerly await the pooling of their data at the next meeting.

Two days later the instructor rushes into the classroom and puts all the results from 35 students on the board. The students sit on the edge of their seats in awe as the numbers accumulate. The final tally results in 245 out of 350 replications ending on an odd trial. Zowie! THAT IS 70%! Something is wrong. The pennies must have been seriously flawed.

The instuctor showing no emotion on his face allows the buzzing and chattering to go on for several minutes. Finally he cracks a grin and informs the students that this result is a very good estimate although it is a tad too high. He proudly states that the actual answer is P=2/3 or 67%. The students are dumbfounded and become quite excitable. They actually all cheer for the instructor and demand a formal proof (Did I say “cheer” in a stat class? I must be delirious from a high fever!).

Here is what the instructor wrote on the board:

The solution involves the sum of the first n terms of a geometric series expressed as:

S = a + ar + ar^2 + … + ar^(n-1)

Where
a = first term of the series
n = number of terms
r = the common ratio
S = the sum of the first n terms calculated by

S = a (1 – r^n) / (1 – r)

In our case, a =1/2 = .5 and r = (1/2)(1/2) = .5^2 or .25 and using the first expression for S we have:

S = .5 + .5^3 + .5^5 + …

In words, the above is stating that the probability of getting the first head on an odd trial is the probability of getting a head on the first trial (.5) plus the probability of geting a head on the third trial (.5)(.5)(.5) plus the probability of getting a head on the fifth trial (.5)(.5)(.5)(.5)(.5) plus etc.,etc. for n trials.

Now to compute what this sum would be for n trials we calculate using the second formula:

S = .5 (1 – .25^n) / (1 – .25)

Finally taking the limit of this calculation as n approachs infinity, we arrive at

S = (.5) / (1 – .25) = (1/2) / (3/4) = 2/3 or .67

I didn’t actually laugh at many of the jokes on that webpage, but this demo looks fun and educational.

P.S. Typo fixed (see Bob’s comment).