Consider a pinball machine like the one shown in Figure 8. Launch 32 balls, assuming a well-balanced board so that the ball has equal odds of falling right or left at any juncture when hitting a pin. Your expected outcome is that many balls will land in the center columns and that the number of balls will decrease as you move to the columns away from the center.
Next, consider a gedanken, a thought experiment. A man flips a coin and after each toss he takes a step to the left or a step to the right, depending on whether the coin came up heads or tails. This is called the random walk, but it does not necessarily concern itself with walking. You could identically say that instead of taking a step to the left or to the right, you would win or lose $1 at every turn, and you will keep track of the cumulative amount that you have in your pocket.
Assume that I set you up in a (legal) wager where the odds are neither in your favor nor against you. Flip a coin. Heads, you make $1, tails, you lose $1.
At the first flip, you will either win or lose.
At the second flip, the number of possible outcomes doubles. Case one: win, win. Case two: win, lose. Case three: lose, win. Case four: lose, lose. Each of these cases has equivalent odds, the combination of a single win and a single loss has an incidence twice as high because cases two and three, win-lose and lose-win, amount to the same outcome. And that is the key for the Gaussian. So much in the middle washes out—and we will see that there is a lot in the middle. So, if you are playing for $1 a round, after two rounds you have a 25 percent chance of making or losing $2, but a 50 percent chance of breaking even.
FIGURE 8: THE QUINCUNX (SIMPLIFIED)—A PINBALL MACHINE
Drop balls that at every pin, randomly fall right or left. Above is the most probable scenario, which greatly resembles the bell curve (a.k.a. Gaussian disribution).
Let us do another round. The third flip again doubles the number of cases, so we face eight possible outcomes. Case 1 (it was win, win in the second flip) branches out into win, win, win and win, win, lose. We add a win or lose to the end of each of the previous results. Case 2 branches out into win, lose, win and win, lose, lose. Case 3 branches out into lose, win, win and lose, win, lose. Case 4 branches out into lose, lose, win and lose, lose, lose.
We now have eight cases, all equally likely. Note that again you can group the middling outcomes where a win cancels out a loss. (In Galton’s quincunx, situations where the ball falls left and then falls right, or vice versa, dominate so you end up with plenty in the middle.) The net, or cumulative, is the following: 1)
Out of the eight cases, the case of three wins occurs once. The case of three losses occurs once. The case of one net loss (one win, two losses) occurs three times. The case of one net win (one loss, two wins) occurs three times.
Play one more round, the fourth. There will be sixteen equally likely outcomes. You will have one case of four wins, one case of four losses, four cases of two wins, four cases of two losses, and six break-even cases.
The quincunx (its name is derived from the Latin for five) in the pin-ball example shows the fifth round, with sixty-four possibilities, easy to track. Such was the concept behind the quincunx used by Francis Galton. Galton was both insufficiently lazy and a bit too innocent of mathematics; instead of building the contraption, he could have worked with simpler algebra, or perhaps undertaken a thought experiment like this one.
Let’s keep playing. Continue until you have forty flips. You can perform them in minutes, but we will need a calculator to work out the number of outcomes, which are taxing to our simple thought method. You will have about 1,099,511,627,776 possible combinations—more than one thousand billion. Don’t bother doing the calculation manually, it is two multiplied by itself forty times, since each branch doubles at every juncture. (Recall that we added a win and a lose at the end of the alternatives of the third round to go to the fourth round, thus doubling the number of alternatives.) Of these combinations, only one will be up forty, and only one will be down forty. The rest will hover around the middle, here zero.