Pascal's Marble Run - this example has six rows of switches and seven collecting bins.
Pascal's Marble Run - or the Deterministic Galton Board
by Karl Sims
This marble run can be used to help understand the mathematics of probability in a physical and interactive way. Marbles rolling through a triangular array of switches and collecting in a series of bins can demonstrate concepts such as bell shaped distributions, powers of two, binary numbers, coin flipping odds, and Pascal's triangle.
Blaise Pascal was a French mathematician and inventor of mechanical calculating machines. In 1653 he described the triangle of binomial coefficients, now called Pascal's triangle, in which each number is the sum of the two above it. As you will see, elements of this triangle can be generated by this marble run.
A "Galton board" is a device in which falling balls bounce randomly left and right off of rows of pins and are collected in a number of bins at the bottom. It was invented around 1860 by Sir Francis Galton, a cousin of Charles Darwin, who used it as a tool for demonstrating the normal distribution or bell curve.
The marble run described here is a non-random variation of the Galton board that has a switch at each juncture to alternate the direction of the marbles passing through it. At each switch, if the first marble falls to the left, the next will fall to the right, and so on. This causes predictable patterns instead of random behavior, and if an even number of marbles pass through each switch, the results should exactly match the probabilities if the bounces had been random.
Here are a few questions to start you thinking.
- How many different paths through the switches could a marble take?Hint Hint: A marble passes through 6 switches, and at each switch it can roll two possible ways, left or right.Answer Answer: 64. After the top switch there are 2 possible paths (L or R). After the next switch there are 4 (LL, LR, RL, RR), and so on. At each level the number of possible paths doubles, so after 6 levels there are 2 x 2 x 2 x 2 x 2 x 2 = 64 possible paths.
- Tilt all the switches to the right so the first marble lands in the left-most bin. How many marbles need to roll down for one to land in the right-most bin?Answer Answer: 64. The last path taken will be all rights, and then the cycle starts over again.
- If the switches start tilted in random directions, can you predict what state they will all be in after 64 marbles fall through?Hint Hint: think about how many times each switch will flip.Answer Answer: Yes. After 64 marbles, every switch should change state an even number of times, so they will all return to their initial states.
- You can experiment with smaller triangles by inserting marbles in the middle of the board above a specific switch instead of starting them at the top. What sequences of marble counts in the bins will you get after you insert: 4 marbles 2 switches above the bottom? 8 marbles 3 switches above the bottom? or 16 marbles 4 switches above the bottom?
Hint Hint: the first one is easy: 1 2 1. Can you figure out the next one by combining that sequence with itself somehow?Answer Answer: 1 2 1, 1 3 3 1, and 1 4 6 4 1. These are the rows of Pascal's triangle! If you add a row to itself shifted over by 1 you get the next row.
- How many possible paths lead to each switch and each bottom bin?Hint Hint: There is only one path that reaches all the switches on the outer edges (either all lefts or all rights) but there are multiple ways a marble can reach the center switches. For example to reach the center of row 2 it can go left-right, or right-left. Can the number of possible paths to the switches on one level help you find the number of possible paths to those on the next level down?Answer Answer: Pascal's triangle! For all the switches on the outer edges, there is only one path (all lefts or all rights). For the other switches, a marble can come from either the right or left above, so the number of possible paths to one location is just the number of possible paths to both of the locations above it. You can find each new number in the triangle by adding the two numbers above it.
- What fraction of marbles pass through each switch?Answer Answer: This is similar to Pascal's triangle again, but you need to turn the number of possible paths into fractions. For each row, divide the number of possible paths to the switch by the total number of possible paths for that row. All marbles pass through the top switch. Each switch on the next row gets half of the marbles. For the next row there are 4 possible paths and the switches get 1/4, 2/4, and 1/4 of those. And so on.
- How often do the edge switches (the first and last in each row) get flipped?Answer Answer: It takes double the number of marbles to flip each next switch on either edge. The top switch is flipped by every marble, and the edge switches in the other rows are flipped by every 2nd, 4th, 8th, 16th, and 32nd marble.
- If you start with all the switches tilted to the left, and somebody secretly lets some number of marbles through (between 0 and 63), can you tell how many marbles fell through just by looking at the switches?
Hint Hint: look at just the 6 switches on the left edge.Answer Answer: Yes. The switches on the left edge are counting the marbles in binary. If leaning left=0 and right=1, the switches count: 000000, 000001, 000010, 000011, 000100, etc. After 111111 is reached they all flip together to start over again at zero, so it can't count any higher than 63 marbles. To convert from binary to decimal you can add the appropriate power of 2 for each switch that is leaning to the right. Note that the right edge of six switches are also counting the marbles in binary, but with the switch left/right reversed, and delayed by one.
- For each bin at the bottom, how many times does a marble need to fall to the right to land in that bin?
Answer Answer: If the bins are numbered left to right from 0 to 6, that is also the number of rights the marble needs to make to get there. The total number of rights vs lefts (in any order) determine which bin the marble lands in.
- If you flip 6 coins what are the chances they will all be heads? Note that this is the same as the chances a marble will fall left all 6 times.
Answer Answer: 1/64. Only one of the 64 possible combinations gives all heads, or all lefts.
- If you flip 6 coins what are the chances that only 1 of them will be a tail? Note that this is the same as the chances a marble will fall right only once.
Hint Hint: at how many switches could the marble make the one right?Answer Answer: 6/64 (or 3/32 reduced). Of the 64 possible combinations, there are 6 ways to get 1 tail (each of the 6 coins could be that tail). Of the 64 possible marble paths, there are 6 of them that have 1 right (each of the 6 switches could be that right).
- If you flip 6 coins what are the chances that you'll get 3 heads and 3 tails? Note that this is the same as the chances that a marble will take 3 lefts and 3 rights, in any order.
Hint Hint: you may want to refer to Pascal's triangle or the number of marbles in your bins after dropping 64 of them.Answer Answer: 20/64 (or 5/16 reduced). If a marble takes 3 lefts and 3 rights in any order, it will always end up in the middle bin. If you drop 64 marbles to traverse each possible path, 20 of those land in the middle bin.
- If you have 6 items how many ways can you choose 2 of them? Note that this is the same as how many ways a marble can take 2 rights.
Hint Hint: there are 6 ways to choose 1, then there are 5 left to choose the 2nd from, but the order doesn't matter.Answer Answer: 15. There are 6 ways to choose the first (or take a right) times 5 ways to choose the 2nd (or take a 2nd right), but that includes both orderings of the 2 items so we divide by 2 for the answer: 6x5/2 = 15. If you drop 64 marbles, 15 of those land in the third bin (or bin 2 if you start at 0).
How to build a Pascal's Marble Run
For the 6 level version shown here, you will need:
- 64 or more 1" marbles (or 15/16" or 25mm)
- 1 piece of smooth plywood for the base 19" x 36" x 1/2"
- 21 T-shaped switches 1/2" thick
- 21 bumpers 2" x 1½" x 1/2" with corners cut diagonally and 6 with bottoms cut off as shown
- 6 bin dividers 12" x 1" x 1/2"
- 4 sides 1½" x 1/2", two each of 19" and 37" lengths, one with a 7" gap for marbles to exit
- 2 removable sticks to hold back marbles 3/4" x 1/2" x 19"
- Various other guides and ramps cut from 1/2" thick wood
- Leg to keep board tilted: 19" x 5" x 1½" and optional front leg 19" x 1" x 1½" or similar
- Misc nails, tiny washers, dowels, glue, and tools
This schematic shows the dimensions and layout of the small parts. You can make the T shaped switches in two parts and glue them together, or you might be able to make them in one piece if you have tools that can machine small parts well. Drill a small hole in the center of each switch. We used a band saw with templates, drill press, and disc sander, to carefully replicate copies of each shape.
Refer to the schematic and images above for placement. Glue or nail the bumpers and guides to the board. Attach the switches with thin nails and tiny washers so they can rotate freely. Test each level to make sure the marbles can pass through the switches to the next level without getting stuck.
Note that some changes would be needed if this device were used in a public setting. The table should probably be covered so the marbles can't easily be reached, and some technique such as a conveyer belt or a pinball shooter would need to return the marbles to the top. This should also limit the rate at which the marbles are dropped to prevent them from jamming. Currently a person needs to nudge the marbles now and then to unclog them where they enter the marble run at the top.
Thanks to my sons Arlo and Felix for help building this prototype, and to the MIT Hobby Shop for the use of their tools.
Pascal's triangle of numbers can be constructed by starting with a single 1 at the top and then growing downwards using a simple rule: each new number is the sum of the two numbers above it. On the edges we assume the empty space is 0 so the 1's are just copied down each side. Many interesting patterns and mathematical connections are hidden in Pascal's triangle...
Bell shaped curves
If you graph the values from a single row of Pascal's triangle, it forms a bell shaped curve. This "normal distribution" shows up all over the place when a number of random events are combined. Stock market returns, test scores, human heights, life spans for a given species, and so on, all tend to have bell shaped probability distributions.
Graphs showing row 6 and row 30 of Pascal's triangle (with different scales)
Pascal's triangle also provides a quick way to look up binomial coefficients. For example, to expand (x+y)6 the number of each x and y power combination follows row 6 of Pascal's triangle:
(x+y)6 = 1 x6 + 6 x5 y + 15 x4 y2 + 20 x3 y3 + 15 x2 y4 + 6 x y5 + 1 y6
Powers of 2
Sum each row of Pascal's triangle to get the powers of two: 1, 2, 4, 8, 16, 32, 64...
Powers of 11
If you collapse each row into a single number by taking each element as a digit (and carrying over to the left if the element has more than one digit) you get the powers of eleven: 1, 11, 121, 1331, 14641, 161051...
Added adjacent triangulars give squares
The 3rd diagonal contains the "triangular numbers" (1, 3, 6, 10, 15, 21, 28, 36, 45...) and if you add adjacent pairs of these, you get the perfect squares: 1+3=4 3+6=9 6+10=16 10+15=25 and so on.
The Choose function
Say you have 6 items and you want to know how many different ways you can choose 2 of them. The answer of "6 choose 2" is 6 x 5 / 2 = 15. There are 6 ways to choose the 1st item and then 5 left to chose the 2nd item from, but that includes both orderings of the 2 items so we divide by 2 since the order doesn't matter. The general equation for "N choose K" is N! / (N-K)! / K! and this formula can be used to calculate an arbitrary binomial coefficient or element of Pascal's triangle. Row N number K (starting at 0) in Pascal's triangle equals "N choose K".
12 Days of Christmas gifts
In the "12 Days of Christmas" song, the number of gifts can be found in Pascal's triangle. The 3rd diagonal contains the "triangular numbers" which are each the sum of the first N integers. The 4th diagonal contains the "tetrahedral numbers" (1, 4, 10, 20, 35, 56, 84, 120...) which are each the sum of the first N triangular numbers. If you stack triangles of sizes 1 to N you get a 3D tetrahedron with N per edge. In the classic Christmas song "my true love gave to me" N new gifts each day plus all the gifts from previous days are repeated, so the number of gifts received on each day are the triangular numbers, and the cumulative total gifts are the tetrahedral numbers. The grand total gift count is the 12th tetrahedral number: 364.
Triangular numbers Tetrahedral numbers
If N people are at a party and all click their glasses with each other, how many total clinks are there? Each person clinks N-1 others, so avoiding double counting, the answer is N(N-1)/2. This is also "N choose 2" and is also the sum of 1 to N-1. It is also the Nth triangular number found in the 3rd diagonal of Pascal's triangle.
If the 2nd number in a given row is a prime number, all the numbers in that row, except the 1's, are multiples of that same prime number. All the elements in the downward pointing triangle from this row are also multiples of that same prime, since when you add two multiples of the same number you get another multiple of that number. It also appears that if the 2nd number in a given row is any power of a prime number, all the numbers in that row except the 1's are multiples of that prime number.
Rows with prime multiples