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By George Hart for the Museum of Mathematics

Many science museums feature a Galton board in which balls branch randomly left or right through an array of pegs and end up binned at the bottom in an approximation of a normal distribution. Here is a deterministic variation on that idea, in which there is a two-position gate at each juncture that causes each ball to go the opposite direction from the previous ball to visit the gate. If you use 64 balls, can you figure out how many will end up in each of the bins at the bottom? Karl Sims built this marble run and gives instructions on how to make your own, along with a clear explanation of its operation.

The side view below shows the tilt of the board and the final distribution of 64 marbles in the seven bins: 1, 6, 15, 20, 15, 6, 1. The mechanism forces each marble to take a different combination of the six left/right choices.

## 14 thoughts on “Math Monday: Make a Marble Run”

Comments are closed.

(as i’m sure all and various know well) the basis of the normal curve, as demonstrated by the Galton board is that there are N binary choices that are

independentof all previous events. So what exactly is the distribution here?! this is a very impressive thought provoking contrari-verse device.I think that it is still a normal distribution (binned). the individual probability of the marble going left/right is independent of what occurred on the previous level, not the linear position. Basically, each level represents a separate, non-deterministic random event.

Of course, I am not a mathematician, just a lowly physicist :-) I would be happy to hear another explanation, particularly if I am incorrect.

I agree with Jason about it still being a normal distribution for a galton board. That is the def’n of a distribution, independent events can still be distributed in some regular way. Each event doesn’t give you knowledge about the next but the distribution of each event is very predictable.

One thing to note of course is that the above board is not a galton board and each event is NOT independent of the earlier events.

As a note one of the answers that he gives is wrong.

Question #2 if you drop enough marbles, will each of those possible paths be taken?

He lists the answer as yes for this board if you drop 64 marbles every path will be taken.

Depending on the initial configuration of the switches that is not true. There are paths that will never be taken given a certain initial setup. This doesn’t change the distribution of marbles though.

For example, given the diamond of items at the top. Lets assume every switch is set

so that the first ball to hit that peg falls left.

I will step through a few runs listing the choice made so you can see the history of each switch.

1.

– L

– L –

–

2.

– LR

– L – L

– L

3.

– LRL

– LR – L

– LR

4.

– LRLR

– LR – LR

– LR

5.

– LRLRL

– LRL – LR

– LR

The 5th step is now repeating the 1st and from now on will just repeat that pattern. Notice however that a LRL run or RLR run is non existent and will never happen. When the top switch is let to go left and the left-most switch of the diamond is set to go right, you are guaranteed that the bottom switch is also set to go right, at least given this initial configuration.

Oh dang I didn’t think about the spacing getting messed up. I intended those to be diamonds as follows with S standing in for the switch.

……..S

..S……..S

…….S

This is a binomial Distribution – the motion of the marbles isn’t completely random – if you notice the numbers listed – they match up to pascal’s triangle – this is a simple binomial distribution (i.e. the marble has a 50% chance of going left or right – chances of going left 3 times and right 3 times and so on).

Link: http://en.wikipedia.org/wiki/Binomial_distribution

It would be normal if it wasn’t for the gates, and if perfectly constructed. Note that a binomial distribution appears to look like a normal one, and by the Central Limit Theorem, will look exactly like one at high enough n.

I agree this is a Binomial Distribution, but don’t you think in your second case where it was perfectly constructed without the paddles, there would still be a 50/50 chance of taking either path and it would still, in fact, be a Binomial distribution?