probability – Montessori Muddle

Planting Probabilities

The Gardening Department of our Student-Run-Business sowed seeds in little coconut husk pellets. The question was: how many seeds should we plant per pellet.

Since we’ll only let one seedling grow per pellet, and cull the rest, the more seeds we plant per pellet, the fewer plants we’ll end up with. On the other hand, the fewer seeds we plant (per pellet) the greater the chance that nothing will grow in a particular pellet, and we’ll be down a few plants as well. So we need to think about the probabilities.

Fortunately, I’d planted a some tomato seeds a couple weeks ago that we could use for a test case. Of the 30 seeds I planted, only 20 sprouted, giving a 2/3 probability that any given seed would grow:

$P[\text{grow}] = \frac{2}{3}$

So if we plant one seed per pellet in 10 pellets then in all probability, only two thirds will grow (that’s about 7 out of 10).

What if instead, we planted two seeds per pellet. What’s the probability that at least one will grow. This turns out to be a somewhat tricky problem–as we will see–so let’s set up a table of all the possible outcomes:

Seed 1	Seed 2
grow	grow
grow	not grow
not grow	grow
not grow	not grow

Now, if the probability of a seed growing is 2/3 then the probability of one not growing is 1/3:

$P[\text{not grow}] = 1 - P[\text{grow}] = 1 - \frac{2}{3} = \frac{1}{3}$

So let’s add this to the table:

Seed 1	Seed 2
grow (2/3)	grow (2/3)
grow (2/3)	not grow (1/3)
not grow (1/3)	grow (2/3)
not grow (1/3)	not grow (1/3)

Now let’s combine the probabilities. Consider the probability of both seeds growing, as in the first row in the table. To calculate the chances of that happening we multiply the probabilities:

$P[(\text{seed 1 grow}) \text{ and } (\text{seed 2 grow})] = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$

Indeed, we use the ∩ symbol to indicate “and”, so we can rewrite the previous statement as:

$P[(\text{seed 1 grows}) \cap (\text{seed 2 grows})] = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$

And we can add a new column to the table giving the probability that each row will occur by multiplying the individual probabilities:

Seed 1	Seed 2	And (∩)
grow (2/3)	grow (2/3)	4/9
grow (2/3)	not grow (1/3)	2/9
not grow (1/3)	grow (2/3)	2/9
not grow (1/3)	not grow (1/3)	1/9

Notice, however, that the two middle outcomes (that one seed grows and the other fails) are identical, so we can say that the probability that only one seed grows will be the probability that the second row happens or that the third row happens:

$P[\text{only one seed grows}] = P[(\text{Row 2}) \text{ or } (\text{Row 3})$

When we “or” probabilities we add them together (and we use the symbol ∪) so:

$P[\text{only one seed grows}] = P[(\text{Row 2}) \cup (\text{Row 3}) \\ = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}$

You’ll also note that the probability that neither seed grows is equal to the probability that seed one does not grow and seed 2 does not grow:

$P[\text{neither seed grows}] = P[(\text{seed 1 does not grow}) \cap (\text{seed 2 does not grow}) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

So we can summarize our possible outcomes a bit by saying:

Outcome	Probability
both seeds grow	4/9
only one seed grows	4/9
neither seed grows	1/9

What you can see here, is that the probability that at least one seed grows is the probability that both seeds grow plus the probability that only one seed grows, which is 8/9 (we’re using the “or” operation here again).

In fact, you can calculate this probability by simply taking the opposite probability that neither seeds grow:

$P[\text{neither seed grows}] = 1 - P[\text{neither seed grows}]$

Generalizing a bit, we see that for any number of seeds, the probability that none will grow is the multiplication of individual probability that one seed will not grow:

Probability that no seeds will grow

Number of seeds	Probability they wont grow
1	1/3	(1/3)¹
2	(1/3)×(1/3) = 1/9	(1/3)²
3	(1/3)×(1/3)×(1/3) = 1/27	(1/3)³
n	(1/3)×(1/3)×(1/3)×…	(1/3)ⁿ

So to summarize, the probability that at least one plant will grow, if we plant n seeds is:

$P[\text{at least one seed grows}] = 1 - P[\text{no seeds grow}]$

which is:

$P[\text{at least one of n seeds grows}] = 1 - P[\text{1 seed grows}]^n$

Which is something we may have seen before: What are the odds?

Finally to answer our question: how many seeds we should plant, we need to decide how high a probability we need of success:

Probability that at least one seed will grow

Number of seeds	Probability that at least one seed will grow	%
1	2/3	67%
2	8/9	89%
3	26/27	96%
4	80/81	99%
n	1-(1/3)ⁿ

The Head of Gardening leads the planting of seedlings.

DNA Models and Games

The DNA strand in this board game splits as players end up on different paths based on their choices in the game.

Ms. Mertz’s class is studying DNA–replication, translation, transcription, etc.–and she gave them the option of making a model or creating a game to test each others knowledge. There were some interesting projects:

A DNA strand modeled with the bases represented by colored marshmallows. Toothpicks connect the marshmallows along the backbone of the helices, while Twizzlers are used to show the bonding across the two DNA strands.

The quiz questions in this board game are visible to all the players, which allows for more in-depth discussion of the answers.

Helicase (the cotton ball) splits the DNA double helix.

Ms. Mertz helps a student with their research project on cloning.

The study of DNA and heredity offer great opportunities to study probability. In this case, a player has to traverse 80 squares by rolls of a single dice, and then answer a question from a card. If they get the wrong answer they do not advance, but if they pull a Go to Jail card they have to go back to Jail. If there are 20 cards and two of them send you back to Jail, what's the probability of anyone finishing the game? — The study of DNA and heredity offer great opportunities to study probability. In this case, a player has to traverse 80 squares by rolls of a single dice, and then answer a question from a card. If they get the wrong answer they do not advance, but if they pull a Go to Jail card they have to go back to Jail. If there are 20 cards and two of them send you back to Jail, what’s the probability of anyone finishing the game?

How to be Lucky

The lucky try more things, and fail more often, but when they fail they shrug it off and try something else. Occasionally, things work out.

— McRaney, 2013: Survivorship Bias on You Are Not So Smart.

David McRaney synthesizes work on luck in an article on survivorship bias.

… the people who considered themselves lucky, and who then did actually demonstrate luck was on their side over the course of a decade, tended to place themselves into situations where anything could happen more often and thus exposed themselves to more random chance than did unlucky people.

Unlucky people are narrowly focused, [Wiseman] observed. They crave security and tend to be more anxious, and instead of wading into the sea of random chance open to what may come, they remain fixated on controlling the situation, on seeking a specific goal. As a result, they miss out on the thousands of opportunities that may float by. Lucky people tend to constantly change routines and seek out new experiences.

McRaney goes also points out how this survivorship bias negatively affects scientific publications (scientists tend to get successful studies published but not ones that show how things don’t work), and in war (deciding where to armor airplanes).

What are the odds?

My middle school class stumbled upon a nice probability problem that I might just use in my pre-calculus exam.

Oregano

You see, four of my students took cuttings of oregano plants from the pre-school herb gardens when we were studying asexual reproduction, but only three of them grew roots.

That gives a 75% success rate.

$P[success] = 0.75$

The next time I do it, however, I want each student to have some success. If I have them take two cuttings that should increase the chances that at least plant will grow. And we can quantify this.

If the probability of success for each cutting is 75% then the probability of failure is 25%.

P[failure] = 0.25

Given a probability of one plant failing of 25%, the probability of both plants failing is the probability of one plant failing and the other plant failing, which, mathematically, is:

P[2 failures] = P[failure] x P[failure]

P[2 failures] = 0.25 x 0.25

P[2 failures] = 0.25²

P[2 failures] = 0.0625

So the probability of at least one plant growing is the opposite of the probability of two failures:

P[at least 1 success of two plants] = 1 – P[2 failures]

P[at least 1 success of two plants] = 1 – 0.0625

P[at least 1 success of two plants] = 0.9375

which is about 94%.

If I had the students plant three cuttings instead, then the probability of at least one success would be:

P[at least 1 success of 3 plants] = 1 – P[3 failures]

P[at least 1 success of 3 plants] = 1 – 0.25³

P[at least 1 success of 3 plants] = 1 – 0.02

P[at least 1 success of 3 plants] = 0.98

So 98%.

This means that if I had a class of 100 students then I would expect only two students to not have any cuttings grow roots.

An Equation

In fact, from the work I’ve done here, I can write an equation linking the number of cuttings (let’s call this n) and the probability of success (P). I did this with my pre-calculus class today:

$P = 1 - 0.25^n$

so if each student tries 4 cuttings, the probability of success is:

P = 1 – 0.25⁴

P = 1 – 0.004

P = 0.996

Which is 96.6%, which would make me pretty confident that at least one will grow roots.

However, what if I knew the probability of success I needed and then wanted to back calculate the number of plants I’d need to achieve that probability, I could rewrite the equation solving for n:

Start with:
$P = 1 - 0.25^n$

isolate the term with n by moving it to the other side of the equation, and switching P as well:
$0.25^n = 1 - P$

take the natural log of both sides (we could take the log to the base 10 if we wanted, or any other base log, but ln should work just as well).

$\ln{(0.25^n)} = \ln{(1 - P)}$

now use the rules of logarithms to bring the exponent down into a multiplication:

$n \ln{(0.25)} = \ln{(1 - P)}$

solving for n gives:

$n = \frac{\ln{(1 - P)}}{\ln{(0.25)}}$

Note that the probability has to be given as a fraction (between 0 and 1), so 90% is P = 0.9. A few of my students made that mistake.

Lavender

roots-3628 — Roots on two of three plants.

At the same time my students were making oregano cuttings, six of them were making cuttings of lavender. Only one of the cuttings grew.

So now I need to find out how many lavender cuttings each student will have to make for me to be 90% sure that at least one of their cuttings will grow roots.

Students can do the math (either by doing the calculations for each step, or by writing and solving the exponential equation you can deduce from the description above), but the graph below shows the results:

probability-cuttings — Probabilities of successfully growing a lavender cutting.

Beating the Odds: The Sheer Improbability of Being Here

visual.ly posts and hosts some excellent graphics. The one below, calculates the nearly infinitesimal probability of just being born. There’s hardly a better argument for appreciating life.

It’s also a good example of working with probabilities [and] exponents. Very large exponents.

by visually via

Beating probability

Since we just finished doing a bit of probabilities in math, here’s an article about how one guy figured out how to beat the lottery.

The first lottery Mohan Srivastava decoded was a tic-tac-toe game run by the Ontario Lottery in 2003. He was able to identify winning tickets with 90 percent accuracy.
–Lehrer (2010) in Cracking the Scratch Lottery Code

However, he decided not to just try to get rich of what he’d discovered. It’s an example of using the power of math for good:

“People often assume that I must be some extremely moral person because I didn’t take advantage of the lottery,” [Srivastava] says. “I can assure you that that’s not the case. I’d simply done the math and concluded that beating the game wasn’t worth my time.”

As a side note, my philosophy about the lottery is that it’s basically a tax on the poor:

[H]igh-frequency players tend to be poor and uneducated, which is why critics refer to lotteries as a regressive tax. (In a 2006 survey, 30 percent of people without a high school degree said that playing the lottery was a wealth-building strategy.)
–Lehrer (2010): Cracking the Scratch Lottery Code

Boredom in a fractal world

Doxocopa_laurentia_male_face_sup — Brazilian butterfly Doxocopa laurentia (from Wikipedia)

A few of my students have been complaining that we don’t do enough different things from week to week for them to write a different newsletter article every Friday. PE, after all, is still PE, especially if we’re playing the same game this week as we did during the last.

So I’ve been thinking of ways to disabuse them of the notion that anything can be boring or uninteresting in this wonderful, remarkable world. A world of fractal symmetry, where a variegated leaf, a deciduous tree and a continental river system all look the same from slightly different points of view. A counterintuitive world where the smallest change, a handshake at the end of a game, or a butterfly flapping its wings can fundamentally change the nature of the simplest and the most complex systems.

“Chaos is found in greatest abundance wherever order is being sought. It always defeats order, because it is better organized.”
— Terry Pratchett (Interesting Times)

Dragon_trees — Fractal trees (from Wikimedia Commons)

There are two things I want to try, and I may do them in tandem. The first is to give special writing assignments where students have to describe a set of increasingly simple objects, with increasingly longer minimum word limits. I have not had to impose minimum word limits for writing assignments because peer sharing and peer review have established good standards on their own. Describing a tree, a coin, a 2×4, a racquetball in a few hundred words would be an exercise in observation and figurative language.

To do good writing and observation it often helps to have good mentor texts. We’re doing poetry this cycle and students are presenting their poems to the class during our morning community meetings. It had been my intention to make this an ongoing thing, so I think I’ll continue it, but for the next phase of presentations, have them chose descriptive poems like Wordsworth’s “Yew Trees“*.

The world is too interesting a place to let boredom get between you and it.

* An excellent text for a Socratic dialogue would be the first page of Michael Riffaterre’s article, Interpretation and Descriptive Poetry: A Reading of Wordsworth’s “Yew-Trees”. It’s testing in its vocabulary but remarkably clear in thought if you can get through it.