Learn machine learning · part 3 · session 1
Many weak voters
Ask twenty five people who are each right sixty percent of the time, take the majority answer, and the majority does better than any of them. The catch is in one word, and this page has a dial for it.
After this session you will be able to say why a crowd of models that copy each other is worth no more than one of them, and put a number on what independence is buying.
The gate is three unseen configurations. Call where each one's majority lands before the engine reveals it, and two of three passes.
01 · The question
Your tree from 2.4 changed its root split when I reshuffled the training rows. That looked like a fault. In this part it becomes the method. The idea is old and it is not about machine learning: ask twenty five people who are each right sixty percent of the time, take the majority answer, and the majority does better than any of them. A group of separate answers pooled into one is an . The catch is in one word. Independent.
The dial is called correlation. At zero, each voter makes up their own mind. At one, every voter is copying the same source, so when it is wrong they are all wrong together. On this page there is no model, only a simulation of voters and that dial. Before you move it, write down what you think the crowd is worth at both ends of it.
02 · Predict
Before anything runs
Twenty five people each get it right 60 times in 100. First case: they all decide independently. Second case: they all follow the same one source. What does the majority score in each case?
Two numbers: the independent case first, then the copying one.
03 · Try it
What the simulation does
illustrativen = 25 voters. Each voter is right on an item with probability p = 0.60. Correlation enters through one shared latent draw. For each item, draw Z from a standard normal once and an independent E for every voter. Voter i is right when sqrt(rho) * Z + sqrt(1 - rho) * E_i > t, with t = -0.2533, the value that holds every single voter's own accuracy at exactly 0.60 for every setting of rho. The ensemble answer is the majority of the 25, and the dial is the correlation of the shared latent variable.
The curve is computed by quadrature over the shared latent with a binomial tail inside, so the figure is identical on every device. The sampled run beside it draws real items from 7.
Twenty five voters, one item
illustrativethe 3.1 simulation, seed 7Correlation 0, item 1. 15 of 25 right, and the majority is right.
The crowd
illustrativen 25, p 0.60These voters are a simulation, not a model. Nothing is trained here. Each voter is right on an item with probability 0.60 at every setting of the dial, and the dial changes only how much of each answer comes from the one shared draw.
Odd numbers only, so the majority always settles.
How much of each answer comes from the shared draw.
Majority accuracy against correlation
25 voters, shared-latent correlation 0.00, majority accuracy 0.8462. A single voter is 0.60. That is 0.2462 above one voter and 0.0000 below the same crowd with no shared latent.
The filled dot is the dial position the controls are on.
A sampled run beside the exact figure
The curve above is what the maths says happens on average. This button runs 2000 real items from seed 7 at the current dial position and shows what actually happened, so you can see the two land in the same place.
No sampled run yet. The exact figure at this setting is 0.8462.
Adding voters to a crowd that copies itself
Pin the dial at 0.90 and raise the crowd. The comparison is 25 voters against however many the control is set to.
Three voters. The majority is right when at least two are. The four ways that happens are 0.6 × 0.6 × 0.6 plus three lots of 0.6 × 0.6 × 0.4. Now do the arithmetic.
To three decimals, within 0.001.
Attempts: 0
04 · Compare
The verdict
Record a position and run the sampled draw at least once, then read the verdict against what you committed to.
Waiting on a recorded position and one run of the simulation.
05 · Check your understanding
Your check: predict results on new examples
not yetThree unseen configurations. Call where each majority lands before the engine runs it. Two of three passes, and every reveal is the real quadrature on that configuration.
Variant 1 of 3
51 voters, each right 0.60 of the time, shared-latent correlation 0.00. Where does the majority land?
Call it before you run it.
Variant 2 of 3
25 voters, each right 0.60 of the time, shared-latent correlation 0.20. Where does the majority land?
Call it before you run it.
Variant 3 of 3
25 voters, each right 0.45 of the time, shared-latent correlation 0.00. Where does the majority land?
Call it before you run it.
Called 0 of 3 variants so far.
Ask for a hint
Rungs open as you run the session, not as time passes. You have run it 0 times.
1. A question: available after 3 evaluated runs.
2. A narrowing: available after 5 evaluated runs.
3. A worked fragment on a different example: available after 7 evaluated runs.
4. The answer: available after 9 evaluated runs.
Save your takeaway
Your takeaway
Edit it into your own words, or keep the line as it is.
This session is a concept session, so nothing goes on the shelf. The artifact arrives in 3.2, where the voters become trees and the shared draw becomes two settings you control.