Tuesday

"Ten laps today, ten flips," Comet says, smoothing the tally sheet. "Before we flip, what should we expect?"

"Half of 10 is 5 heads," Wren says. "But not exactly 5 every time. What do you notice about the bar graph?"

Nova lifts the graph for 10 flips. "Would you like a hint? Most of the bars sit between 2 and 8."

"So the model expects about 5, give or take," Comet says. "How much give or take?"

"The spread is the square root of n times p times 1 - p," Wren says. "√(10 × 1/2 × 1/2) ≈ 1.58."

"Two of those on each side," Comet says. "1.8 to 8.2. So 2 through 8 heads is ordinary."

They flip. T, T, T, T, T, H, T, H, H, T. "3 heads," Wren says. "Inside the band. The coin is behaving."

Write all 32 orders of 5 flips and sort them by how many heads they hold. The counts are 1, 5, 10, 10, 5, 1.

Each order has probability 1/32, so the probability of k heads is the count of orders over 32.

This way shows where the probabilities come from. It is slow for ten flips, so the crew has a second way.

P(exactly k heads in n flips) = C(n, k) × p^k × (1 - p)^(n - k).

C(n, k) counts the orders with k heads. The powers multiply p once for each head and 1 - p once for each tail.

For 5 flips and 2 heads: C(5, 2) × (1/2)^2 × (1/2)^3 = 10/32 = 5/16. Both ways agree.

A random variable gives each outcome a number. Here X is the number of heads. Its distribution is the table from Monday.

The expected value of X is each value times its probability, added up. For 5 flips it comes to 2.5, which is n × p.

The spread is the standard deviation √(n p (1 - p)). For 10 flips: √(10 × 1/2 × 1/2) ≈ 1.58.

The crew's rule: a count within 2 standard deviations of the expected count is ordinary. For 10 flips, that is 1.8 to 8.2.

Two ways that agree. Tomorrow is Data Lab: you run the pacing trial ten times and see the distribution for yourself.