Six Heads That Are Never Repaid, and a Proportion That Converges Anyway

You'll learnwhy a long-run proportion settles without any result ever compensating for another, and how to design a simulation whose chances match the situation and report what it produces as the estimate it is.

A fair coin has landed heads six times. Now suppose every toss after that splits exactly evenly — one head, one tail, forever — so the streak is never once repaid. The running proportion of heads goes 1.0000, 0.8000, 0.6500, 0.5600, 0.5300, 0.5030, 0.5003. It converges to 0.5, and the surplus of three extra heads is still sitting there, unchanged, at every single checkpoint. The even split is imposed on purpose: this is the extreme case, and even in it the proportion comes back. That is what the law of large numbers actually says, and it is why the seventh toss is still 0.5. The other half of the topic is the design: 70% needs seven of the ten digits, not the digits 0 through 7, and a simulated probability is an estimate whose spread shrinks like one over the square root of the number of trials.

8 STEPS · 6 QUICK CHECKS · LONG-RUN RELATIVE FREQUENCY · DILUTION · DIGIT ASSIGNMENT · STANDARD ERROR · v1

surplus 3 at every checkpoint · gap = 3 / n · 0.5003 at n = 10000

Before you start

What you're looking at

The running proportion of heads for one fair coin, plotted against the number of tosses so far on a logarithmic axis so that 10 and 10,000 both fit.

The question

Six heads have already landed. If no later toss ever repays them, how does the proportion get back to 0.5 — and what does that say about the next toss?

Watch for

The vertical gap between each dot and the 0.5 line. The surplus that produces it never changes; only the number underneath it does.

Step 1 / 8