Problem 1 · AMC 8 Stretch
Core
Counting & Probability
logical-reasoningaccounting-for-all-possibilities
You flip a fair coin and get 5 heads in a row. (1) What is the probability that the very next flip is also heads? (2) Separately, before you start flipping at all, what is the probability of getting 6 heads in a row? To warm up, list all the outcomes of just 2 flips and find the probability of getting 2 heads in a row.
Show answer
Answer: Next flip: \(\frac{1}{2}\). A streak of 6 heads from the start: \(\frac{1}{64}\)
Show hints
Hint 1 of 4
These are two different questions! One asks about a single next flip; the other asks about a whole streak from the start. Don't let them blur together.
Still stuck? Show hint 2 →
Hint 2 of 4
A coin has no memory. It doesn't know it just landed on 5 heads. So for the single next flip, the past flips change nothing.
Still stuck? Show hint 3 →
Hint 3 of 4
For a streak, list the equally likely outcomes. Two flips give HH, HT, TH, TT — four equally likely cases, so 'two heads' happens 1 time out of 4.
Show solution
Approach: Accounting for all possibilities — independence vs. counting a whole streak
- The single next flip: coin flips are independent, so nothing that already happened can change a future flip. The coin has no memory of those 5 heads. So the chance the next flip is heads is just \(\frac{1}{2}\), the same as always.
- A streak from the start: warm up with 2 flips. The four equally likely outcomes are \(HH, HT, TH, TT\). Only 1 of these 4 is two-heads, so the probability is \(\frac{1}{4}=\frac{1}{2}\times\frac{1}{2}\).
- Each extra flip multiplies by another \(\frac{1}{2}\). For 6 heads in a row, \(\left(\frac{1}{2}\right)^6=\frac{1}{64}\).
- So both are true at once: a long streak really is unlikely as a whole, but that does NOT make the single next flip anything other than \(\frac{1}{2}\).
Mark:
· log in to save