Puzzle 14 of 26 · Medium

A spinner race

A spinner has 7 equal spots numbered 1 to 7. You spin it again and again. What’s the chance a 5 comes up twice before either a 4 or a 6 comes up even once?

Two 5s before any 4 or 6.

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Answer: 1/9 ≈ 11.1%

Step 1 · Ignore spins that aren’t 4, 5 or 6
$\displaystyle \text{they don't change anything}$
A 1, 2, 3 or 7 just means you spin again. Only spins of 4, 5 or 6 move the race on.
Step 2 · A spin that counts is a 5
$\displaystyle 1 \div 3 = \frac{1}{3}$
Among 4, 5 and 6, each is equally likely, so a spin that counts is a 5 with chance 1/3.
Step 3 · The first two spins that count are both 5s
$\displaystyle \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$
You win only if the first two spins that count are 5s. Those spins don’t affect each other, so multiply.

Answer $\displaystyle \frac{1}{9} \approx 11.1\%$

Step 3 of 3

The method · Unit 3, Conditional probabilitySplitting into casesFind a chance by adding up every case

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