Puzzle 20 of 26 · Medium

An ant on a tetrahedron

An ant stands on one corner of a tetrahedron (4 corners, each joined to the other 3). Each step it walks to one of the other 3 corners at random. On average, how many steps until it has visited every corner and is back where it started?

Each step goes to one of the 3 other corners, each with chance 1/3.

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Answer: 8.5

Average steps for each stage.

Step 1 · Wait for the 2nd corner
$\displaystyle \text{every neighbour is new } = 1 \text{ step}$
From the start, all 3 corners it can walk to are new, so the first step always finds one.
Step 2 · Wait for the 3rd corner
$\displaystyle 1 \div \frac{2}{3} = 1.5$
With 2 corners seen, 2 of its 3 moves reach a new corner. A chance of 2/3 each step means $1 \div 2/3$ steps on average.
Step 3 · Wait for the 4th corner
$\displaystyle 1 \div \frac{1}{3} = 3$
With 3 corners seen, only 1 of its 3 moves is new.
Step 4 · Wait to get back to the start
$\displaystyle 1 \div \frac{1}{3} = 3$
It’s now away from the start, and from any other corner 1 of its 3 moves goes home.
Step 5 · Add the waits
$\displaystyle 1 + 1.5 + 3 + 3 = 8.5$
The four waits happen one after another, so the average total is their sum.

Answer $\displaystyle 8.5$

Step 5 of 5

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