One bubble-sort pass

1. Step 1 · In one pass, a number moves left at most 1 place

   $\displaystyle \text{it only moves left when } a \text{ bigger number passes it}$

   The pass carries the biggest number so far to the right. Anything it passes shifts left by just one place.

2. Step 2 · So number k must start in the first k + 1 places

   $\displaystyle 1 \text{ in place } 1 \text{ or } 2,\, 2 \text{ in places } 1 \text{ to } 3,\, \text{ and so on}$

   To end in place $k$, number $k$ can start at most one place to its right. That’s the only thing that can go wrong.

3. Step 3 · Place 1, then 2, then 3 …

   $\displaystyle 2 \times 2 \times 2 \times 2 \times 1 = 2^{4} = 16$

   Each number has 2 allowed spots left when you place it (the last has 1). So 16 orders get sorted.

4. Step 4 · Chance

   $\displaystyle 16 \div 5! = 16 \div 120 = \frac{2}{15}$

   All 120 orders are equally likely, and 16 of them get sorted.

Answer$\displaystyle \frac{2}{15} \approx 13.3\%$