Two random breaks, three pieces, and three inequalities that collapse into one. The quarter falls out of a square with no integral at all, and the average longest piece, 11/18, explains why the answer feels too low but isn't.
Also asked asbroken stick · broken stick problem · break a stick into three pieces · stick breaking triangle probability
Read the full explanation — Break a Stick Twice, and the Pieces Form a Triangle ¼ of the Time
Wind appears nowhere in the winning condition, so delete it: three cards in six equally likely orders, and you win in the two where fire comes last. One in four is the correct probability that fire is last of all four cards, a strictly smaller event, and the gap is exactly one twelfth.
Every family averages exactly one girl and contains exactly one boy, so the ratio of expected counts is exactly one half and a large town splits evenly. The expected share inside a single family is not one half but ln 2, and it is still 0.5249 across ten families, with the excess falling off like one over four m.
A pebble climbing four boxes on coin flips needs 18/5 flips on average, and the two-line renewal argument that gives 4 is wrong. Its premise is true, since half of all games really do end on flip two, but the non-finishing half is two different states: tails-tails sends the pebble home while heads-heads leaves it on box 3, one flip from the exit and worth only 14/5.