We have 3 coins: 🪙 A = HT 🪙 B = HH 🪙 C = TT First pick You pick a coin and get H . So the first coin cannot be TT . There are two possibilities: Possibility 1: You picked HT Remove it → remaining coins are: H H , T T HH,\ TT Second pick: HH → H TT → T So chance of H = 1/2 . Possibility 2: You picked HH Remove it → remaining coins are: H T , T T HT,\ TT Second pick: HT → H or T, each with probability 1/2 TT → T So chance of H = 1/4 . Now, because you already got H on the first pick, HH was twice as likely as HT : HT chance = 1/3 HH chance = 2/3 Therefore: 1 3 × 1 2 + 2 3 × 1 4 \frac13\times\frac12+\frac23\times\frac14 = 1 6 + 1 6 = 1 3 =\frac16+\frac16 =\boxed{\frac13} Summary: First H → remove that coin → depending on whether it was HT or HH, the remaining coins are different → combining those two cases gives 1 / 3 1/3 . So the answer is 1 / 3 \boxed{1/3} .