Three things can happen with equal probability on the fateful toss when "at least one" is heads:
Coin A heads; Coin B heads
Coin A heads, Coin B tails
Coin A tails; Coin B heads
That makes one out of three with both heads
I think it would be 1/2 but the answer is 1/3 and I have no idea where you would get that answer.
Coin A heads; Coin B heads
Coin A heads, Coin B tails
Coin A tails; Coin B heads
That makes one out of three with both heads
1. (Heads, Heads)
2. (Heads, Tails)
3. (Tails, Heads)
4. (Tails, Tails)
Out of these four outcomes, only one outcome has both coins coming up heads. Therefore, the probability of getting both coins heads is 1 out of 4, which simplifies to 1/4 or 0.25.
However, note that the process stops as soon as at least one coin comes up heads. This means that the only outcomes we are interested in are the ones that have at least one head. So in this case, the outcomes (Tails, Heads) and (Tails, Tails) are not considered. Therefore, out of the three remaining possible outcomes, only one has both coins coming up heads.
If we only consider the outcomes (Heads, Heads), (Heads, Tails), and (Tails, Heads) and give equal weightage to each, the probability of getting both coins heads is now 1 out of 3 possible outcomes. Hence, the probability of both coins coming up heads in this scenario is 1/3 or approximately 0.33.
So, while the probability of getting both coins heads in a single toss is 1/4, the probability in this specific scenario is 1/3 because not all outcomes are considered due to the stopping rule.