# Reason why a European binary call should be worth half of its American counterpart when driftless and out-of-the-money

Exercise 11 of chapter 8 of Mark Joshi's "The concepts and practice of mathematical finance", asks to compare prices of an American and a European digital (binary) calls when out-of-the-money. The options have same strike and expiry, and are cast on an asset following Brownian motion with no interest rates.

Before peeking at the solution, I had argued that due to the fact we are out-of-the-money, American optionality plays less of a role and the two would be worth about the same.

However, the answer to the exercise states that, given $$r = 0$$ and the symmetry of paths of a Brownian motion, when the American call pays off there is a 50% chance the European also will, so the latter is worth about half the former.

Now, I don't understand the logic of this answer: since we have no knowledge of how bad out-of-the-money we are, I'll concede there is a 50% chance that the asset will end up being worth more than its current value, yes, and also that the value of the European digital is its current probability to end up in-the-money, but how can it be justified that the European is worth half of the American digital?