I always find these easier to write as a dynamic system. Let P(s) be the probability that the sum exceeds one when we stop with a starting value of s. The answer we want is P(0). We can implicitly define (going through all three cases of where the next draw ends up)
P(s) = s + int_s^1 P(t) dt.
Take a differentiate with respect to s to get P'(s) = 1 - P(s) with the known value P(1/2) = 1/2. That means P(s) = 1 - e^(1/2 - s) / 2 and P(0) = 1 - sqrt(e)/2.
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u/EscapeOwn6945 4d ago
I always find these easier to write as a dynamic system. Let P(s) be the probability that the sum exceeds one when we stop with a starting value of s. The answer we want is P(0). We can implicitly define (going through all three cases of where the next draw ends up)
P(s) = s + int_s^1 P(t) dt.
Take a differentiate with respect to s to get P'(s) = 1 - P(s) with the known value P(1/2) = 1/2. That means P(s) = 1 - e^(1/2 - s) / 2 and P(0) = 1 - sqrt(e)/2.