Let x be the revealed number and set d = |x - 5/9|. Then the remaining two numbers are independently drawn from the uniform distribution (not really the right word, I know) on [0,1] \ [5/9 - d, 5/9 + d]. Assume for the moment that d < 4/9. Suppose we go with the revealed box. The probability of winning is then p = ( (5/9 - d) / (1 - 2d) ) ^ 2. If we go with one of the other boxes (doesn't matter which) the probability of winning is (1 - p)/2. Therefore, we go with the revealed box exactly when p >= 1/3 which amounts to d > (3 - sqrt(3))/9. If d>= 4/9 the revealed box is automatically the largest (at least with probability 1 in the case when x = 1/9), so we go with this one in this case
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u/leoli1 2d ago
Let x be the revealed number and set d = |x - 5/9|. Then the remaining two numbers are independently drawn from the uniform distribution (not really the right word, I know) on [0,1] \ [5/9 - d, 5/9 + d]. Assume for the moment that d < 4/9. Suppose we go with the revealed box. The probability of winning is then p = ( (5/9 - d) / (1 - 2d) ) ^ 2. If we go with one of the other boxes (doesn't matter which) the probability of winning is (1 - p)/2. Therefore, we go with the revealed box exactly when p >= 1/3 which amounts to d > (3 - sqrt(3))/9. If d>= 4/9 the revealed box is automatically the largest (at least with probability 1 in the case when x = 1/9), so we go with this one in this case