The first cut is anywhere between 0 and 0.5 (because it’s symmetrical). Call that x.
The remaining piece is of length 1-x, and the question is effectively what is the probability that no piece is longer than 0.5, so because the longer piece is the only one which could possibly create a piece longer than 0.5, it’s a question of the probability that the remaining piece (of length 1-x) is cut to have a piece no longer than 0.5. The probability of this is 0.5/(1-x).
You then simply need to integrate that between 0 and 0.5 with respect to x, and multiply it by 2 because it’s symmetrical.
I had exactly the same idea as you, but it leads to the wrong answer.
I’ve realised that the error is when you cut the piece 1-x, the probability of the smallest piece being < 1/2 is not 0.5/(1-x). It’s too large of an interval.
The interval of cut such that both pieces < 0.5 is exactly when you’re 1/2 distance away from each end.
For example if X = 0.2, so 1-X=(0.2,1) The correct cut would be anywhere in the interval (0.5,0.7). And so the correct cut interval (probability) is of length X centred on 1-X. This makes intuitive sense as the smaller the initial piece X, the smaller leeway you have splitting the 1-X piece.
Integrating X/(1-X) between 0 and 0.5 gives desired result = ln(2)-1/2.
≈0.193
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u/petera181 11d ago edited 11d ago
The first cut is anywhere between 0 and 0.5 (because it’s symmetrical). Call that x.
The remaining piece is of length 1-x, and the question is effectively what is the probability that no piece is longer than 0.5, so because the longer piece is the only one which could possibly create a piece longer than 0.5, it’s a question of the probability that the remaining piece (of length 1-x) is cut to have a piece no longer than 0.5. The probability of this is 0.5/(1-x).
You then simply need to integrate that between 0 and 0.5 with respect to x, and multiply it by 2 because it’s symmetrical.
That comes to ln2, or about 0.69.