One by one straws are placed on a mathematical camel’s back.

The straws have random weight, being drawn independently from the uniform distribution 0 to 1.

The camel’s back breaks when the total weight of the straws exceeds 1.

What’s the expected weight of the straw that breaks the camel’s back?

Source: abridged from "Mind your decisions" blog.

(In reply to

Analytic for 2 straws; spreadsheet by Larry)

Nice analysis, Larry!

By the way, you wrote :"I find it curious that the higher the straw number, the lower the weight of the final straw."

This makes sense to me.

If the penultimate total weight is x, then the straw that breaks the camel's back will be between (1-x) and 1, with an expected value of 1- (x/2). So, the higher the penultimate total weight, the lower the expected weight of the last straw.

Ans since the expected total weight goes up with each straw, one would expect that the second straw, if it was the last, would have a higher expected value than the third straw, if it was last.

In fact, you can work backwards.

When the 2nd straw is the ultimate one, since it has an average value of 2/3, then the average weight before that straw is drawn is 2*(1-2/3) = 2/3.

When the 3rd straw is the ultimate one, since it has an average value of .625, then the average weight before that straw is drawn is 2*(1-.625) = 3/4.

When the 4th straw is the ultimate one, since it has an average value of .6, then the average weight before that straw is drawn is 2*(1-.6) = 4/5.

Hmm... That looks like a pattern. If the 5th straw was the breaking one, did it by any chance have an average value of 1-(5/6)/2 = 1-(5/12) = 7/12 = .583...

That would make the first 4 average weights 4/6, 5/8, 6/10 and 7/12.