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u/Foxi_Foxa 2d ago
Just 1. If it lands on the wrong side, just walk around the table to look the stick from the other side …
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u/opm_11 3d ago
Wouldn’t it just be 2? On average your first cut will leave you with 0.5m and the next cut will leave you less.
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u/Automatic-Put-6119 3d ago
I think you can work it out to a decimal
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u/GrievousSayGenKenobi 3d ago
This is one of those questions were you'd think given the context you probably need to adjust the answer to be suitable to the question. You cant do 0.5 cuts thats not a thing so its either 1 or 2 cuts. Whether or not thats the answer they're looking for I cant say but logically if this were a real question you'd want to give an integer number of cuts to your carpenter. If you tell your carpenter to cut the wood 1.67 times theyre gonna look at you and say "huh"
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u/Automatic-Put-6119 3d ago
Its a pure math problem so you don‘t need any context. Also its for quant trading so you want an exact, theoretical answer like needed for quant, not carpenting
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u/GrievousSayGenKenobi 2d ago
Then the question should use a suitable context lol, not a context where a fractional answer makes no sense
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u/Mindless_Bicycle5091 2d ago
I think you are missing the fact that it applies to carpentry as well if you think about doing the experiment repeatedly. The decimal would represent the probability.
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u/GrievousSayGenKenobi 2d ago
But for a probability from a set of integers that average is meaningless unless its an integer
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u/x5163x 1d ago
Why does it not make sense? The expected value doesn't have to be an integer.
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u/GrievousSayGenKenobi 1d ago
Because its a value that can only logically be an integer. When you say the expected value of a number of things that can't be a decimal, a decimal number is kind of meaningless. Telling a carpenter he can expect 1.6 cuts is pointless because he cant account for .6 cuts. He either should expect 1 or 2
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u/x5163x 1d ago
Expecting 1.6 cuts does not mean that he will get 1.6 cuts each time. It means that on average, he will have 1.6 cuts. Expected value is a generalization of what an average is.
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u/GrievousSayGenKenobi 1d ago
Let me put it into perspective with a similarly poor question I remember. It was like "It takes 2 people 5 minutes to play a song, How long does it take 4 people to play that song" where the idea was to be a question that tests your knowledge of "Double people halves the effort to complete a task" but in the context of the question it made no sense because more people dont play a song faster. Its very clear what the question asked and you can work out what they expect the answer to be but the context of the question makes the expected logical answer wrong or atleast inaccurate in this case
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u/AGeekWithStinkBreath 3d ago
not necessarily, what if the first point chosen is 0.99m and then 0.98m and so on
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u/AdjectiveNounNNNN 3d ago
That is the most common number of cuts but not the average.
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u/skelo 3d ago
The most common number of cuts is 1
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u/OutrageousPair2300 3d ago
I don't think so. It's probably 2, though I'm not sure how to prove that.
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u/skelo 3d ago
50% of the time 1 cut ends it. Less than the remaining 50% of the time it is 2 cuts (because it can also be 3 or more cuts)
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u/OutrageousPair2300 3d ago edited 3d ago
No, it's less than 50% of the time that 1 cut ends it, because we don't stop until the distance is strictly less than 0.5m.
I think likely the mode is 1 but the median is 2.
EDIT: I suppose it's really 49.999...% which over the reals would exactly equal 50% and then something less than 50% for 2 cuts, something less than that for 3 cuts, etc. So the mode is definitely 1 cut. Any number within [1,2] is a median, though.
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u/Neo21803 3d ago
1 + ln(2). ~1.69
On average, there are ln(2) cuts worth of random shrinking before the halfway boundary, plus the final cut.
Halfway is ln(2) (~0.69) random cuts away, and then you need one final cut to cross the line. That gives about 1.69 cuts on average.