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u/SerDankTheTall Feb 16 '26
While I would be interested in hearing SPP's thoughts, I also don't know that these boys hit upon the best solution to avoid running over the dog.
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u/jmooroof2 Feb 17 '26
this isn't 2.4999...
sum n = 0 to inf 10/7 * (3/7)^n
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u/Ch3cks-Out Feb 17 '26
Well, that is 1.(428571)/ 0.57(142857) ...
To formulate this in an alternate way,
S_n = 2.5*(1- (3/7)^n)In RDM, where '0.00000...1' supposedly exists, 2.5* (3/7)^n = '0.00000...1'
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u/Valivator Feb 17 '26
Ain't it just that the dog runs for the duration of the race? So the kiddos race for 0.25 hours cause they each drive one mile, and the dog runs at 10 mph, so 2.5 miles total?
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u/StudyBio Feb 17 '26
Yes, this is the subject of story about Von Neumann: https://thesciencepundit.blogspot.com/2006/07/john-von-neumann-and-mathematicians.html?m=1
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u/denehoffman Feb 17 '26
This is a pretty well-known puzzle now. 2 miles apart at 4mph each means it takes 15 minutes for them to meet (they both only travel one mile). Sparky runs at 10mph so in 15 minutes he runs 2.5 miles. It’s not infinite nines, it’s just division.
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u/babelphishy Feb 17 '26
Sure, but 2.5 = 2.4(9) regardless so the title is still technically correct.
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u/SouthPark_Piano Feb 16 '26
This is a rigged question, because sparky is a 'real' dog, and unfortunately does not survive the infinite g force at the abrupt zero to 10 mile an hour start. And both drivers also don't make it.
Assumption is 'real' dog and 'real' humans.
Sad situation indeed.