I throw a die 10 times. The longest run (consecutive 1's or 2's or ...) is on average of length 2 or so? What is the exact estimate or expected value for the LLS (Length of the Longest Streak)?
Reminder to mathematically inclined rec.puzzlers to check IBM's monthly puzzle:
https://www.research.ibm.com/haifa/ponderthis/challenges/January2020.html This month's is about average difficulty ... which means: NOT too easy.
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On Friday, January 3, 2020 at 4:08:06 AM UTC+7, Hen Hanna wrote:
I throw a die 10 times. The longest run (consecutive 1's or 2's or ...) is on average of length 2 or so? What is the exact estimate or expected value for the LLS (Length of the Longest Streak)?
I don't think there's any easy exact formula but simulation finds that
a simple formula gives a fairly good approximation:
log_6(n) + 0.715
is approximately the expected value of the longest run when tossing a
fair 6-sided die n times.
Here are the results of a simulation.
(n was always 10 times a power of 6.)
10 --> 2.02
60 --> 3.00
360 --> 4.00
2160 --> 5.00
12960 --> 6.01
James Dow Allen
On Friday, January 3, 2020 at 7:50:52 AM UTC-8, James Dow Allen wrote:
On Friday, January 3, 2020 at 4:08:06 AM UTC+7, Hen Hanna wrote:
I throw a die 10 times. The longest run (consecutive 1's or 2's or ...) is on average of length 2 or so? What is the exact estimate or expected value for the LLS (Length of the Longest Streak)?
I don't think there's any easy exact formula but simulation finds that
a simple formula gives a fairly good approximation:
log_6(n) + 0.715
That's great! Did you have a hunch that it'd be like log_6(n)... ?
So i've been wondering about this for 30+ years, and this was the
first impressive demonstration that the MC simulation method works.
What other great demonstrations of the MC simulation method do you know of ?
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