• #### throw a die 10 times. The longest run is of length 1.8 or so ?

From Hen Hanna@21:1/5 to All on Thu Jan 2 13:08:04 2020
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)?

If I throw a die 100 times, would the ave. LLS be 3 or so ?

Harder to guess for 1000 times (or 1000 dice) -- 5 or so?

Thank you. HH

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• From James Dow Allen@21:1/5 to Hen Hanna on Fri Jan 3 07:50:51 2020
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.

~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

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• From Hen Hanna@21:1/5 to James Dow Allen on Mon Jan 6 08:26:43 2020
On Friday, January 3, 2020 at 7:50:52 AM UTC-8, James Dow Allen wrote:
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.

~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

That's great! Did you have a hunch that it'd be like log_6(n)... ?

Textbooks often use [estimation of Pi] as a demonstration
of MC (Monte Carlo) simulation method, and that's a really lame,

first impressive demonstration that the MC simulation method works.

i hope i'll know soon
1. how to arrive at the [ Log_6(n) + C ] formula, and
2. if Gauss, Newton, ... knew this formula

What other great demonstrations of the MC simulation method do you know of ?

___________________________

I'm sure i saw this in a Martin Gardner book... Does anyone remember ?

There's a 3x4 (?) grid, & each cell contains a Letter. The reader is
asked to find the longest word by going Zigzag... and the answer is
the surprising (unobvious) [Switzerland], which hits all or most of the cells.

Thank you. HH

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• From James Dow Allen@21:1/5 to Hen Hanna on Tue Jan 7 07:56:53 2020
On Monday, January 6, 2020 at 11:26:46 PM UTC+7, Hen Hanna wrote:
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)... ?

The intuition is pretty easy. If your total sequence is K times as long
as some other sequence, you'll K times as many of any particular string.
So, with K=6, if the shorter sequence is just long enough to have about 1
run of, e.g., ten sixes, you'd expect the 6-times-longer sequence to
have about 6 runs of ten sixes. On average, one of those 6 runs will
be a run of eleven sixes.

It's more fun to come up with an exact formula but there are at least
two reasons that's likely to be VERY difficult for your problem.

first impressive demonstration that the MC simulation method works.

What other great demonstrations of the MC simulation method do you know of ?

Weather and climate forecasting use the Monte Carlo method, I think.
Historical linguistics is a topic that interests me; MC is used to
guesstimate the clading diagram for Indo-European!

I've been doing Monte Carlo simulations (many in response to rec.puzzles!)
for so long, I get air-ticket discounts when I fly to Monte Carlo! :-)

I guess that the only such simulation there is for Maverick solitaire (discussed in 2 or 3 threads here at rec.puz).

But I've seen people resort to simulations when straightforward calculation
is easier? Odds at Texas Hold'em? There's only about a few million
possible flop/turn/rivers -- at today's speeds, just iterate through them!!

Here's a puzzle (due to J.J. Sylvester?):
Pick three points at random inside a parallelogram of area 1.
What is the expected area of the triangle formed by the three points?
(First demonstrate, using intuition of affine transforms, that the
shape of the parallelogram is irrelevant.)

Three ways to proceed are:
(1) Monte Carlo simulation.
(2) Submitting a sextuple definite integral to MatLab!
(3) Proceeding via very simple steps.

I posed the problem elsewhere and the only response was the sextuple
integral. But the simple step-by-step is much more satisfying.

James Dow Allen

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