10 ppl come in , and each gets a random number in [1 -- 10] inclusive.
---> Expected # of distinct Numbers is 6.5

100 ppl ---> 63 Expected distinct Numbers

The part that's surprising to me is how it (63...) scales up linearly...
----------- the most significant digits == 632........ always

Does anyone have a simple explanation for what ths would be the case ?

Is there an analogous (or corresponding) invariant (ratio) in all the rows of Pascal's Triangle ?

10 --- 6.513215599
100 --- 63.396765872677086
1000 --- 632.3045752290362
10000 --- 6321.389535670295
100000 --- 63212.23982317534
1000000 --- 632120.7427789335

100000000000 --- 63212058926.887726
1000000000000 --- 632112420612.6902
10000000000000 --- 6322349313875.826
100000000000000 --- 63182640340411.53
1000000000000000 --- 631826403404113.6

# of (expected) distinct Birthdays for N people
22 --- 21.37
50 --- 46.78
1000 --- 341.51

( With 1000 ppl, (only) 341 out of 365 days would be taken)

---------- " Perhaps it is quite surprising that you should expect there to be so many (24) unclaimed birthdays with so many people ! "

------------ this (24 unclaimed) sounds about right to me.

On Wednesday, 14 September 2022 at 04:56:32 UTC+1, henh...@gmail.com wrote:
> 10 ppl come in , and each gets a random number in [1 -- 10] inclusive.
> ---> Expected # of distinct Numbers is 6.5
>
> 100 ppl ---> 63 Expected distinct Numbers
>
>
> The part that's surprising to me is how it (63...) scales up linearly...
> ----------- the most significant digits == 632........ always
>
> Does anyone have a simple explanation for what ths would be the case ?

(1 - 1/10)^10 = 0.6513215599

On 14/09/2022 04:56, henh...@gmail.com wrote:
>
> 10 ppl come in , and each gets a random number in [1 -- 10] inclusive.
> ---> Expected # of distinct Numbers is 6.5
>
> 100 ppl ---> 63 Expected distinct Numbers
>
>
> The part that's surprising to me is how it (63...) scales up linearly...
> ----------- the most significant digits == 632........ always
>
>
> Does anyone have a simple explanation for what ths would be the case ?
>

[snip]

The probability that a given number is allocated to at least one person
is 1-(1-1/n)^n, and by linearity of expectation this also equals the
expected proportion of the available numbers allocated to at least one
person. L'Hopital's rule can be applied to prove that as n approaches
infinity this proportion approaches 1-1/e = 0.6321205588285577.

Duncan

On Wednesday, September 14, 2022 at 8:57:28 AM UTC-7, duncan smith wrote:
> On 14/09/2022 04:56, henh...@gmail.com wrote:
> >
> > 10 ppl come in , and each gets a random number in [1 -- 10] inclusive.
> > ---> Expected # of distinct Numbers is 6.5
> >
> > 100 ppl ---> 63 Expected distinct Numbers
> >
> >
> > The part that's surprising to me is how it (63...) scales up linearly...
> > ----------- the most significant digits == 632........ always
> >
> >
> > Does anyone have a simple explanation for what ths would be the case ?
> >
> [snip]
>
> The probability that a given number is allocated to at least one person
> is 1-(1-1/n)^n, and by linearity of expectation this also equals the
> expected proportion of the available numbers allocated to at least one
> person. L'Hopital's rule can be applied to prove that as n approaches
> infinity this proportion approaches 1-1/e = 0.6321205588285577.
>
> Duncan

thank you .... i didn't realize that 0.63212... was such a special (or common) number.

henh...@gmail.com <henhanna@gmail.com> wrote:
> 10 ppl come in , and each gets a random number in [1 -- 10]
> inclusive. ---> Expected # of distinct Numbers is 6.5
> 100 ppl ---> 63 Expected distinct Numbers
> The part that's surprising to me is how it (63...) scales up
> linearly... ----------- the most significant digits == 632........
> always Does anyone have a simple explanation for what ths would be
> the case ?
....
> 10 --- 6.513215599
> 100 --- 63.396765872677086
> 1000 --- 632.3045752290362
> 10000 --- 6321.389535670295
....

`duncan smith` has already noted "this proportion approaches 1-1/e =
0.6321205588285577"; here's a slightly different calculation to the
same end: If m integers are uniformly-randomly chosen from among
1...m, the chance of a given number not being chosen in the process is
(1-1/m)^m, independently for all m numbers. Then the expected number
of not-chosen numbers, say NC(m), is m*(1-1/m)^m, and the expected
number of chosen numbers, say CC(m), is m-NC(m). Because
lim{n->oo}((1-1/m)^m) is 1/e (see ref) we have NC(m) ~ m/e and CC(m) ~
m*(1-1/e) ~ m * 0.6321205588... The following python code illustrates
this for several values of m and compares m/NC(m) to e. (Ref
<https://socratic.org/questions/how-do-you-find-the-limit-of-1-1-x-x-as-x-approaches-infinity>)

from math import exp
def nc(m): return m*(1-1/m)**m
def cc(m): return m-nc(m)
def sho(b,kmax=9):
for k in range(1,kmax):
m=b**k; c=cc(m); ne=m/nc(m); de=exp(1)-ne
print(f'{m:13.2f} {c:17.5f} {ne:12.9f} {de:12.9f}')
print()
sho(10)
sho(9)
sho(7.7)

Anyhow, I think of 0.6321205588... as recognizable mostly as the
complement of the more-prominent value, 1/e, ~ 0.36787944117...

> Is there an analogous (or corresponding) invariant (ratio) in all
> the rows of Pascal's Triangle ?
....

Sort of. See eg <https://online.stat.psu.edu/stat414/lesson/28/28.1>
about "Normal Approximation to Binomial" and many other pages about
that approximation. Expected outcomes of experiments having
Binomial(m,0.5) distribution are in direct proportion to numbers in
the mth row of Pascal's Triangle. As m gets large the approximation
improves steadily.