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David Hayden · 07-30-2020, 08:00 PM

PPC Journal V5N4 Page 13 poses a question about a deck of cards. Choose a card at random from a deck of 52 and put the card back. What is the average number of choices required to see all 52 cards?

I simulated the experiment in C++ and got the right answer (236) but I can't figure how to do this analytically. Is it possible?

If you simplify the problem down to a deck of 2 cards then it's easier: you choose the first card, and then it's just a matter of the average number of choices to get the second: 1*1/2 + 2*1/4 + 3*1/8 ..., but can that be extended to more cards?

Dave

Albert Chan · 07-30-2020, 08:58 PM

(07-30-2020 08:00 PM)David Hayden Wrote: Choose a card at random from a deck of 52 and put the card back.
What is the average number of choices required to see all 52 cards?

First card picked must never been seen, thus only 1 try needed to get 1st card.
Probability of getting the next "new" card = 51/52, or averaged 52/51 tries to get 2nd card
...

We can get averaged picks either way:

XCas> sum(1/t, t=1 .. 52) * 52.0 → 235.978285436
XCas> int((t^52-1)/(t-1), t=0 .. 1) * 52.0 → 235.978285436

→ averaged 236 picks to see all 52 cards

Jim Horn · 07-30-2020, 09:49 PM

So, it approaches x*(ln(x)+0.57721), where the latter constant is the Euler-Mascheroni constant. This isn't as accurate as doing it analytically but is good for fast calculations. For instance, if you picked a random second of a year every second, how many years would it take? About 17.84 total.

Albert Chan · 07-30-2020, 10:21 PM

Here is simulations, using Lua
Note: math.random(n) return 1 .. n, each with same probability.

Code:

function run(tests,cards)   -- averaged picks

    local n = 0             -- bad tries

    for i=1,tests do

        for t=1,cards do

            while math.random(cards) < t do n=n+1 end

        end

    end

    return n/tests + cards

end

lua> run(1e6, 52)
235.970842

John Keith · (This post was last modified: 07-31-2020 12:51 AM by John Keith.)

(07-30-2020 09:49 PM)Jim Horn Wrote: So, it approaches x*(ln(x)+0.57721), where the latter constant is the Euler-Mascheroni constant.

Or in other words, n*H(n), where H(n) is the nth harmonic number. (Albert's first formula in post #2)

(07-30-2020 09:49 PM)Jim Horn Wrote: This isn't as accurate as doing it analytically but is good for fast calculations.

Actually it is very accurate as the first image in the Wikipedia link shows: LN(n)+gamma is the asymptotic limit of H(n) as n approaches infinity.

Albert Chan · 08-21-2020, 11:00 PM

(07-30-2020 08:58 PM)Albert Chan Wrote: First card picked must never been seen, thus only 1 try needed to get 1st card.
Probability of getting the next "new" card = 51/52, or averaged 52/51 tries to get 2nd card
...

I might jumped too quickly to say averaged 1/p tries to get the next card.

Let probability of getting the card = p ; of not getting it = q = 1-p

Expected tries
= p + 2qp + 3q²p + 4q³p + ...
= (1-q) (1 + 2q + 3q² + 4q³ + ...)
= (1 + 2q + 3q² + 4q³ + ...) - (q + 2q² + 3q³ + ...)
= 1 + q + q² + q³ + ...
= 1/(1-q)
= 1/p

For n > 0, we have: \(\Large {1 \over (1-x)^n}\normalsize = \displaystyle \sum_{k=0}^∞ \binom{-n}{k}(-x)^k = \sum_{k=0}^∞ \binom{k+n-1}{n-1} x^k \)

Examples:

XCas> series(1/(1-x)^2, x, polynom) → 1+2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5
XCas> series(1/(1-x)^3, x, polynom) → 1+3*x + 6*x^2+10*x^3+15*x^4+21*x^5 // triangular numbers coefficients
XCas> series(1/(1-x)^4, x, polynom) → 1+4*x+10*x^2+20*x^3+35*x^4+56*x^5 // tetrahedral numbers coefficients

Gerson W. Barbosa · (This post was last modified: 08-24-2020 02:07 PM by Gerson W. Barbosa.)

(07-31-2020 12:24 AM)John Keith Wrote:
(07-30-2020 09:49 PM)Jim Horn Wrote: So, it approaches x*(ln(x)+0.57721), where the latter constant is the Euler-Mascheroni constant.

Or in other words, n*H(n), where H(n) is the nth harmonic number. (Albert's first formula in post #2)

The H(n) function can be easily defined on the HP 50g:

« 1. + Psi 1. Psi - »

'H' STO

For instance,

52 ENTER ENTER H × → 235.978285436

pinkman · 08-24-2020, 09:49 PM

Thanks for all those formulas and explanations.

Here is an implementation of the brute force algorithm for the experimental analysis:

Code:

// Forward declaration

CRDS(N);

// Main func:

// N is the number of cards

// L is the number of tests

// Returns average number of peeks

EXPORT CARDS(N,L)

BEGIN

 LOCAL I, S := 0;

 FOR I FROM 1 TO L DO

  S := S + CRDS(N); 

 END;

 RETURN S/L;

END;

// Peek a card until all cards

//   are peeked at least once

EXPORT CRDS(N)

BEGIN

 LOCAL LL, C := 0;

 LL:=MAKELIST(0,A,1,N);

 WHILE ΠLIST(LL)=0 DO

  LL(RANDINT(1,N)):=1;

  C := C+1;

 END; 

 RETURN C;

END;

Usage:
CARDS(number-of-cards, number-of-analysis)

Example:
CARDS(52, 1000)
Returns: 235.951

Albert Chan · (This post was last modified: 08-28-2020 01:23 AM by Albert Chan.)

(07-30-2020 09:49 PM)Jim Horn Wrote: So, it approaches x*(ln(x)+0.57721), where the latter constant is the Euler-Mascheroni constant.

Replacing ln(x) by ln(x+0.5) gives much better estimate for ψ(x+1).
ψ(x+1) = slope of ln(gamma(t)) at t = x+1. When x is large:

\(\psi(x+1) ≈ {\ln\Gamma(x+2)\;-\;\ln\Gamma(x)\over 2}=\ln\sqrt{(x+1)(x)} ≈ \ln(x+0.5)\)

>>> x = 52
>>> H = log(x+0.5) + 0.57722
>>> print format(x*H, 'g')
235.978

see https://en.wikipedia.org/wiki/Digamma_fu...roximation

Gerson W. Barbosa · (This post was last modified: 08-26-2020 12:00 AM by Gerson W. Barbosa.)

FWIW,

« 1. Psi + EXP 6. OVER INV 12. + / + 1. - »

'H-¹' STO

Example:

1100 H → 7.58073560027

H-¹ → 1100

H-¹ should be accurate to 11 or 12 digits for arguments equal or greater than 7.5

Albert Chan · (This post was last modified: 09-01-2020 02:22 PM by Albert Chan.)

(08-25-2020 11:41 PM)Gerson W. Barbosa Wrote: FWIW,

« 1. Psi + EXP 6. OVER INV 12. + / + 1. - »

'H-¹' STO

Example:

1100 H → 7.58073560027

H-¹ → 1100

H-¹ should be accurate to 11 or 12 digits for arguments equal or greater than 7.5

Here is another version, using asymptotic estimate: \(\quad\exp(\psi(x+0.5)) ≈ x + \large{1\over24x} \)

Let k = exp(ψ(x+0.5)), n = x - 0.5 → k = exp(H(n) + ψ(1)) ≈ x + 1/(24x)

If x is big, x ≈ k, we have: \(n ≈ k - \large{1\over 24k} - {1\over2}\)

HP50g inverse harmonic function: « 1. Psi + EXP DUP 24. * INV - .5 - »

Both formulas perform well for large H, but this new version is also good for smaller H

>>> from mpmath import exp, euler, harmonic
>>> f1 = lambda k: k + 6/(1/k+12) - 1
>>> f2 = lambda k: k - 1/(24*k) - .5 # new version
>>> for n in range(50,100,10):
... h = harmonic(n)
... k = exp(h - euler)
... e1, e2 = n - f1(k), n - f2(k) # under-estimated errors
... print '%d\t%.10f\t%.10f\t%g' % (n, e1, e2, abs(e1/e2))
...
50 -0.0000013228 0.0000000364 36.3546
60 -0.0000009261 0.0000000212 43.7606
70 -0.0000006844 0.0000000134 51.1669
80 -0.0000005263 0.0000000090 58.5736
90 -0.0000004172 0.0000000063 65.9804

Gerson W. Barbosa · 08-26-2020, 08:23 AM

(08-26-2020 03:06 AM)Albert Chan 0.247866177136� 54 Wrote: HP50g inverse harmonic function: « 1. Psi + EXP DUP 24. * INV - .5 - »

Both formulas perform well for large H, but this new version is also good for smaller H

For H less than 2.0701076 the first formula is slightly better:

« DUPDUP H H¹ - SWAP DUP H H¹2 - / ABS »

50 → 36.4435261708
10 → 6.75851700917
2.0701076 → 1.
1 → 0.290013346857
0.5131 → 2.30966647121E-5
0.5 → 7.2984301462E-3
0.2 → 0.162258684815
0 → 0.247866177136

Albert Chan · (This post was last modified: 08-31-2020 08:31 PM by Albert Chan.)

I added a tiny correction to compensate errors for smaller H.

HP50g H^-1 : « 1. Psi + EXP DUPDUP SQ 24. * 2.7 + / - .5 - »

Code:

from mpmath import exp, euler, harmonic, psi

f1 = lambda k: k + 6/(1/k+12) - 1

f2 = lambda k: k - 1/(24*k) - .5

f3 = lambda k: k - k/(24*k*k+2.7) - .5    # new formula

def test(lst):

    print 'n\t err(f1)\t err(f2)\t err(f3)'

    for n in lst:

        h = harmonic(n)

        k = exp(h - euler)

        print n,'\t% .10f\t% .10f\t% .10f' % (n-f1(k), n-f2(k), n-f3(k))

>>> test(range(10) + range(10,60,10))

n err(f1) err(f2) err(f3)
0 0.0031607566 0.0127518672 -0.0067666262
1 -0.0003177730 0.0010957186 -0.0001620997
2 -0.0002588484 0.0002719596 -0.0000171164
3 -0.0001712995 0.0001037256 -0.0000035307
4 -0.0001178786 0.0000497922 -0.0000010529
5 -0.0000852046 0.0000275594 -0.0000003957
6 -0.0000641850 0.0000167992 -0.0000001742
7 -0.0000499820 0.0000109785 -0.0000000859
8 -0.0000399758 0.0000075616 -0.0000000463
9 -0.0000326776 0.0000054263 -0.0000000266
10 -0.0000271983 0.0000040243 -0.0000000162
20 -0.0000076840 0.0000005432 -0.0000000006
30 -0.0000035570 0.0000001651 -0.0000000001
40 -0.0000020419 0.0000000705 -0.0000000000
50 -0.0000013228 0.0000000364 -0.0000000000

If needed, we could apply Newton's method to get better estimate.

>>> h = harmonic(10) # h = 2.9289682539682538
>>> x = f3(exp(h-euler)) # x = 10.000000016205583
>>> x - (harmonic(x) - h) / psi(1, x+1)
mpf('10.0')

Gerson W. Barbosa · 08-26-2020, 06:13 PM

D’oh, my latest improvement ended up being equivalent to your f2 program, same number of bytes as yours (56.5).

That’s an old attempt at exact results, but I wasn’t pleased with it. Perhaps I can make it shorter some day:

https://hpmuseum.org/forum/thread-6157-post-55046.html

Gerson W. Barbosa · 08-27-2020, 10:07 PM

How about « 1. Psi + EXP DUP 1. + 4. * INV OVER 1. EXP / + INV OVER 24. * + INV - .5 - » ?

Again, best for smaller H. A previous attempt ( « 1. Psi + EXP DUP SQ 24. * 1. EXP + INV NEG 1. + * .5 - » ) gave the same results as yours when I replaced 1 EXP ( e ) with 2.7.

Albert Chan · (This post was last modified: 08-29-2020 02:51 AM by Albert Chan.)

(08-26-2020 02:13 PM)Albert Chan Wrote: I added a tiny correction to compensate errors for smaller H.

HP50g H^-1 : « 1. Psi + EXP DUPDUP SQ 24. * 2.7 + / - .5 - »

(08-27-2020 10:07 PM)Gerson W. Barbosa Wrote: How about « 1. Psi + EXP DUP 1. + 4. * INV OVER 1. EXP / + INV OVER 24. * + INV - .5 - » ?

Just to be clear, both corrections are not based on actual continued fraction of e^ψ(x+1/2)
My correction (and probably Gerson's version) are based on curve fitting, for small H.
I picked 2.7 instead of e to get better accuracy at bigger H, it is simply a trade-off.

To build e^ψ(x+1/2) continued fraction with XCas:

XCas> k := exp(Psi(x+0.5))
XCas> limit(k/x, x=inf) → 1
XCas> limit(1/(k -x)/x, x=inf) → 24
XCas> limit(1/(1/(k -x) -24x)/x , x=inf) → 10/37
XCas> limit(1/(1/(1/(k -x) -24x) -10/37*x)/x , x=inf) → 0 ???

The last one is likely a XCas bug, Mathematica get the coefficient: 689976/74381

\(\qquad\qquad\exp( \psi(x+1/2)) = x
+\frac{1}{24 \cdot x
+\Large\frac{1}{\frac{10}{37} \cdot x
+\frac{1}{\frac{689976}{74381} \cdot x \;
+\; \cdots}}} \)

To confirm, lets try H^-1(H(20)):

XCas> f(x) := 1/(24*x + 1/(10/37*x + 1/(689976/74381*x)))
XCas> h := float(harmonic(20)) → 3.59773965714
XCas> k := exp(h - euler_gamma) → 20.5020317757
XCas> fsolve(x+f(x) = k, x=k) - 0.5 → 20.0

If f(x) is small, x ≈ k. We can also iterate for x, like this:

XCas> x0 := k
XCas> x0 := k - f(x0) → 20.5000002012
XCas> x0 := k - f(x0) → 20.5
XCas> n := x0 - 0.5 → 20.0

Gerson W. Barbosa · 08-28-2020, 11:39 PM

For the inverse digamma function I’ve been using a continued fraction with increasing number of terms. I am sure about only the first half of them, though. It’s difficult to be sure about the other half when one is limited to only 12 digits.

\(\rm{e}^{{x}}+\frac{1}{2+\frac{1}{6\rm{e}^{{x}}-\frac{1}{2+\frac{1}{\rm{e}^{{x}-1}-\frac{1}{2+\frac{1}{\rm{e}^{{x}}+\frac{1}{2}}}}}}}\)

Albert Chan · 08-29-2020, 04:02 PM

(08-27-2020 10:07 PM)Gerson W. Barbosa Wrote: How about « 1. Psi + EXP DUP 1. + 4. * INV OVER 1. EXP / + INV OVER 24. * + INV - .5 - » ?

FYI, above H^-1 is equvialent to Ψ^-1 (previous post)
Since H(n) = Ψ(n+1) - Ψ(1), we flip the sign of -0.5 to +0.5

XCas> invPsi1(k) := k - 1/(24k + 1/(k/e + 1/(4k+4))) + 1/2

\(k \rightarrow k-\frac{1}{24\cdot k+
\large\frac{1}{\frac{k}{e}+\frac{1}{4\cdot k+4}}}+\frac{1}{2}\)

XCas> invPsi2(k) := k + 1/(2 + 1/(6k - 1/(2 + 1/(k/e - 1/(2 + 1/(k+1/2))))))

\(k \rightarrow k+\frac{1}{2+\frac{1}{6\cdot k-\frac{1}{2+\frac{1}{\frac{k}{e}-\frac{1}{2+\frac{1}{k+\frac{1}{2}}}}}}}\)

XCas> simplify(invPsi1(k) - invPsi2(k)) → 0

Let's try a few x, to see if it round-trip.

XCas> invPsi(x) := invPsi1(exp^float(x)) // Note: k = e^x
XCas> map(range(1,6), x->[x, Psi(invPsi(x))])

\(\qquad\left(\begin{array}{cc}
1 & 0.999999666188 \\
2 & 2.00000000607 \\
3 & 3.00000000018 \\
4 & 4.0 \\
5 & 5.0
\end{array}\right)\)

Albert Chan · 06-23-2021, 12:08 AM

An integral, from a very old (2006) thread: So your HP can INTEGRATE ...
I revived the old thread recently, for an integral that HP71B failed, see HP71B IBOUND fooled

Code:

            / Inf

            |     FRAC(x)

     I7 =   |     -------  .dx

            |       x²

           / 1

I put the solution here because integral is closely related to harmonic number

I7 = ∫((x-1)/x^2, x=1..2) + ∫((x-2)/x^2, x=2..3) + ∫((x-3)/x^2, x=3..4) + ...
= (ln(2) - ln(1) - 1/2) + (ln(3) - ln(2) - 1/3) + (ln(4) - ln(3) - 1/4) + ...

\(I_7 - 1 = \displaystyle \lim_{n \to ∞} (\,\ln(n) - H_n\,) = - \gamma\)

\(I_7 = 1 - \gamma ≈ 0.422784\)