Cut the Cards

[Albert Chan]

(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