Coupon collector's problem

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In probability theory, the problem of coupon collectors describes the contest "collect all coupons and win". This asks the following question: Suppose there are urns of different coupons, from which the coupons are collected, the possibility is the same, with replacement. What is the probability that more than t a sample experiment is needed to collect all the coupons? An alternative statement is: Given a coupon, how many coupons would you expect you to draw with a replacement before withdrawing each coupon at least once? The mathematical analysis of the problem reveals that the number of expected experiments required grows as $Â\; Â{\displaystyle Â\; Â\; Â\; Â\; Â\; Â\; Â\mathrm{?}Â\; Â\; Â\; Â\; Â\; Â\; Â(Â\; Â\; Â\; Â\; Â\; Â\; Â\; Â\; ÂnÂ\; Â\; Â\; Â\; Â\; Â\; Â\; Â\; Â\; Â\; Â}$ log                (          n        )        )               {\ displaystyle \ Theta (n \ log (n))}   . For example, when n Ã, = Ã, 50 it takes about 225 experiments on average to collect all 50 coupons.

Video Coupon collector's problem

Solution

Calculate expectations

Di sini H _n adalah nomor harmonik n . Menggunakan asimtotik dari bilangan harmonik, kita memperoleh:

$            {\displaystyle         \mathrm{E}                (        T        )        =        n        ?        H_{            n          }        =        n        }$ log                   n                  ?          n                                           1              2                                       O          (          1                    /                   n         )         ,                  {\ displaystyle \ operatorname {E} (T) = n \ cdot H_ {n} = n \ log n \ gamma n {\ frac {1} {2}} O (1/n),}   

di mana ${\ displaystyle \ gamma \ approx 0.5772156649}  ÂÂ$ adalah konstanta Euler-Mascheroni.

Sekarang seseorang dapat menggunakan ketidaksetaraan Markov untuk mengikat probabilitas yang diinginkan:

${\ displaystyle \ operatorname {P} (T \ geq cnH_ {n}) \ leq {\ frac {1} {c}}.}  ÂÂ$

Menghitung varians

sejak ${\ displaystyle {\ frac {\ pi ^ {2}} {6}} = {\ frac {1} {1 ^ {2}}} {\ frac { 1} {2 ^ {2}}} \ cdots {\ frac {1} {n ^ {2}}} \ cdots}  ÂÂ$ (lihat masalah Basel).

Sekarang seseorang dapat menggunakan ketidaksetaraan Chebyshev untuk mengikat probabilitas yang diinginkan:

${\ displaystyle \ operatorname {P} \ left (| T-nH_ {n} | \ geq cn \ right) \ leq {\ frac {\ pi ^ {2}} {6c ^ {2}}}.}  ÂÂ$

Taksiran ekor

Batas atas yang berbeda dapat berasal dari pengamatan berikut. Biarkan ${\ displaystyle {Z} _ {i} ^ {r}}  ÂÂ$ menunjukkan peristiwa bahwa ${\ displaystyle i}   Kupon$ -tidak dipilih di ${\ displaystyle r}  ÂÂ$ uji coba. Kemudian:

${\ displaystyle {\ begin {aligned} P \ left [{Z} _ {i} ^ {r} \ right] = \ kiri (1 - {\ frac {1 } {n}} \ right) ^ {r} \ leq e ^ {- r/n} \ end {aligned}}}  ÂÂ$

Source of the article : Wikipedia