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I need to know if there is a formula or asymptotic approximation for the following coupon collector's problem involving very large numbers of coupons.

  1. Coupons are arranged in a circle
  2. It doesn't matter which coupon is collected first
  3. Coupon exploration is by random walk, one step at a time, either clockwise or counter-clockwise i.e., $\pm1$ step from the current position
  4. You collect a coupon the first time that you visit its location
  5. Exploration stops only after collecting all coupons

I don't get a simple curve that fits Monte Carlo results up to 300 coupons, and Monte Carlo trials are out of the question when it comes to large numbers like $10^{30}$ coupons. An asymptotic lower bound would be OK.

vvg
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user14343099
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  • So, you want to know number of steps in the random walk to collect $N$ coupons? – vvg Jul 22 '23 at 14:34
  • Yes, that is the statistic I am looking for. – user14343099 Jul 22 '23 at 23:23
  • This is the coupon collector problem without replacement. A previous MSE question and answers are here: https://math.stackexchange.com/questions/2172876/coupon-collector-without-replacement – vvg Jul 23 '23 at 02:00
  • You are mistaken. The standard coupon collectors problem is not random walk as I specified in item 3 above. Random walk does not allow you to draw coupons more than one position away. The number of required trials will be higher. Please read the question carefully. Thanks. – user14343099 Jul 23 '23 at 18:00
  • Also note that items 1 and 3 specify that the random walk is circular. This is not the usual coupon collector's problem. – user14343099 Jul 23 '23 at 18:07
  • Suppose you label the urns containing the coupon as $1, 2, \cdots, N$, place them in any order and use a $N$-faced die to generate the urn number to pick the coupon from, how is it different from a circular random walk? – vvg Jul 24 '23 at 01:37
  • Do you mean to say, only the first choice is random and after that you collect $N$ coupons from the sequential positions on the circle one at a time? If that is what you meant, I didn't get that from the question. You may want to edit the question to clarify that. – vvg Jul 24 '23 at 01:43
  • Thank you for your inputs, vvg. The experiment you defined is completely random selection from an urn. That's not what random walk does. In random walk you cannot use the die to select one of N urns. The die would be replaced with a coin. If you get heads you step left one urn and if you get tails you step right one urn. In a strictly random walk you would use a 3-sided die allowing you to remain in place. What makes this different from the experiment you defined is that in your experiment you can move any distance from 0 to N-1. In my question you can move at most 1 step left/right. – user14343099 Jul 24 '23 at 03:22
  • One last question: Suppose $N$ is the total number of coupons and $M$ is the number of distinct coupons (i.e., coupon types). Is $M = N$ or $M \lt N$? If it is the latter, is there a distribution of the coupon types that should be assumed? – vvg Jul 24 '23 at 09:00
  • In this case M=N. – user14343099 Jul 24 '23 at 16:42
  • The best way to visualize this is to think of it as a broken clock with an arbitrary number of seconds on the dial. Each second, the second hand moves clockwise or counter-clockwise one second. The direction is random and it always moves. So the question is, how many seconds would it take to have visited all of the second positions on the dial? – user14343099 Jul 25 '23 at 13:15
  • I do believe the "random walk" has to be interpreted as a dynamics on the power set of ${1,2,\ldots,N}$, starting with the empty set. There are some french authors talking about this interpretation (if that's interesting i can try find the article). It is called cover time and one can have some approximations on that. – Peter Strouvelle Aug 03 '23 at 12:03
  • See also https://math.stackexchange.com/questions/116446. – joriki Feb 10 '24 at 00:51

1 Answers1

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  • It is well known that with a simple random walk starting at $0$, the expected number of steps to reach the first of the positions $-a$ and $+b$ is $ab$.
  • So in your circle, as soon as you have collected the $k$th coupon, the expected number of steps to reach the $k+1$th coupon you will collect is $k$.
  • So the total expected number of steps to collect all $n$ coupons in the circle is $1+2+\cdots +(n-1)= \frac12n(n-1)$.

As a check for very small $n$,

  1. for $n=1$ this gives $0$ corresponding to a single coupon which you collect at the start before moving;
  2. for $n=2$ this gives $1$ corresponding to you starting with the initial coupon and finishing with the other coupon after a single step in either direction;
  3. for $n=3$ this gives $3$ which calculated another way would be $2 \times \frac12+3\times \frac14+4 \times \frac18 +5\times \frac1{16}+\cdots$.
Henry
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