Average Throws Needed for All Dice Points to Appear

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Discussion Overview

The discussion revolves around the average number of throws required for all points (2 through 12) to appear when rolling a pair of 6-sided dice. Participants explore both simulation results and theoretical calculations, focusing on probability derivations and comparisons to related problems.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant shares simulation results indicating that between 55 and 60 throws are needed on average for all points to appear, expressing interest in a probabilistic derivation.
  • Another participant identifies the problem as a variation of the Coupon Collectors Problem, noting the differing probabilities for each outcome when rolling two dice.
  • A different participant proposes a calculation method based on the distribution of outcomes from rolling two dice, suggesting that the average number of throws required is 36, derived from the least likely outcomes (2 and 12).
  • One participant challenges the reasoning of the previous calculation, suggesting that it lacks rigor and invites a simpler analogy involving coin flips to illustrate the point.
  • A later reply acknowledges a mistake in the overall argument but maintains that the distribution presented is correct.

Areas of Agreement / Disagreement

Participants express disagreement regarding the average number of throws required, with competing views on the validity of the proposed calculations and the underlying assumptions.

Contextual Notes

Some assumptions regarding the distribution of outcomes and the application of the Coupon Collectors Problem remain unresolved, and the discussion reflects differing interpretations of probability in this context.

Ed Wiggins
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Dice---How many throws?

How many throws of a pair of 6-sided dice are required (on average) for all points (2 through 12) to appear?

My simulation program says between 55 and 60, but I am more interested in knowing how to derive the answer by calculating probabilities.
 
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This is a variation of the so-called Coupon Collectors Problem (see, e.g. http://rec-puzzles.org/sol.pl/probability/coupon ). Your problem is harder because the probabilities are not all the same for the different "coupons."
 
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Ed Wiggins said:
How many throws of a pair of 6-sided dice are required (on average) for all points (2 through 12) to appear?

My simulation program says between 55 and 60, but I am more interested in knowing how to derive the answer by calculating probabilities.

I do not think that the average is so high.If X is the event representing the number of points which appear when a pair of dice is thrown then we have the distribution (it is the sum of 2 distributions for single die throws) :

i

2 ---> P[2]=1/36

3 ---> P[3]=2/36

4 ---> P[4]=3/36

5 ---> P[5]=4/36

6 ---> P[6]=5/36

7 ---> P[7]=6/36

8 ---> P[8]=5/36

9 ---> P[9]=4/36

10 ---> P[10]=3/36

11 ---> P[11]=2/36

12 ---> P[12]=1/36

If N is the number of throws the mean value for the apparition of different outcomes 'i';i=from 2 to 12 is N*P.Thus the average number of throws required for the appearance,at least once,of all outcomes is given by the outcomes having the least chance of appearance (sum 2 and 12) from the inequation: N/36 ≥ 1,that is the required N is 36.
 
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There's a lot of handwaving in that argument, and it's wrong. Try applying it to a simlpe case: on average, how many flips of a coin before you get both a heads and a tails?
 
My time,not yours,I do not appreciate the side comments.And yes you are right,I'm wrong overall (but not the distribution I have written).
 
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