Average Throws Needed for All Dice Points to Appear

[Ed Wiggins]

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.

 

[Tide]

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."

 

[metacristi]

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.

 

[Hurkyl]

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?

 

[metacristi]

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).