- #1
musicgold
- 304
- 19
Hi,
I am stuck with the following problem and not sure where I am going wrong. I know that I can solve this problem by simulation, but I want to find a more elegant way of solving it.
Problem : Suppose a cereal company is running a promotion on its cereal by including one of six different colored pens in each box. Assume that you have an equal chance of getting any of the six colored pens when you buy a box. How many boxes should one buy to be 50% sure that he will have at least one complete set of the six colored pens?
My approach: The minimum number of boxes one needs to buy to get the full set = 6.
Then I tried to calculate the chance of getting a full set on purchasing only six boxes.
6 boxes
Permutations = 6 x 6 x 6 x 6 x 6 x 6 = 46,656
Chance of getting a full set = 6 P 6 x (1 / 46,656) = 720 x (1 / 46,656) =0.01543
I extended the same logic further, however I am not sure about the results. For example, see the following calculations for 7 and 11 boxes. I am not sure why the chance of getting a full set declines in the case of 11 boxes.
7 boxes
Permutations = 6 x 6 x 6 x 6 x 6 x 6 x 6 = 279,936
Chance of getting a full set = 7 P 6 x (1 / 279936) = 5040 x (1 / 279936) = 0.0180
11 boxes
Permutations = 362,797,056
Chance of getting a full set =11 P 6 x (1 / 362797056) = 332640 x (1 / 362797056) = 0.0009
Thanks,
MG.
I am stuck with the following problem and not sure where I am going wrong. I know that I can solve this problem by simulation, but I want to find a more elegant way of solving it.
Problem : Suppose a cereal company is running a promotion on its cereal by including one of six different colored pens in each box. Assume that you have an equal chance of getting any of the six colored pens when you buy a box. How many boxes should one buy to be 50% sure that he will have at least one complete set of the six colored pens?
My approach: The minimum number of boxes one needs to buy to get the full set = 6.
Then I tried to calculate the chance of getting a full set on purchasing only six boxes.
6 boxes
Permutations = 6 x 6 x 6 x 6 x 6 x 6 = 46,656
Chance of getting a full set = 6 P 6 x (1 / 46,656) = 720 x (1 / 46,656) =0.01543
I extended the same logic further, however I am not sure about the results. For example, see the following calculations for 7 and 11 boxes. I am not sure why the chance of getting a full set declines in the case of 11 boxes.
7 boxes
Permutations = 6 x 6 x 6 x 6 x 6 x 6 x 6 = 279,936
Chance of getting a full set = 7 P 6 x (1 / 279936) = 5040 x (1 / 279936) = 0.0180
11 boxes
Permutations = 362,797,056
Chance of getting a full set =11 P 6 x (1 / 362797056) = 332640 x (1 / 362797056) = 0.0009
Thanks,
MG.
Last edited: