The Christmas cookie dice game
Solution
Answer 7: The elves ate on average 44.6 cookies.
First of all, it doesn’t matter how many elves play the game because only
the total number of cookies was asked, not who eats them.
The minimal number of eaten cookies required to end the game is obviously six (answer 1); this happens, e.g., when the numbers 1, 2, 3, 4, 5, 6 are
rolled. The maximal number of eaten cookies required to end the game is
unlimited (answer 10). This happens, e.g., when only ones are rolled. The
average number is somewhere in between these extremes. It doesn’t have to
be an integer. (The average number to be rolled is 3.5 which is also not an
integer).
Let’s assume for the moment the game ends after m die rolls. How many
cookies have been eaten? At least six dice rolls are required to fill every plate
once. At the end of the game, the six plates are emptied at once. Should a
plate be emptied during the game, it has to be filled using a further dice roll.
Thus, (m − 6)/2 + 6 cookies are consumed in total.
It remains to compute how many dice rolls are required on average to end
the game. Let’s assume that at some point in the game, five plates are filled.
If the number of the empty plate shows up in the next dice roll, the game
ends after this dice roll. This happens with probability 1/6. If, on the other
1
hand, the number of a filled plate shows up, the number of filled plates drops
to four. This happens with probability 5/6. Here it doesn’t matter which five
of the six plates were filled.
Let m5 denote the average number of remaining turns once there are five
filled plates. Then we have
m5 = 1 +
5
1
· 0 + · m4 ,
6
6
where m4 denotes the average number of remaining turns once there are four
filled plates.
What happens with four filled plates? If the number of an empty plate shows
up in the next dice roll, the number of filled plates rises to five. This happens
with probability 2/6. If, on the other hand, the number of a filled plate shows
up, the number of filled plates drops to three. This happens with probability
4/6. It doesn’t matter which four of the six plates were filled. As before, we
have
2
4
m4 = 1 + · m5 + · m3 .
6
6
and by analogous considerations
3
3
· m4 + · m2 ,
6
6
4
2
m2 = 1 + · m3 + · m1 ,
6
6
1
5
m1 = 1 + · m2 + · m0 .
6
6
m3 = 1 +
Here m2 , m1 , and m0 denote the average number of remaining turns once
there are two, one, or no filled plates, respectively. The sought for number is
thus m0 .
Finally, we assume that at some point in the game all plates are empty, which
is the case at the beginning of the game. After one dice roll there will be one
filled plate, no matter what number shows up. Hence
6
m0 = 1 + m1 .
6
2
Summarizing, we have the following linear system of equations (multiplying
all equations by six)
6m5 − 5m4 + 0m3 + 0m2 + 0m1 + 0m0
−2m5 + 6m4 − 4m3 + 0m2 + 0m1 + 0m0
0m5 − 3m4 + 6m3 − 3m2 + 0m1 + 0m0
0m5 + 0m4 − 4m3 + 6m2 − 2m1 + 0m0
0m5 + 0m4 + 0m3 − 5m2 + 6m1 − 1m0
0m5 + 0m4 + 0m3 + 0m2 − 6m1 + 6m0
= 6,
= 6,
= 6,
= 6,
= 6,
= 6.
Its solution (obtained e.g. by Gaussian elimination) is
[m5 , m4 , m3 , m2 , m1 , m0 ] =
1
[630, 744, 786, 808, 822, 832].
10
Thus, the game ends on average after m0 = 83, 2 dice rolls. (This is answer
9.) Hence the number of consumed cookies is
(m0 − 6)/2 + 6 = (83, 2 − 6)/2 + 6 = 44, 6.
Therefore, answer 7 is correct.
About the remaining wrong answers:
2. 2π,
3. twice the minimal number,
4. 10e,
5. 10π,
6. the Answer to the Ultimate Question of Life, The Universe, and Everything,
8. the most random whole number below 100,
9. the average number of dice rolls.
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