If it is possible, then define a simply connected set in 3-dimensional Euclidean space that has volume 1 and infinite surface area. Alternately, if it is impossible then prove that such a set is impossible.
In this puzzle, a die with k sides, numbered from 1 to k, has an equal probability (1/k) of landing with any one of these numbers up whenever it is thrown. The questions were: On average, how many times do you have to throw the die to generate n sequential outcomes of the number 1? Which takes more throws on average, k and n or k-1 and n+1 ?
Following is an outline for a solution that was verified independently by Bob Conger.
Let AWT(k,n) be the average “waiting” time it takes to roll n sequential 1s given k.
AWT(k,1)=p(1+2q+3q2+…)=p/(1-q)2,
where p = 1/k and q = 1 – p and hence AWT(k,1)=k.
Immediately prior to rolling n+1 consecutive 1s first n consecutive 1s must be rolled. However, there could be any number of failed attempts to roll the n+1 consecutive 1. In each of these cases, after failure is like starting over. Therefore:
AWT(k,n+1)=(AWT(k,n)+1) p(1+2q+3q2+…)=(AWT(k,n)+1) k.
The solution to this recursion is:
AWT(k,n)=k+k2+…+kn=(k–kn+1)/(1-k).
Obviously, for the second question to be meaningful, k≥2.
k=2
AWT(2,n)=2n+1-2
AWT(1,n+1)=n+1
-
- n=1, AWT(2,1)=2= AWT(1,2)
- n≥2, , AWT(2,n)>2n> AWT(1,n+1)=n+1
k≥3
Denote the difference between AWT(k,n) and AWT(k-1,n+1) to be D(k,n).
D(k,n)=AWT(k,n)-AWT(k-1,n+1)
=((k–kn+1)/(1-k))-(((k-1)-(k-1)n+2)/(1-(k-1))).
After putting the fractions above under a common positive denominator, we get the numerator. Let it be N(k,n):
N(k,n)=(2-k)(k–kn+1)+(1-k)2-(k-1)n+3
=-2kn+1+kn+2+1-(k-1)n+3
=(k-2) kn+1+1-(k-1)n+3.
Whether AWT(k,n) is greater than, equal to, or less than, respectively, AWT(k-1,n+1) is dependent on whether the numerator above is greater than, equal to, or less than 0, respectively.
For n=1, note that AWT(k-1,2)=(k-1)+(k-1)2=k(k-1)>k=AWT(k,1).
As n becomes larger, eventually (k-2) kn+1 becomes arbitrarily larger than (k-1)n+3.
Consequently, although D(k,1) < 0, eventually, for large enough n, N(k,n)>0.
Solutions were also submitted by Eamonn Long, Jerry Miccolis, John Noble, Anthony Salis and Chris Terrill.
