Solution to Previous Puzzle—Odds and Ends
2 is interesting in many ways.
In the introduction to the puzzle last time, I
asked if anyone could suggest why 2 would
be interesting or odd as a prime number.
One common answer received is that 2 is
odd among primes because it is not odd; it
is the only even prime. A less mathematical
answer is that 2 the only prime not spelled
with an “e.” Finally, there were answers that
related to the fact that 2 is the only prime
that when added to another prime might
give a prime producing two primes whose
difference is 2; they are called twin primes.
Problem 1 can be solved using twin primes.
Problem 1: N is an age and the product
of two primes. N+ 5 is also the product of
two primes. The sum of the two primes
in the second product is a prime. Find N.
Answer: N = 77
Solution: Let the second product be
ab. Since a+b is prime, one of a or b must
be 2. Let a= 2. Thus, b and b+ 2 are twin
primes. The pairs of twin primes (p, p+ 2)
in which 2p is in a reasonable range of ages
are: ( 3, 5), ( 5, 7), ( 11, 13), ( 17, 19), ( 29, 31),
( 41, 43), and ( 59, 61). The list of numbers
2p– 5 from these pairs is: 1, 5, 17, 29, 53,
77, 113. Of these only 77 is the product of
Problem 2: Three actuaries are sitting at
one end of a bar. Three accountants are
sitting at the other end. The bar has seven
bar stools, leaving one empty between the
two groups. Customers are only allowed to
switch stools using the following two moves.
■ ■ Move 1) A person may move to an adjacent empty bar stool.
■ ■ Move 2) A person may move to an empty
bar stool that is adjacent to someone
adjacent to them.
Can the two groups switch places by
using a sequence of allowed moves? If
so, what is the minimum number of
allowed moves to do it? (On each move,
the empty stool moves one or two spaces
to the right or the left denoted by 1, 2, – 1,
– 2 respectively.) Answer: Yes, 15 moves.
Solution: 1, - 2, - 1, 2, 2, 1, - 2, - 2, - 2, 1,
2, 2, - 1, - 2, 1. Thanks to John Snyder for
proving that this sequence and its negative
are the only two solutions. The sequences
are found by never letting the actuaries
and accountants backtrack.
Problem 3: You and four friends are sitting
around a table at a bar. Everyone volunteers
to pay the bill. You propose the following
game to decide who will pay. You will flip a
coin. If it comes up heads, you will pass the
coin to the right, tails to the left. Whoever
gets the coin will be excluded from paying,
will proceed to toss the coin again, and pass
the coin to the left or right as before. The
process continues. When four people have
been excluded, the remaining tippler will
get to pay. What is each person’s probability
of paying? Why? (You are person 1, on your
right is person 2, etc. up to person 5 who is
on your left.)
Answer: (0.25, 0.125, 0.25, 0.25, 0.125)
where P(person 1 pays) = 0.25, P(person
2 pays) = 0.125, etc.
Solution: Some thought that once
people were excluded from payment
they would get up from the table. That
is understandable because it would be a
practical way to limit the number of coin
tosses to four, and actuaries are nothing if
not practical. But this is a math problem,
not an actuarial problem, and the statement of the problem said nothing about
people getting up from the table or doing
anything else but continuing the process.
There were a variety of correct solutions.
Some involved evaluating geometric se-
ries. One began, “The essential principle
in solving this is a well-known fact about
random walks with absorbing barriers.”
Two solutions were based on very pretty
simulations. The following seems to be a
clear and quick solution that combines the
ideas of a few solvers. Let Qi(j|k) be the
probability that person j will be exclud-
ed before person i, given that the coin is
currently held by person k. So P(person 1
pays) =0.5Q1( 5| 2) + 0.5Q1( 2| 5) says that
at the beginning there is an even chance
that the coin will go right or left; and if
it goes right to 2, and then proceeds to 5
before hitting 1, person 1 will pay; and
similarly if it goes left to 5.
We now show that once the coin goes
to person 2, the probabilities that persons 1, 3, 4, and 5 will pay are equal. Let
X= Q1( 5| 2) which = 0.5Q1( 5| 3) ➔■2X =
Q1( 5| 3) = 0.5Q1( 5| 2) + 0.5Q1( 5| 4)➔3X
= Q1( 5| 4) = 0.5Q1( 5| 3) + 0.5× 1 ➔4X = 1
or Q1( 5| 2) = 0.25. By symmetry Q3( 4| 2)
= 0.25 and Q4( 3 and 5| 2) = Q5( 1 and 4| 2).
And since all four must sum to 1, the last
2 must both = 0.25 also.
By symmetry, we get the same results
on the other side for Q1( 2| 5) and related
probabilities. Putting all this together, we
have the following table for the probabilities
of each person paying.
Jason Ash, Robert Bartholomew, Bob
Byrne, William Carroll, Samantha
Casanova, Bob Conger, Andrew Dean,
Bernie Erickson, Bill Feldman, Yan
Fridman, Steve Gallancy, Edgar Gonal,
Rui Guo, Robert Johnson, Clive Keatinge,
David Lovit, Jerry Miccolis, Donald
Onnen, John Pauly, David Promislow,
Anthony Rubiano, Noam Segal, Tomasz
Serbinowski, John Snyder, Mark Spong,
Al Spooner, Ronald Stokes, Yevgeny
Person 1 2 3 4 5 Total
1stflipH 0.25 0 0.25 0.25 0.25 1
1stflipT 0.25 0.25 0.25 0.25 0 1
Sum/2 0.25 0.125 0.25 0.25 0.125 1
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list, your solutions must be
received by Nov. 30, 2017.