I ONLY GET A MEASLY $200?

by Greg Brume

Solution: UNIQUELY TRANSFORM EACH SET

The logic puzzle hints that it's a game of Monopoly.
While it has the same spaces as the most common US version (as Dave
notes), the *arrangement* of the spaces is quite
different.

After the game, Charon's notes indicate letters to write on the game board. The first part indicates which square write on, and the second indicates the letter of the property that the indicated player lands on in the numbered turn. Namely, "On A1, D3-3" means that on the 1st square Azrael lands on, write the 3rd letter of the 3rd property Dave lands on.

Doing this with a regular board gives the phrase CALL IN ANSWER "PPHFBBTFTBLT." Doing so gets some taunting from the hosts. But doing this with the alternate board configuration gives, reading from Go! clockwise, UNIQUELY TRANSFORM EACH SET.

This a step-by-step progression of the game, with each player noted by an initial and a number indicating the round.

In round 1, A, B, and C rolled six dice altogether
that include all six possible numbers. This totals 21. Since
B_{1}=C_{1}+1 and C_{1}=A_{1}+1,
B_{1} = 8, C_{1} = 7, and A=6. Since A_{1}
used odd numbers only, A_{1} was {1,5}, B_{1} was
{2,6}, and C_{1} was {3,4}.

A_{1}: 6 {1,5} B_{1}: 8 {2,6}
C_{1}: 7 {3,4} D_{1}: 11 {5,6}

In round 2, A_{2} = B_{2}, and was
greater than 7. The only possible die rolls for A_{2} and
B_{2} are {1,2}, {1,6}, {2,5}, {5,6}. Only {5,6} is greater
than 7. A_{2} = B_{2} = 11, putting A_{2}
on square #17, and B_{2} on square #19. If D_{2}
was 6, he would end with A on #17, which is not possible. He rolled
5, and ended on #16. (Note that we can't always definitively pin
down his dice, which may be another reason for the demons to take
out their anger on him.) C_{2} included a 4, and either 2
or 6. If 6, then C_{2} would end with A on #17, which is
impossible, so C_{2} rolled {2,4} and ended on #13.

A_{2}: 17 {5,6} B_{2}: 19 {5,6}
C_{2}: 13 {2,4} D_{2}: 16 {5}

In round 3, C ended on #18, so he rolled {2,3}.
D_{3} = 8, so he ended on #24. A_{3} = 10, so he
rolled {6,4} ending on #27. B_{3} = 7, so he rolled
{5,2}.

A_{3}: 27 {4,6} B_{3}: 26 {5,2}
C_{3}: 18 {2,3} D_{3}: 24 {8}

In round 4, C_{4} = {3,X} and more than 6
spaces. {3,5} and {3,6} both put C on a space someone else had
already been on (A_{3} and B_{3}), so C_{4}
was {3,4}. A_{4}=B_{4}, so, needing a die each from
the previous round, they each rolled {2,4}, ending on 33 and 32
respectively. D_{4} was at least 11, and could not be 12 (a
double), so it was 11.

A_{4}: 33 {2,4} B_{4}: 32 {2,4}
C_{4}: 25 {3,4} D_{4}: 35 {5,6}

In round 5, A_{5} and B_{5} shared a
die in common, and D_{5} was the average of these two
rolls. Since A_{5}={2,X} and B_{5} = {4,X},
D_{5}=3+X. D_{5} was not 5. If X = 5 or 6, B would
have gone past Go, a contradiction of the clue. If X = 1, D could
not have gone past Go. X=3, and D rolled a 6 (either 1,5 or 2,4}.
If C rolled {4,6}, he would have landed on #35, which is impossible
(D_{4}). He rolled {4,5} and D rolled {2,4}.

A_{5}: 38 {2,3} B_{5}: 39 {4,3}
C_{5}: 34 {4,5} D_{5}: 1/41 {2,4}

In round 6, the only four consecutive squares that are reachable are #2-#5. Charon finished on #5, Dave #4, Baalzebub #3, and Azrael #2.

A_{6}: 2/42 {3,1} B_{6}: 3/43 {3,1}
C_{6}: 5/45 {5,6} D_{6}: 4/44 {2,1}

In round 7, A, B, and C must each roll {1,6}. D therefore rolled 10.

A_{7}: 9/49 {1,6} B_{7}: 10/50 {1,6}
C_{7}: 12/52 {1,6} D_{7}: 14/54 {10}

In round 8, D moved 3 times as many spaces as C. This
can only be 9 and 3 (since 2&6 and 4&12 involve doubles).
From the next clue, A and B moved the same number of spaces, and
more than 9. They did not move 10 (which would put A on #19, an
impossibility from B_{2}), so they moved 11.

A_{8}: 20/60 {6,5} B_{8}: 21/61 {6,5}
C_{8}: 15/55 {2,1} D_{8}: 23/63 {9}

In round 9, A and B moved the same number of spaces,
so they will continue to occupy adjacent squares. They each rolled
{5,X}, and X can't be 5 or 6, so they moved between 6 and 9 spaces.
Moving 6 or 7 puts A on a space that has previously been occupied
(B_{3} or A_{3}). Moving 9 puts A on Go To Jail
(which C lands on at the end of the game). They move 8 spaces ahead
each. C moves from 3 to 8 spaces; only 7 puts C on a square that
has not been previously occupied. D moved 8 spaces.

A_{9}: 28/68 {5,3} B_{9}: 29/69 {5,3}
C_{9}: 22/62 {2,5} D_{9}: 31/71 {8}

In round 10, D moves to Go! C moves to Go To Jail. B and A swap positions, so A moves 9 spaces to #37, B moves 7 spaces to #36.

A_{10}: 37/77 {3,6} B_{10}: 36/76
{3,4} C_{10}: 30/70 {5,3} D_{10}: 40 /80 {9}