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 B₁=C₁+1 and C₁=A₁+1, B₁ = 8, C₁ = 7, and A=6. Since A₁ used odd numbers only, A₁ was {1,5}, B₁ was {2,6}, and C₁ was {3,4}.

A₁: 6 {1,5} B₁: 8 {2,6} C₁: 7 {3,4} D₁: 11 {5,6}

In round 2, A₂ = B₂, and was greater than 7. The only possible die rolls for A₂ and B₂ are {1,2}, {1,6}, {2,5}, {5,6}. Only {5,6} is greater than 7. A₂ = B₂ = 11, putting A₂ on square #17, and B₂ on square #19. If D₂ 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.) C₂ included a 4, and either 2 or 6. If 6, then C₂ would end with A on #17, which is impossible, so C₂ rolled {2,4} and ended on #13.

A₂: 17 {5,6} B₂: 19 {5,6} C₂: 13 {2,4} D₂: 16 {5}

In round 3, C ended on #18, so he rolled {2,3}. D₃ = 8, so he ended on #24. A₃ = 10, so he rolled {6,4} ending on #27. B₃ = 7, so he rolled {5,2}.

A₃: 27 {4,6} B₃: 26 {5,2} C₃: 18 {2,3} D₃: 24 {8}

In round 4, C₄ = {3,X} and more than 6 spaces. {3,5} and {3,6} both put C on a space someone else had already been on (A₃ and B₃), so C₄ was {3,4}. A₄=B₄, so, needing a die each from the previous round, they each rolled {2,4}, ending on 33 and 32 respectively. D₄ was at least 11, and could not be 12 (a double), so it was 11.

A₄: 33 {2,4} B₄: 32 {2,4} C₄: 25 {3,4} D₄: 35 {5,6}

In round 5, A₅ and B₅ shared a die in common, and D₅ was the average of these two rolls. Since A₅={2,X} and B₅ = {4,X}, D₅=3+X. D₅ 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 (D₄). He rolled {4,5} and D rolled {2,4}.

A₅: 38 {2,3} B₅: 39 {4,3} C₅: 34 {4,5} D₅: 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.

A₆: 2/42 {3,1} B₆: 3/43 {3,1} C₆: 5/45 {5,6} D₆: 4/44 {2,1}

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

A₇: 9/49 {1,6} B₇: 10/50 {1,6} C₇: 12/52 {1,6} D₇: 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 B₂), so they moved 11.

A₈: 20/60 {6,5} B₈: 21/61 {6,5} C₈: 15/55 {2,1} D₈: 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 (B₃ or A₃). 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.

A₉: 28/68 {5,3} B₉: 29/69 {5,3} C₉: 22/62 {2,5} D₉: 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.

A₁₀: 37/77 {3,6} B₁₀: 36/76 {3,4} C₁₀: 30/70 {5,3} D₁₀: 40 /80 {9}