T H _ R _
_ R _
N _ M _ R _ _ S
P _ S S _ B L _
S _ L _ T _ _ N S .
S _ M _
_ F
T H _
P _ S S _ B L _
S _ L _ T _ _ N S
C _ N
B _
F _ _ N D
B _ L _ W .
1 3
>
9
>
1 3
>
1 4
>
1 0
>
6
>
5
>
1
>
2
>
3
>
7
>
1 1
>
1 5
>
1 6
>
1 2
>
8
>
4
1 3
>
9
>
1 3
>
1 4
>
1 0
>
1 1
>
1 5
>
1 6
>
1 2
>
8
>
7
>
6
>
5
>
1
>
2
>
3
>
4
1 3
>
1 4
>
1 3
>
9
>
1 0
>
1 1
>
1 5
>
1 6
>
1 2
>
8
>
7
>
6
>
5
>
1
>
2
>
3
>
4
A
F _ W
_ T H _ R
P _ S S _ B L _
S _ L _ T _ _ N S
_ R _ :
1 3
>
9
>
1 3
>
1 4
>
1 5
>
1 6
>
1 2
>
1 1
>
1 0
>
6
>
5
>
1
>
2
>
3
>
7
>
8
>
4
1 3
>
1 4
>
1 3
>
9
>
5
>
1
>
2
>
6
>
1 0
>
1 1
>
1 5
>
1 6
>
1 2
>
8
>
7
>
3
>
4 Clue
THERE ARE NUMEROUS POSSIBLE SOLUTIONS. SOME OF THE POSSIBLE SOLUTIONS CAN BE FOUND BELOW. 13 > 9 > 13 > 14 > 10 > 6 > 5 > 1 > 2 > 3 > 7 > 11 > 15 > 16 > 12 > 8 > 4 13 > 9 > 13 > 14 > 10 > 11 > 15 > 16 > 12 > 8 > 7 > 6 > 5 > 1 > 2 > 3 > 4 13 > 14 > 13 > 9 > 10 > 11 > 15 > 16 > 12 > 8 > 7 > 6 > 5 > 1 > 2 > 3 > 4 A FEW OTHER POSSIBLE SOLUTIONS ARE: 13 > 9 > 13 > 14 > 15 > 16 > 12 > 11 > 10 > 6 > 5 > 1 > 2 > 3 > 7 > 8 > 4 13 > 14 > 13 > 9 > 5 > 1 > 2 > 6 > 10 > 11 > 15 > 16 > 12 > 8 > 7 > 3 > 4 THE CRIMINAL SHOULD OPEN DOOR NUMBER 2. IF THE 1ST STATEMENT IS TRUE, THEN THE 2ND STATEMENT WILL ALSO BE TRUE. SO WE CAN CONCLUDE THAT THE 2ND STATEMENT IS TRUE AND THE 1ST STATEMENT IS FALSE. FROM THE 2ND STATEMENT, WE KNOW THAT THERE IS A LADY IN ONE OF THE ROOMS AND A TIGER IS IN THE OTHER ROOM. THE LADY HAS TO BE BEHIND DOOR 2 AS THE FIRST STATEMENT IS NOT TRUE TAKE THE FIRST 13 CARDS OFF THE TOP OF THE DECK AND FLIP THEM OVER. THIS IS THE FIRST PILE. THE SECOND PILE IS JUST THE REMAINING 39 CARDS AS THEY STARTED. THE LOGIC CAN BE A BIT DIFFICULT TO DIGEST, SO LET'S LOOK AT AN EXAMPLE. LET'S ASSUME THAT FROM THE FIRST 13 CARDS, 5 OF THEM ARE FACE-UP. SO FROM THE REMAINING 39 CARDS, 8 CARDS WILL BE FACE-UP. WHEN WE FLIP THE CARDS FROM THE FIRST PILE OF 13 CARDS, 8 OF THEM WILL BECOME FACE-UP AND 5 WILL BE FACE-DOWN YOU SPLIT THE COINS INTO A GROUP OF NINETY AND A GROUP OF TEN. YOU THEN FLIP ALL OF THE COINS IN THE GROUP OF TEN. WHEN THE LIGHTS ARE TURNED ON YOU'LL FIND THAT THERE ARE AN EQUAL NUMBER OF HEADS IN BOTH GROUPS. THIS METHOD WILL ALWAYS WORK. IF IT'S DIFFICULT TO COMPREHEND HOW THIS WORKS, LET'S LOOK AT AN EXAMPLE. ASSUME THAT THERE ARE 3 COINS THAT HAVE HEADS IN THE GROUP OF NINETY. SO THE GROUP OF TEN WILL HAVE 7 COINS WITH HEADS AND 3 COINS WITH TAILS. WHEN ALL THE COINS ARE FLIPPED IN THE GROUP OF TEN, THE NUMBER OF COINS WITH HEADS WILL BECOME THREE