Joe placed his pencil on point A of the hexagon below. He flipped a coin 4 times and recorded the sequence of heads and tails. Each time he flipped heads, he moved his pencil along one edge clockwise to the next vertex. Each time he flipped tails, he moved his pencil along one edge counter-clockwise to the next vertex. How much greater is the probability that he will end up with his pencil back on point A than the probability that he will end up with his pencil on point C? Express your answer as a common fraction.
Hexagon
D._______.E
/ \
C. / \.F
\ /
\ _________ /.A
B.
Whoops
I tried showing the hexagon
Well clockwise it’s like this, E,F,A,B,C,D
To get to C he needs H twice in a row
P(H,H) = 1/4
To determine the probability of Joe ending up with his pencil back on point A, we need to calculate the favorable outcomes over the total possible outcomes.
Let's analyze the movement of the pencil for each flip:
- When Joe flips heads, he moves his pencil along one edge clockwise to the next vertex. This rotation can be represented as a clockwise movement of 1 vertex.
- When Joe flips tails, he moves his pencil along one edge counter-clockwise to the next vertex. This rotation can be represented as a counter-clockwise movement of 1 vertex.
For Joe to end up back on point A, he needs to complete a full revolution around the hexagon, which means he needs to encounter a series of 6 movements (3 clockwise and 3 counter-clockwise) that cycle back to the starting point.
Now, let's consider flipping the coin 4 times. We need to count the number of favorable outcomes for ending up on point A, and the number of favorable outcomes for ending up on point C.
For Joe to end up on point A, he can have the following combinations of movements (where C represents a clockwise movement and CC represents a counter-clockwise movement):
1. CCCC (4 clockwise movements): Joe ends up back on point A.
2. CCCCT (3 clockwise movements and 1 counter-clockwise movement): Joe ends up back on point A.
3. CCCTC (2 clockwise movements and 2 counter-clockwise movements): Joe does not end up on point A.
4. CCTCC (2 clockwise movements and 2 counter-clockwise movements): Joe does not end up on point A.
5. CTCCC (2 clockwise movements and 2 counter-clockwise movements): Joe ends up back on point A.
6. TCCCT (1 clockwise movement and 3 counter-clockwise movements): Joe does not end up on point A.
7. TCCTC (1 clockwise movement and 3 counter-clockwise movements): Joe ends up back on point A.
8. TCTCC (1 clockwise movement and 3 counter-clockwise movements): Joe does not end up on point A.
9. TTCCC (0 clockwise movements and 4 counter-clockwise movements): Joe ends up back on point A.
Out of these 9 combinations, 6 result in Joe ending up back on point A.
Now let's consider the combinations that result in Joe ending up on point C:
1. CCCTC (2 clockwise movements and 2 counter-clockwise movements): Joe ends up on point C.
2. CCTCC (2 clockwise movements and 2 counter-clockwise movements): Joe ends up on point C.
3. CTCCC (2 clockwise movements and 2 counter-clockwise movements): Joe does not end up on point C.
4. TCCCT (1 clockwise movement and 3 counter-clockwise movements): Joe ends up on point C.
5. TCCTC (1 clockwise movement and 3 counter-clockwise movements): Joe ends up on point C.
6. TCTCC (1 clockwise movement and 3 counter-clockwise movements): Joe does not end up on point C.
7. TTCCC (0 clockwise movements and 4 counter-clockwise movements): Joe does not end up on point C.
Out of these 7 combinations, 4 result in Joe ending up on point C.
Therefore, the probability of Joe ending up with his pencil on point A is 6/9 = 2/3, and the probability of Joe ending up with his pencil on point C is 4/9.
To find the difference in probability, we subtract the probability of ending up on point C from the probability of ending up on point A:
2/3 - 4/9 = 6/9 - 4/9 = 2/9
Hence, the probability that Joe will end up back on point A is 2/9 greater than the probability that he will end up on point C.