I have found a counter intuitive puzzle. I have read the answer given at the source and understand it completely. But, what I am unable to understand is why my intuition turned out to be wrong. Following is the question:
Alice has a dozen cartons, arranged in a 3x4 grid, which for convenience we have labeled A through L:
A B C D E F G H I J K L
She has randomly chosen two of the cartons and hidden an Easter egg inside each of them, leaving the remaining ten cartons empty. She gives the dozen cartons to Bob, who opens them in the (row-wise) order A, B, C, D, E, F, G, H, I, J, K, L until he finds one of the Easter eggs, whereupon he stops. The number of cartons that he opens is his score. Alice then reseals the cartons, keeping the eggs where they are, and presents the cartons to Chris, who opens the cartons in the (column-wise) order A, E, I, B, F, J, C, G, K, D, H, L, again stopping as soon as one of the Easter eggs is found, and scoring the number of opened cartons. Whoever scores lower wins the game; if they score the same then it's a tie.
For example, suppose Alice hides the Easter eggs in cartons H and K. Then Bob will stop after reaching the egg in carton H and will score 8, while Chris will stop after reaching the egg in carton K and will score 9. So Bob wins in this case.
Who is more likely to win this game, Bob or Chris? Or are they equally likely to win?
Following is a rephrased version of the official solution (rephrased to make it easier to understand):
Label the cartons "A" and "L" with "xx" since both players reach those cartons simultaneously.
As far as labelling all the other cartons are concerned, label a carton with a "b", if Bob will reach it before Chris would. Similarly, label a carton with a "c", if Chris will reach it before Bob would. Also record Bob's and Chris' score upon reaching that carton.
For instance, Bob will reach the carton "B" before Chris would and Bob's score would be 2, therefore, carton B will be labelled as b2.
Therefore, the labelling of the 12 cartons would be as follows:
xx b2 b3 b4
c2 c5 b7 b8
c3 c6 c9 xx
Now, note that there are five b cartons and five c cartons. So the cases in which Alice selects carton A or carton L are equally split between Bob and Chris. Similarly if Alice selects two b cartons then Bob necessarily wins, but these are balanced out by an equal number of cases in which Alice selects two c cartons and C necessarily wins.
The crucial cases occur when Alice selects one b carton and one c carton. Bob wins if the b carton has a lower score than the c carton:
b2 and (c3 or c5 or c6 or c9)
b3 and (c5 or c6 or c9)
b4 and (c5 or c6 or c9)
b7 and c9
b8 and c9
Chris wins if the c carton has a lower score than the b carton:
c2 and (b3 or b4 or b7 or b8)
c3 and (b4 or b7 or b8)
c5 and (b7 or b8)
c6 and (b7 or b8)
Since this yields 12 cases in Bob's favor and only 11 cases in Chris's favor, Bob has the advantage.
Now, my first reaction after reading the question was that both of them will have an equal chance of winning. But, it turned out that I was wrong.
While I understand the answer, I am looking for a simpler and intuitive way to understand why it wasn't necessary that both of them would have an equal chance. And why it was necessary to look at all the possible combinations, to determine the solution.