This was a very counterintuitive puzzle, and a relatively difficult one, though one does not need to know any specific mathematical construct for it. Only two people (father – daughter pair) got the correct answer – Mahi and Suman Saraf! Very well done!
Bob will win more.
I am reproducing two different answers, one from the source, and the other one from Mahi. I like both the answers, both very intuitive.
Here is the one from Mahi.
Here is the answer from the source, which I find equally appealing:
Label a carton with a “b” (respectively, a “c”) if Bob (respectively, Chris) reaches that carton more quickly, and also record Bob’s (respectively, Chris’s) score upon reaching that carton. Label cartons A and L with “xx” since both players reach those cartons simultaneously.
xx b2 b3 b4
c2 c5 b7 b8
c3 c6 c9 xx
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.
Hope you all enjoyed the puzzle!