I like the
double-elimination format the World Baseball Classic adopted for this year's tournament, but I'm confused with the opportunity for a loser-bracket team to beat an undefeated team just once to earn the top seed coming out of the pool. Japan beat South Korea 14-2 in pool play, sending South Korea to the losers bracket, but now that SK has earned the chance to play Japan in the tournament finale, they only need one victory to earn first-place honors over the Japanese? That doesn't seem fair to Japan, especially if South Korea wins by fewer than 12 runs.
Furthermore, Japan won 14-2 in 7 innings, a mercy rule victory. If a tiebreaker is applied to a SK victory in the final game of the pool, should Korea be forced to win by more runs in just as many innings?
The obvious remedy, which would avoid the need for a tiebreaker altogether, is to schedule a double-header if the Koreans happen to beat the Japanese in game six, thereby creating a one-off championship for the top seed out of the bracket. I'm guessing this was not explored because of overuse of pitchers, especially when the tournament mandates strict pitch-count limits.
That said, there are some big games Monday in pool play:
Pool A: South Korea vs. Japan, 5:30 am EST
*Winner earns top seed from pool & likely avoids Cuba in game one of Pool 1
Pool B: Mexico vs. South Africa, 10 am EST
*Loser becomes first to exit pool; winner meets Cuba or Australia
Pool C: Italy vs. Canada, 5:30 pm EST
*Loser becomes first to exit pool; winner meets Venezuela to qualify for Pool 2
Pool D: The Netherlands vs. Puerto Rico, 6:30 pm EST
*Winner advances to Pool 2 and likely avoids USA in game one; loser meets D.R.
My picks:Japan over South Korea - Japan earns first place in Pool A
Mexico over South Africa - SA heads home while Mexico meets Australia for survival
Canada over Italy - The Italians exit pool play, but Canada gets a shot at Venezuela
Puerto Rico over The Netherlands - The Dominican Republic gets a chance for revenge