If all three teams tie doe it matter that Canada gave up eight runs to S.Africa? I thought it just mattered how many runs the tied teams gave up between themselves?

Mexico has to beat canada and give up at least four runs to Canada. US is sitting -8 to Canada's -6 to Mexico's -2
in runs allowed between the tied teams. But USA is done with a -8 while Mex/Cn play each other.

Is that what you said? or am I wrong, once again. or is total runs in the whole tourney the last tie breaker before drawing straws?

I meant Mexico has to win and give up at least four runs for US to win the group.

According to my calculations (you may want to double check me), assuming the US and Mexico both beat South Africa, and assuming Mexico and Canada play a nine-inning game:

1) If Canada beats Mexico, USA and Canada advance.

2) If Mexico beats Canada:
Mexico advances if they allow five or fewer runs.
Canada advances if they allow two or fewer runs.
USA advances if Mexico allows six or more runs, or if Canada allows three or more runs.

Extra innings could screw up those calculations though...

So technically, Mexico and Canada could agree to let Mexico win 1-0, and they'd both advance, and USA would be out.

The games against RSA count even though they won't be in the tie.

You can't say automatically that the USA will advance if Canada wins because the USA still has to beat RSA. It's very likely they will, but they still have to do it.

Canada gave up more runs against RSA than the USA despite:
1) playing in a bigger park
2) against inferior hitters
3) using supposedly better pitchers

Baseball is difficult to understand all over.

OK, Bob, you're confusing me.

To quote, the second tiebreaker:

The tied teams shall be ranked in the standings for that Round according to fewest runs allowed divided by the number of innings (including partial innings) played in defense in the games in that Round between the teams tied.

To me, that means that only the games which involve the tied teams count in the tiebreaker. So the RSA games wouldn't count in the tiebreaker, because they're not included in the phrase between the teams tied. Doncha think?

5You can't say automatically that the USA will advance if Canada wins because the USA still has to beat RSA

You better hope USA wins, I was on Dodger Thoughts when you made your $100 payment pledge if RSA beats USA.

(Ok, that is what I meant. USA has to beat RSA in all scenarios but in the end it doesn't matter that Canada gave up eight runs to them)

No need to speculate, here are the official tiebreaker scenarios according to MLB.com:

Canada advances if:
• If it wins Thursday in its game against Mexico, it advances to the second round with a perfect 3-0 mark.
• Canada can lose Thursday's game and still advance, but only if it allows one or two runs in that loss to Mexico. If Canada allows three or more runs to Mexico, Team USA and Mexico will advance based on the tiebreaker of runs allowed and Canada would be the odd-team out.

Team USA advances if:
• Canada beats Mexico on Thursday, and the U.S. defeats South Africa on Friday, as expected. If the U.S. loses to South Africa, then it's a three-way tiebreaker between Team USA, Mexico and South Africa, with only one team advancing. Of course, if the U.S. loses to South Africa, it might have bigger issues to address than tiebreaker rules.
• Mexico beats Canada on Thursday and Mexico scores three or more runs in that victory. Mexico and the U.S. would then advance based on the first tiebreaker, runs allowed for the teams tied. The U.S. has allowed eight runs its games vs. Mexico and Canada. Canada would then have allowed nine or more runs and would be the odd-team out.

Mexico advances if:
• If it defeats Canada at 8 p.m. Thursday, Mexico advances to the second round

To get a little more specific, there are 2 scenarios in which the US moves on:
1. They win and either Canada wins OR Mexico scores 3 or more runs (3 teams would be 2-1).
2. They lose and Canada wins and South Africa scores 6 or fewer runs (3 teams would be 1-2).

News accounts today seem unanimous that if Mexico defeats Canada by a score of 2-1 or 2-0 the USA is eliminated. Since (barring extra innings) that would leave the USA and Canada with identical RA/9 in games among the three teams, I do not see how it eliminates the USA. (Each team would have given up 8 runs in 18 innings.) The assumption (explicit in some stories) seems to be that the USA would lose out to Canada based on the result of the head-to-head game. But head-to-head is the first tiebreaker; when we apply RA/9 we are already at the second tiebreaker (which we reached because, assuming the USA defeats South Africa, the first tiebreaker does not help decide which two the three teams with 2-1 records should advance). Having reached the second tiebreaker, and finding two teams (USA and Canada) still tied, why would we go back to the first tiebreaker? It seems to me we should go on to the third tiebreaker (ERA) and if necessary the fourth (BA).

I admit this is somewhat academic, since moving on to the remaining tiebreakers would be unlikely to help the USA. If one or both of Mexico's runs were unearned, Canada would win the ERA tiebreaker. Otherwise that tiebreaker would also be tied, and the final tiebreaker (well, drawing lots is the final tiebreaker, so I should say the fourth tiebreaker) would apply. Since a mere three hits would likely be enough for Canada to best the USA's paltry .231 BA, only a Mexico two-hitter would help the USA. But the assumption that Canada's head-to-head victory over the USA would come into play once the second tiebreaker has been reached still seems wrong to me.