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  • April 15, 2008 12:30 PM ET

Cueto will win the NL Rookie of the Year Award

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Johnny Cueto is the obvious front runner for rookie of the year. Despite being 1-1, his other numbers are amazing. In 19.1 innings pitched, he has a 3.72 ERA, 11 hits, 1 walk, and 24 strike outs. He is a true rookie unlike Fukudome and has posted outstanding numbers and should get the Rookie of the Year Award.


Don't you think it's a bit early to be predicting the end-of-season Rookie of the Year?

But I'll play along. I'll go with Dodgers starting pitcher Hiroki Kuroda. Like Cueto, he is 1-1 in 3 starts, but in the most important pitching statistic (ERA), he has a 2.89 ERA, to Cueto's 3.72 ERA (a difference of 83 points). Kuroda is also good in other areas. His WHIP is impressive, and he also has a good BAA and a low number of walks.

Cueto is a good strikeout pitcher, and he doesn't give up a lot of hits, but in his 3 starts, he has given up 4 home runs, with at least one in each game. Just two days ago, he gave up 5 ER in 6 innings. Kuroda, on the other hand, has given up 2 HR, and his worst start was 3 ER in 6 IP.


Sure its a bit early but that doesn't matter. Aren't the Tigers 3-10?
Anyway, you picked another Japanese players who have possibly more experience than Cueto. Kuroda is like Ichiro, Daisuke, and Fukudome. They have played on a professional level despite it being in Japan, while Cueto has done what he has in his first 3 career MLB starts. He had no experience against pro hitters, while Kuroda has. Cueto also had a slightly tougher schedule having pithced to Milwaukee while Kuroda has pitched to the Padres, who are worse hitting wise than the Brewers. In Cueto's first outing he fanned 10. Kuroda just reached that mark after 18 innings. Cueto is more of a rookie than Kuroda and is off to a better start for a worse team.


Uhh...Rookie of the Year is Rookie of the Year, it doesn't matter if the guy has played 20 seasons in Japan. He is still a rookie, who is eligible for the award. If you want to argue who should win the award not including Japanese players, feel free to start another TD. According to Major League Baseball, Kuroda and Cueto are both equal rookies.

How is Cueto playing for a worse team make him a better candidate? The only statistic that is actually affected by how his own team plays is the Win-Loss column. Everything else depends on him. And besides, the Dodgers have a worse record than the Reds.

Kuroda has actually faced tougher batting teams. The D'Backs, Padres and Pirates are (respectively) 2nd, 3rd and 14th in batting average. Cueto has faced the D'Backs, Pirates and Brewers, who are (respectively) 2nd, 14th and 18th in batting average. So, the Padres are actually a better hitting team than the Brewers? Thus far, yes.

In the end, what matters is giving up as few runs as possible. Strikeouts are fine, but if we judged who was best on K/9, the greatest three pitchers of all time are Randy Johnson, Kerry Wood and Pedro Martinez.


Look, I'm saying Kuroda has the upper hand playing in Japan. But Cueto still seems to have more potential than Kuroda. Cueto is currently among the top in strikeouts. Kuroda is a cellar dwellar in strikeouts despite pitching about the same amount of innings. Cueto has possibility to lead his team, Kuroda doesn't. Cueto is the better prospect, thus the better rookie.


"Cueto is the better prospect, thus the better rookie."

Talk about asinine. We're not talking about the better prospect, or the better rookie, we're talking about who will win NL Rookie of the Year. I don't care if Kuroda is 50 years old, if he deserves to win, he will win it.

You also put way too much stock in strikeouts. In 2006, Chien-Ming Wang finished 2nd in AL Cy Young voting. Want to know what his strikeout total was? 76. That's right, 76 Ks. It was good for 61st in the just the AL, behind even relievers. Yet he still finished 2nd in Cy Young voting...

Cueto has "the possibility to lead his team", and Kuroda doesn't? Well, last time I checked, both have the same record and both are on 6-8 teams, so...I guess neither is doing a whole lot of "leading" thus far, am I correct?

April 15, 2008  01:10 PM ET

Doesn't matter if he's had no experience against pro hitters and Kuroda has... they're both rookies.

April 15, 2008  01:48 PM ET

McClouth

April 15, 2008  02:06 PM ET

Rookie=Rookie in MLB. It doesn't take into account a player's time in Japan or whereever. Both Sasaki and Suzuki won it in 2000 and 2001 each with the Mariners with experience in the Japanese leagues.

April 15, 2008  02:07 PM ET

FUKUDOME!!

April 15, 2008  02:59 PM ET

McClouth isn't a rookie.

April 15, 2008  03:18 PM ET

Honestly Cueto isnt the ROOK TO BEAT thats Fukudome and most everyone knows it. I find the fact that you think that just because a player has played in Japan its unfair..Honestly do you think Sandaharo Oh whould have reached over 800 homeruns in the MLB?

April 15, 2008  05:39 PM ET

what was I smoking?

April 16, 2008  12:01 PM ET

I just wanted to point out that if all we're going on is the absudrdly small sample size of their perfomance thus far, Cueto is MILES ahead of Kuroda.

Predictive stats for a pitcher's success include K:BB, K/9, and WHIP. ERA and Wins are less useful for predicting future success.

Cueto's K:BB ration is 24:1 while Kuroda's is 5:1.
Cueto's K/9 is 11.3 while Kuroda's is 4.8
Cueto's WHIP is .62 while Kuroda's is 1.13

Clearly it's way too early to predict, and LIFER has better arguments, but statistically, what Cueto is doing is ridiculously good so far...

 
April 17, 2008  03:59 PM ET

Honestly Cueto isnt the ROOK TO BEAT thats Fukudome and most everyone knows it. I find the fact that you think that just because a player has played in Japan its unfair..Honestly do you think Sandaharo Oh whould have reached over 800 homeruns in the MLB?
Analytical | 04/15/08, 03:18 PM
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